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Multidimensional local skew-fields
DISSERTATION
zur Erlangung des akademischen Grades doctor rerum naturalium (Dr. rer. nat.) im Fach Mathematik eingereicht an der Mathematisch-Naturwissenschaftlichen Fakultи II at Humboldt-Universitи zu Berlin at

von Herr Diplom-Mathematiker Alexander Zheglov geboren am 23.08.1974 in Ivanovo, Russia.

Prи asident der Humboldt-Universitи zu Berlin: at Prof. Dr. Jurgen Mlynek и Dekan der Mathematisch-Naturwissenschaftlichen Fakultи II: at Prof. Dr. Elmar Kulke Gutachter: 1. Prof. Dr. E.W.Zink 2. Prof. Dr. A.N.Parshin 3. Prof. Dr. V.I.Yanchevskii eingereicht am: и Tag der mundlichen Prufung: и 04.04.2002 10.07.2002

Contents
Intro duction. 0 The 0.1 0.2 0.3 structure of two-dimensional lo cal skew fields. General. . . . . . . . . . . . . . . . . . . . . . . . . . . Splittable skew fields. . . . . . . . . . . . . . . . . . . Classification of two-dimensional local splittable skew fi teristic 0. . . . . . . . . . . . . . . . . . . . . . . . . . 0.3.1 The case = Id. . . . . . . . . . . . . . . . . 0.3.2 The general case. . . . . . . . . . . . . . . . . . Splittable skew fields of characteristic p > 0. . . . . . . 0.4.1 Wild division algebras over Laurent series fields 0.4.2 Cohen's theorem . . . . . . . . . . . . . . . . . 0.4.3 Decomposition theorem . . . . . . . . . . . . . Classes of conjugate elements . . . . . . . . . . . . . . New equations of KP-type on skew fields . . . . . . . . automorphisms ......... theorems I and II theorem III . . . of .. . .. a . . . two-dimensional ........... ........... ........... ... ... elds ... ... ... ... ... ... ... ... ... . . of . . . . . . . . . ..... ..... charac..... ..... ..... ..... ..... ..... ..... ..... ..... 2 6 6 12 22 23 29 33 34 35 36 61 70 72 72 84 92 107

0.4

0.5 0.6

1 Classification of 1.1 Basic results. 1.2 Proof of the 1.3 Proof of the Bibliography

lo cal ... ... ...

field. ...... ...... ......

1

Intro duction.
In this work we study local skew fields, which are natural generalisation of n-dimensional local fields, and their applications to the theory of central division algebras over henselian fields. Local fields appear in a natural way in algebraic geometry and algebraic number theory if anyone try to find a connection between local and global properties of such ob jects like algebraic number fields, arithmetic schemes and algebraic varieties. Historically the first examples of 1-dimensional local fields appeared in the 19 century in complex analisys and in algebraic number theory. These examples are known fields C((z )) and Qp . Now we say that 1-dimensional local field is a quotient field of a complete discrete valuated ring. A little bit later the first examples of local skew fields were found. They were finite dimensional division algebras over classical local fields, and they were completely studied by Hasse, Brauer, Noether and Albert. At the same time there were several works of Witt ([34]) about skew fields over discrete valuated fields, which opened researching of skew fields over henselian fields. Basic results about a structure of such skew fields were recently got by Jacob and Wandsworth in ([9]). In the middle of 70-th A.N.Parshin introduced a notion of a multidimensional local field which generalised the notion of a usual local field ([19],[24], [7]). n-dimensional local field is a complete discrete valuated field such that the residue field is a n - 1-dimensional local field. One of the typical examples of such a field is an iterated Laurent series field k ((z1 ))((z2 )) ... ((zn )). Elements z1 ,...,zn are called local parameters of this field. Multidimensional local fields appears also as natural generalisations of local ob jects on 1-dimensional scheme. As an example let us consider the following construction. Consider an algebraic scheme X of dimension n. Let Y0 ... Yn be a flag of closed subschemes in X such that Y0 = X , Yn = x is a closed point on X , Yi is a codimension 1 subscheme in Yi-1 (1 i n), x is a smooth point on all Yi (0 i n). Then there exists a construction which assign in canonical way to any given flag a n-dimensional local field. Moreover, let X be an algebraic variety over a field k , x be a rational point over k , z1 ,z2 ,...,zn k (X ) be fixed local parameters such that zn-i+1 = 0 is an equation of Yi on Yn-1 in a neiborhood of the point x (1 i n). Then our n-dimensional local field can be identified with k ((z1 ))((z2 )) ... ((zn )) ([24], [7]). Using this assignment a number of results known earlier only for the case of a curve was generalised to a higher dimensional case. These are such well-known results as multi-dimensional reciprocity lows of Parshin-Lomadze ([19], [15], [20], [7]). During the last 25 years there was opened another direction in the theory of local fields. This is an application to the theory of integrable systems connected with the Krichever-Sato-Wilson correspondence on a curve (for more details on the Krichever 2

correspondence see [6], [29], [16], [27]). Recently there were issued several papers [17], [18], [23], where the ideas of the Krichever-Sato-Wilson correspondence on a curve were developed to the case of varieties of higher dimension. In particular, A.N. Parshin pointed out one class of noncommutative local fields arising in differential equations and showed that these skew fields possesses many features of commutative fields. He defined a skew field of formal pseudo-differential operators in n variables and studied some of their properties. He raised a problem of classifying non-commutative local skew fields. It was the first argument to begin to study such skew fields. A generalisation of a notion "local field" looks very natural: n-dimensional local skew field is a complete discrete valuated skew field such that the residue skew field is a n - 1-dimensional local skew field. In this work we try to study n-dimensional local skew fields bearing in mind only the definition. Unfortunately, there appear very hard obstructions already on the first steps which leads us to some restrictions. So, we study only skew fields with commutative residue skew field. By the way, a number of results valid in general case (see, for example, proposition 0.7 and corollary 1) and a number of results can be generalised to the case of skew fields with residue skew field finite dimensional over its centre (see, for example, section 1.4). Some applications of developed theory to the Krichever correspondence we get in section 1.6. Namely, we get some generalisations of the classical KP-equations (hierarchy). Surprisingly the studying of local skew fields leads to some new unexpected results in the valuation theory on finite dimensional division algebras. Using general formulas for splittable local skew fields (i.e. for skew fields such that the residue skew field can be embedded into the valuation ring) we get a decomposition theorem for a class of splittable wild division algebras over a Laurent series field with arbitrary residue field of characteristic greater than two. This theorem is a generalisation of the decomposition theorem for tame division algebras given by Jacob and Wadsworth in [9]. An extensive analysis of the wild division algebras of degree p over a field F with complete discrete п rank 1 valuation with char(F ) = p was given by Saltman in [28] ( Tignol in [32] analysed more general case of the defectless division algebras of degree p over a field F with Henselian valuation). In his recent revue [33] Wadsworth pointed out that for most of the specific examples and applications it is suffice to consider Henselian valued fields like iterated Laurent series fields, that is n-dimensional local fields. So, we get in some sense the complete picture of a local structure of the Brauer group over such fields. As a corollary we get the positive answer on the following conjecture: the exponent of A is equal to its index for any division algebra A over a C2 -field F = F1 ((t2 )), where F1 is a C1 -field (see [26], 3.4.5.). From the other hand, the problem of classification of local skew fields leads to the problem of classification of conjugacy classes in the automorphism group of an 3

n-dimensional local (commutative) field. We solve this problem for the group of continuous automorphisms. We note that the automorphism group of a local field of positive characteristic is intensively studied now in the algebraic number theory (we mean recent applications to the problem of description the Galois group of an arithmetically profinite extension). Moreover, the automorphism group of the field Fq ((t)) (so called Nottingham group) is now carefully studied in the group theory (for more details see papers [5], [3], [12], [8], [13], [14], [10], [11], [36]. We hope that our results on the automorphism group will be applied in the future to obtain some useful results about the Galois group of an arithmetically profinite extension. Here is a brief overview of this thesis. It consists of two chapters. The first chapter consists of five paragraphs. In з1 we give general definitions of a local skew field, of a splitness and of an isomorphism of local skew fields. Also we study some general properties of splittable skew fields. Thereafter except з4 we study mostly two-dimensional local skew fields with commutative residue skew field. In з2 we give a sufficient condition for a skew field to be split. Namely, a local skew field splits if a canonical automorphism has infinite order. The canonical automorphism can be defined as a restriction of an inner automorphism ad(z ) on the residue field, where z is any local parameter. We show that there exist counterexamples when this condition does not hold. We note that this condition and counterexamples are true even in more general situation when the skew field is not two-dimensional skew field or the residue skew field is not commutative. We classify all the skew fields which possess this condition up to isomorphism. The results of з2 don't depend on the characteristic of a skew field. In з3 we classify all the local splittable skew fields of characteristic 0 with commutative residue skew field and with the canonical automorphism of finite order. In з4 we study splittable local skew fields of characteristic p > 2 with commutative residue skew field and with the canonical automorphism of finite order. We give a criterium when such a skew field is finite dimensional over its centre. Then we prove that every tame finite dimensional division algebra over a local complete field splits. Using this fact we prove the decomposition theorem for splittable algebras. As a corollary we get the proof of the conjecture mentioned above. In з5 we study some properties of local skew fields described in з3. In particular, we give a criterium when two elements from such a skew field conjugate. This is a generalisation of analogous theorems from [23]. As a corollary we prove that almost all such skew fields are infinite dimensional over their centre. Also we prove that the Scolem-Noether theorem holds only in the case of the classical ring of pseudo-differential operators. In з6 we get new KP-hierarchies for every class of isomorphic two-dimensional local 4

skew fields, which were studied in section 3. We derive new equations of the KP-type for some hierarchies. In the second chapter we classify conjugacy classes in the group of continuous automorphisms of a two-dimensional local field of characteristic zero with the residue field of the same characteristic. Some facts about automorphisms of a local field of characteristic p > 0 one can find in lemma 1.3. Also in this chapter we show how this classification can be generalised to the case of a n-dimensional local field, n > 2. Acknowledgements. At this point, I would like my advisor Prof. Ernst-Wilhelm Zink for the warm atmosphere and valuable discussions, Prof. A. N. Parshin, my scientific advisor during my undergraduate studies in Moscow State University, Prof. V.I. Yanchevskii for stimulating discussions and advises. I would further like to thank Martin Grabitz, who allowed me to more easily orient myself in the new environment, country and language and who spent a lot of time for our interesting discussions. I thank Prof. Helmut Koch for the invitation to HU zu Berlin in 1999 that made it possible for me to stay here for a long time. I would also like to thank my wife Olga for her help in preparing the final version of this thesis and her love, and I thank Graduiertenkolleg "Geometry und Nichtlineare Analisys", which provided the doctoral stipend that allowed me to complete this work.

5

Chapter 0 The structure of two-dimensional lo cal skew fields.
0.1 General.

Definition 0.1 Let K and k be arbitrary skew fields. A skew field K is cal led a complete discrete valuation skew field if K is complete with respect to a discrete valuation. A skew field K is cal led an n-dimensional local skew field if there are skew fields K = Kn ,Kn-1 ,...K0 = k such that each Ki for i > 0 is a complete discrete valuation skew field with residue skew field Ki-1 . The following properties are well known from the valuation theory of division algebras (see for ex. [31]). Lemma 0.2 Let K be a complete diskrete valuation skew field. Then the fol lowing properties hold: i) The valuation ring O is a topological group and a metric space under the natural topology; ii) The ring O is a local ring and a principal ideal domain. For every two-dimensional local skew field we have п п KOKOk п п where O is a valuation ring in K . There are two filtrations K O 2 ... п п п K O 2 ... п пп п where is a maximal ideal in O, is a discrete valuation on K . п 6

Definition 0.3 Two two-dimensional local skew fields K and K are isomorphic if there is an isomorphism which preserves the filtrations above, i.e. it maps OK onto OK , onto . and OK onto OK , K onto K . п п п п Definition 0.4 A two-dimensional skew field K is said to split if there is a section of п the homomorphism OK K . п пп Elements z O, (z ) = 1 and u O K , (u) = 1 are cal led local parameters (or variables) in K . Prop osition 0.5 Suppose K splits. Fix some local parameters z and u. Then K is п isomorphic to a two-dimensional local skew field K ((z )) where za = a z + a1 z 2 + a2 z 3 + ... п п п where a K , is an automorphism, i : K K are linear maps. Pro of. Suppose a K , (a) = j . Then we have (az -j ) = 0 and az -j := п az -j mod K . We will assume that the last element lies in O, since there is a section. Then we have (az -j - az -j ) 1. Continuing this line of reasonings, we get п a = j ai z i , ai K . i= п Now define a = zaz -1 mod , where a K . It's clear that is an automor-1 -1 phism. Since (zaz ) = 0, the element zaz can be written as a series ai z i , i=0 п where ai K . Here we have a0 = a . Now put ai := ai for i 1. It is easy to see that i are linear maps. 2 In fact, the maps i satisfy some identities. To write them we need extra notation. Consider the ring Z < , > of noncommutative polynomials in two variables. Define the map : Z < , > Z < ,,i ; i 1 >, (a1 b1 ...an bn ) = a1 b1 ...b
n-1

an -1 b

n,

where a1 ,bn 0, ai ,bj 1, i > 1, j < n for every word in Z < , >. For example (k ) = k (k l i ) = k l
i-1

where k, l, i are natural numbers, i, l 1. Let Sik Z < , >, i k , i 1 be polynomials given by the following formula: Sik =
Si /G

( ... ... ),
i-k k

where Si is a permutation group and G is an isotropy subgroup. Immediately from the definition we get the following lemma 7

Lemma 0.6 The polynomials Sik satisfy the fol lowing property:Sii = i , Si0 = i , S
k+1 i+1

= Sik

+1

+ Sik

Now we can define the identities for the maps i : Prop osition 0.7 Every map i , i 1 satisfy the identity
i

i (ab) =
k=0

(

i-k

)(a) (Sik )(b),

п a, b K

п Pro of. For any a, b K . We have () (ab) z +(ab)1 z 2 + ... = z (ab) = (a z + a1 z 2 + ...)b

If we represent the right-hand side of () as a series with coefficients shifted to the left and then compare the corresponding coefficients on the left-hand side and righthand side, we get some formulas for i (ab). We have to prove that these formulas are the same as in our proposition. Let z i+1-k b = i+1-k (b)z i+1-k + ... + xk z i+1 + ... and ((a)z + 1 (a)z 2 + 2 (a)z 3 + ...)b = (ab)z + y2 z 2 + y3 z 3 + ...
i-1

Then we have
i

y

i+1

= (a)xi +
k=0

i-k (a)xk =
k=0

(

i-k

)(a)x

k

Note that xk are polynomials which consist of monomials of the type a1 b1 ...an bn
a
n+1

(b),

ak ,bk Z,

ak ,bk 0

(we put 0 to be equal to 1). It is easy to see that these polynomials have integral positive coefficients. We claim that xk = (Sik )(b). To prove this fact it suffice to show that xk contains every monomial from (Sik )(b) and the sum of coefficients in xk is equal to the sum of coefficients in (Sik )(b). By definition every coefficient of (Sik )(b) is equal to 1. It is easy to see that the sum of coefficients is equal to Cik = i!/(i - k )!k !.

8

Let us show that xk contains every monomial from (Sik )(b). By definition, (Sik )(b) consists of monomials ( ( ... ... ))(b), where Si , i.e. it consists of monomials a1 b1 ...an bn n+1 j =1 aj = i - k +1 - n. We have z z
i+1-k-a
n+1

a

i-k
n+1

(b), where aj 0, bj 1,

k

n j =1 bj

= k,

i+1-k a
n+1

b=z =z
+1

i+1-k-a

n+1

a

n+1

(b)z
+1

a

n+1

+ other terms,
a
n+1

a

n+1

(b)z
-1

i+1-k-a a
n+1

n+1

-1

[

a

n+1

(b)z +...+bn
-1

(b)z

bn +1
n+1

+...]z + ...

a

n+1

=

z

i+1-k-a

n+1

a

n+1

(b)z

+1

+z

i+1-k-a

n+1

bn

a

n+1

(b)z

bn +1+a

Now put d1 = bn z
i+1-k-a
n+1

a

n+1

(b). Then we have = ... + z
i+1-k-a
n+1

-1

d1 z

bn +1+a

n+1

-1-an -1

d2 z

bn +1+a

n+1

+an +b

n-1

+1

+ ...,

where d2 = b z
i+1-k- a j -n

n-1

an bn

a

n+1

(b). By induction we get
a
n+1

a1 b1 ...an bn

(b)z

bj +n+

a

j

= a1 b1 ...an bn

a

n+1

(b)z

i+1

+ ...,

that is xk contains any given monomial from (Sik )(b). Let us show that the sum of coefficients of xk is equal to Cik . Denote by sl the sum of coefficients in yl , where n z n az
-n

=
k=0

yk z k ,

п aK
k i+1-k

Then the sum of coefficients of xk is equal to s relation holds
d

. We claim that the following

s=
l=0

d n

s

l n-1

The proof is by induction on n. For n = 1 we have sd = 1 for all d 0, sl = 0 for 1 0 l > 0 and s0 = 1. 0 For arbitrary n put z n-1 az -n+1 = y0 + y1 z + ..., п where y0 K . Then we have z n az Put z az
n -n -n

= zy0 z

-1

+ zy1 z -1 z + ... = [y0 + y01 z + ...]+[y1 + y11 z + ...]z + ...

=
k=0

wk z k .

9

Then we have wd =

d

j (y
j =1

d-j

)+ (yd )
j n-1

Since the sum of coefficients of yj is equal to s
d

, we get

s=
j =0

d n

s

j n-1

Now let us show that sk+1-k = Cik if k < i + 1. The proof is by induction on i. For i i = 0 we have s0 = 1 = Ci0 . For arbitrary i we have 1
k

s (...

k i+1-k

=
l=0

s

l i-k

- = Cik + Cik-11 + ... + Ci0-k =

(((Ci0-k

+ Ci1-
k+2

k+1

)+ Ci2-
k+2

k+2

)+ Ci3-
k+3

k+3

)+ ... + Cik ) =

(... (((Ci1- This completes the proof. 2

)+ Ci2-

)+ Ci3-

)+ ... + Cik ) = Cik+1

Corollary 1 Suppose = Id. Then the fol lowing formula holds
i

i (ab) = i (a)b +
k=1

i-k (a)
(j1 ,...,jl )

Cil-

k+1 j

1

...jl (b)

where 0 = and the second sum is taken over al l the vectors (j1 ,...,jl ) such that jm = k . 0 < l min{i - k +1,k }, jm 1, In the sequel we will need the following definition. Definition 0.8 Let (, ) be endomorphisms of a skew field L. A map : L L , where L L is a subalgebra, is cal led a (, )-derivation if it is linear and satisfy the fol lowing identity (ab) = (a)b + a (b) where a, b L. We wil l say that (, 1)-derivation is an -derivation. For example 1 is an (2 ,)-derivation. 2 If = Id, then 1 is an usual derivation; 2 = 1 + , where is a derivation. 10

Corollary 2 If 1 = ... = k

-1

= 0, then k is an (

k+1

,)-derivation.

The following corollary will be used in з3 of this chapter. п п Corollary 3 Let K be a field, K = k ((u)), k Z (K ) and the maps i , i 1 be continuous if chark = 0. Then

i (
j =N

xj u ) =
j =N

j

xj i (uj ),

xj k

So, for every i the map i is completely defined by elements i (u) and j (u) for j < i. Pro of. If chark = p = 0 and = id the maps i , i 1 are continuous, since i п i (ap ) = 0 for any a K . Since a topology on a 1-dimensional local field is uniquely defined by its local structure, the continuity does not depend on the choice of local parameters (for more information about a relation between a topology and a parametrisation see [35]). If = id one can use lemma 1.29 to reduce this case to the previous one. Let us show that is a continuous map. In our case it suffice to show that preserves the valuation. Our proof will not depend on a characteristic. п It suffice to show that ((u )) = 1 for any u , (u ) = 1. Consider the automorphism п : (a) := z -1 az п where a K (we use the notation from proposition 1.7). It's clear that = -1 . п Let u be an arbitrary parameter. Put = ((u )). We claim that || 1 or q || = p , q N. Assume the converse. Then = mpq , (m, p) = 1, |m| = 1 and there п exist c k , a K such that (u ) = cam . Therefore, we get u = -1 ((u )) = c(-1 (a))m , i.e. (u ) = 1 = (c(-1 (a))m ) = m (-1 (a)), п п п

a contradiction. Let us show that 0. Assume the converse. Consider the element u + u 2 (u + u 3 if chark = 2). Then ((u + u 2 )) = 2 < -1. If chark = 2 we get a contradiction п п with the assertion | ((u ))| = pq or | ((u ))| 1 for any parameter u . If chark = 2 п one can apply the same arguments to the element u + u 3 . п Similarly, for := (-1 (u )) the property 0 1 or = pl holds. Let us show that = pq . Assume the converse. Consider the following two cases: q 1) Suppose 1. There exist r k , a1 k ((u)) such that (u ) = c2 u 2 ap -2 . 1 Therefore, п 1 = 2 (-1 (u )) + (pq - 2) (-1 (a1 )), п 11

i.e. (pq - 2)|1. It is possible only if p = 3, arguments with (u ) = c3 u 5 a-2 . Then we 1 0 1 or = pl ). q 2) Suppose = pl . Let (u ) = cu ap -1 we have п (u ) = 1 = (-1 п

q = 1. In this case one can use the same get (-1 (a1 )) = 2, a contradiction (since п for some c k , a k ((u)), (a) = 1. Then п (u )) + (pq - 1). п

But this contradicts with (-1 (a)) 0. п п п So, = 0 or = 1, i.e. for any parameter u we have ((u )) = 0 or ((u )) = 1. - Suppose = 0. Consider the element x = u + c1 u 2 + c2 (u 3 + c1 u 4 )), where c1 = -w0 1 if (u ) = w0 + ... and c2 is an element such that ((x)) > 1 (it always exists since п п ((u + c1 u 2 )) > 0). But this contradicts with (x) = 1. Therefore, = 1 and is a п continuous map. To complete the proof it suffice to show that the series N xj i (uj ) converges, j= because the topology on k ((u)) is complete and separate. The proof is by induction on i. For i = 0 we have ((uj )) = j and the series converges. For i = 1 we have п п (1 (uj )) = (j - 1) (1 (u)) and again the series converges. п At last, by proposition 0.7 for j > 1wehave i (uj ) = i (uj -1 )y0 + i-1 k (uj -1 )yi-k , k=0 where (yk ) does not depend on j . п п By induction we have min{ (0 (uj -1 )yi ),..., (i-1 (uj -1 )y1 )} > п j -2 j -2 п п min{ (0 (u )yi ),..., (i-1 (u )y1 )} and (y0 ) = 1. So, п j -1 j -1 п п min{ (i (u )y0 ), (0 (u )yi ),..., (i-1 (uj -1 )y1 )} > п п п min{ (i (uj -2 )y0 ), (0 (uj -2 )yi ),..., (i-1 (uj -2 )y1 )}. п Therefore, the series converges. 2

0.2

Splittable skew fields.

п In this section we wil l assume that K is a field. For such a skew field one can define a notion of a canonical automorphism . By definition there exist the following exact sequences: 1 O K - Z 1

where O is a valuation ring; п 1 1+ O K 1 where is a maximal ideal. Consider the map : K Int(K ), (x) = ad(x), 12 ad(x)(y ) = x-1 yx

where Int(K ) is the group of inner automorphisms of the skew field K . Since inner automorphisms preserve the valuation, this group acts on the ring O. Moreover, it п preserve the ideal . Therefore, there exists a map : K Aut(O/) = Aut(K ). Let п us show that the action of (O ) is trivial on K . To show it we use the second exact -1 sequence. Since (1 + ) x(1 + ) = x mod for any x O, the action of (1 + ) п п п on K is trivial. Therefore, there exists an action of K on K . Namely, an element a K пп -1 п п пп acts on x K as a xa mod , where a, x are any lifts of a, x in O. Since K is a п commutative field, this action is trivial. п Definition 0.9 An automorphism of the field K defined by the formula = (z ) where z K and (z ) = 1, is cal led a canonical automorphism. It does not depend on the choice of z . We want to classify all splittable two-dimensional local skew fields which have isomorphic last residue fields up to isomorphism. Let K and K be two splittable skew fields, K K ((z )), K K ((z )). If K K , then one can represent an isomor=п = =п phism : K K as a compositum of an isomorphism : K - K such that (u) = u , (z ) = z , and of an automorphism of the skew field K . Since every isomorphism in our paper preserve the local structure, every automorphism of a splittable two-dimensional local skew field is defined by change of parameters (z ) u u = c0 + c1 z + c2 z 2 + ..., z z = a0 z + a1 z 2 + ..., (c0 ) = 1 п a0 = 0

п where ai ,ci K . It is easy to see that every change of parameters looks like above and can be decomposed into a sequence of changes u u , z z ; u u , z z = a0 z + a1 z 2 + ... (or in a backward order). Also u u can be decomposed into a sequence of changes u u1 = c0 , u1 u2 = u1 + c1 z ,...,ui ui+1 = ui + ci z i ,... and z z can be decomposed into a sequence of changes z z1 = a0 z , z1 z2 = z1 + a1 z 2 ,...,zi zi+1 = zi + ai z i+1 ,.... Remark. We must note that any change of parameters (z ) defines a map f : K K which is not always an automorphism. Indeed, assume the converse. Consider a map which is given by f (z ) = z , f (u) = u, where z is another parameter. Then we must have f (zu) = f (z )f (u) = z u = u z + u1 z 2 + ... f (zu) = f (u z + u1 z 2 + ...) = u z + u1 z 2 + ... 13

Hence, = ; 1 = 1 and so on, i.e. i = i i. Consider the skew field "((u))((z )) with the relation zu = (u + u2 )z and consider a change of parameters z z = z + z 2 . Then z u = (z + z 2 )u = (u + u2 )z + z (u + u2 )z = (u + u2 )z +[(u + u2 )z +(u + u2 )2 z ]z = (u + u2 )z +[u +2u2 +2u3 + u4 - u - u2 ]z 2 = (u + u2 )z +[u2 +2u3 + u4 ]z 2 + ... So, 1 = 1 , a contradiction. Prop osition 0.10 Let K be a splittable two-dimensional local skew field. Suppose the canonical automorphism has infinite order. п Then there exists a parameter z such that z a = a z for any a K . Pro of. We will show that there exists a sequence of parameters {zk } such that the - equality zk azk 1 = a mod k holds and the sequence {zk } converges in K . We need some additional lemmas. Lemma 0.11 Suppose the fol lowing relation holds:= zaz
-1

= a + aj z j + a

j

+1

z

j +1

+ ...,

п aK

where 1 = ... = j -1 = 0, j = 0. Then (i) for z = z + bz q+1 we have z az
-1

= a + ... + a

q

-1

z

q -1

+(aq + ba

q +1

- a b)z q + ...

i.e. aq = aq + ba - a b. (ii) Suppose n = id, n 1. Then for z = z + bz z az (a
q
+j

q +1

q +1

, n|q we have +
j -1

-1

= a + ... + a
j

q

+j -1

z

q +j -1

+ b(a )

j

q

q

-a b

j

+b
k=1

((a ) )

k

j

q -k

-a

j k=0

b )z

k

q +j

+ ...

п (iii) for z = bz , b K , b = 0 we have z az
-1

= a + aj (b-1 ) ... (b-1 ) z j + ...

j

Corollary 4 If = Id, then z az
-1

= a + ... + a

q

+j -1

z

q +j -1

+(a

q

+j

+(q - j )aj b)z

q +j

+ ...

14

Pro of of lemma. (i) We have z az
-1

= (1 + bz q )zaz -1 (1 + bz q )-1 = (zaz (zaz (zaz
-1 -1 -1

-1

+ bz q zaz -1 )(1 - bz q + bz q bz q - ...) =
-1

- zaz -1 bz q + ... + bz q zaz
j

- ...) = z q + ...) =

q +1

- [a + aj z j + ...]bz q + bz q [a + aj z j + ...]+ ...) = - [a b + aj b z j + ...]z q + ba )z q + ...) = a + ... + a
q
-1

(zaz (zaz
-1

q +1

+(-a b + ba

q +1

z

q -1

+(aq + ba

- a b)z q + ...

(ii) We have z az
-1

= (1 + bz q )zaz -1 (1 + bz q )-1 = (zaz
q
+j

-1

+ bz q zaz -1 )(1 + bz q )-1 =

(a + aj z j + ... + a (a +ba

q +1

z

q +j

+ ... + bz q (a + aj z j + ...))(1 + bz q )-1 =
q

z +a z +...+a
q q +j

q

j j

q

+j

z

q +j

+...+b
k=1

((a )j )
q -k

k

q -k

z

q +j

+b(aj ) z
q +j

q

q +j

+...)(1+bz q )-1 =

q

a +(a z +...+a
j j

j j

+j

z

+...+b
k=1 q

((a )j ) ((a )j )
k

k

z

q +j

+b(aj ) z
q

q

+...)(1-bz q +bz q bz q -...) =
j

a +a z +...+a

q

+j

z

q +j

+...+b
k=1

q -k

z
j

q +j

+b(aj ) z
q k=1
k

q +j

+...-aj b z
q -k

q +j

+... =
q +j

a + ... + a

q

+j -1

z

q + j -1

+(a

q

+j

+ b(aj ) - aj b + b + ....
j

q

j -1

((a )j )

-a

j k=0

b )z

k

,

because z j = z j + (iii) We have z az 2
-1

j -1 k=0 b

k

z

q +j

= bz az -1 b

-1

= a + baj (b-1 ) z j + ... = a + aj (b-1 ) ... (b-1 ) z j + ...

j

Lemma 0.12 Let be an (, )-derivation of a field Then is an inner derivation, i.e. there exists d (a) = da - a d п for al l a K .

п K and = . п K such that

15

Pro of. Put d = (a)/(a - a ), where a is any element such that a = a . Put п in (x) = dx - x d. We claim that = in . Indeed, consider the map = - in . It is п п п . Then (ab) = (ba). But we have an (, )-derivation. Take arbitrary b K п п п п (ab) = (a)b + a (b) = a (b), and п п п п (ba) = (b)a + b (a) = a (b) п Therefore, (b) = 0 for any b. 2 Pro of of prop osition. Let zaz

-1

= a + a1 z + a2 z 2 + ...

By proposition 1.7 and corollary 1, 1 is an (2 ,)-derivation. Since 2 = , by lemma 2 0.12 it is an inner derivation, say 1 (a) = d1 a - a d1 . By lemma 0.11, (i) for a parameter z2 = z - d1 z 2 we have
- 2 z2 az2 1 = a + a2 z2 + ....

Note that 1 = 0. By corollary 1, 2 is an (3 ,)-derivation. Since 3 = , by lemma 3 0.12 it is an inner derivation. By lemma 0.11, (i) there exists a parameter z3 = z2 - d2 z2 -1 3 such that z3 az3 = a mod . By induction for arbitrary k N we have
- k zk azk 1 = a + ak zk + ...

and j = 0 for j < k . By corollary 1, k is an (k+1 ,)-derivation. Since k+1 = , it k is an inner derivation. By lemma 0.11, (i) there exists a parameter zk+1 = zk - dk zk +1 -1 such that zk+1 azk+1 = a mod k+1 . n It is clear that the sequence {zn }: zn+1 = zn - dn zn +1 converges in K . Since K is a complete and separate field, there exists a unique limit z . It is clear that zaz -1 = a . The proposition is proved. 2 Theorem 0.13 Let K be a two-dimensional local skew field. If n = id for al l n N then п (i) charK = charK (ii) K splits.

16

Pro of. п п If charK = charK then charK = p > 0. Hence (p) = r > 0. Then for any element -1 п п п t K with (t) = 0 we have ptp = r (t) mod where t is the image of t in K . But on the other hand, pt = tp, a contradiction. The proof of (ii) we will divide in three steps. п Step 1. Let be the prime field in K . Since charK = charK the field is a subring of O. п Lemma 0.14 There exists an element c K such that c

k

= c for al l k N.

п Pro of. We claim that there exists a sequence {cji }, ji ,i N, cji O such that (i) (cji ) > (cji-1 ) i п п (ii) if k = 0 mod j2 ...jl and k cj1 ,...,k (cjl-1 ) = cjl-1 ,k (cjl ) = cjl and = 0 mod j2 ...j
j

l+1

, then k (cj1 ) =

п [(k - Id)(cjl )] < (c п

l+1

)

п Let us construct it. Take an element cj1 such that (cj1 ) = cj1 , and (cj1 ) 1. Such an element always exists. Indeed, consider an element u with (u) = 1. If (u) = u,then п one can put cj1 = u. If (u) = u, then take any element c~1 such that (c~1 ) = c~1 . If j j j п (c~1 ) = 0, then put cj1 = c~1 u. Then we have (cj1 ) = 1 and (c~1 u) = (c~1 )u = c~1 u. пj j j j j Put j1 = 1. Let j2 be a minimal positive integer such that (j1 )j2 (cj1 ) = cj1 , and let k1 = max{ [(j1 )m (cj1 ) - cj1 ],m {1,...,j2 - 1}}. п k1 Take any c~2 such that (j1 )j2 (c~2 ) = c~2 . Put cj2 = c~2 cj1 +1 . Then (j1 )j2 (cj2 ) = cj2 and j j j j п [(j1 )m (cj1 ) - cj1 ] < (cj2 ) m < j2 . п By induction we get a sequence which satisfy (i) and (ii). k Now put c = i=1 cji . Then for all k we have (c) = c. Indeed, let k = 0 mod j2 ...jl and k = 0 mod j2 ...jl+1 . By (ii), k (c) - c = k (cjl ) - cjl + п п п k ( l+1 cjl ) - l+1 cjl . But (k (cjl ) - cjl ) < (cjl+1 ) (k ( l+1 cjl ) - i= i= i= cjl ). Therefore, k (c) - c = 0. i=l+1 2 п Consider the field F = (c) K . Let us show that this field can be embedded in O. Take any lift c O of the element c: c mod = c. It is clear that c commute with any element from . It is easy to see that c is a transcendental element over . Indeed, assume the converse. Then its equation modulo must have infinite number k of solutions, because c = c k , a contradiction. Therefore, [c ] = 0. So, the field п of fractions F can be embedded in O. п п Let L be a maximal field extension of F which can be embedded in O. Denote by п п/п L its image in O. Take a K , a L. We claim that there exists a lifting a O of a п п such that a commutes with every element in L. 17

Step 2. Take any lifting a in O of a. For every element x L we have п axa-1 mod = x. If z is a parameter of K we can write axa
-1

= x + x1 z,

п п where x1 O. The map 1 : x L 1 (x) K is an -derivation. Indeed, a(x1 + x2 )a a(x1 + x2 )a
-1 -1

= (x1 + x2 )+(x1 + x2 )1 z

+ ax2 a-1 = x1 + x11 z + x2 + x21 z = (x1 + x2 )+(x11 + x21 )z п Therefore, (x1 + x2 )1 = x11 + x21 . Then, we have = ax1 a a(x1 x2 )a Hence x1 x2 +(x1 x2 )1 z = (x1 + x11 z )(x2 + x21 z ) = x1 x2 + x1 x21 z + x11 zx2 + x11 zx21 z x1 x2 + x1 x21 z + x11 x z 2 Therefore, (x1 x2 )1 = x11 x + x1 x21 = x11 x + x1 x21 2 2 п п By lemma 0.12, 1 is an inner -derivation, say 1 (x) = d(x - x). Put a1 := ~ (1 + a1 z )a, where a1 mod = -d. Using the same calculations as in lemma 0.11 we have (1 + a1 z )axa-1 (1 + a1 z )-1 = x +(x1 + a1 x - xa1 )z mod 2 Since x1 + a1 x - xa1 = 0 mod , we get a1 xa1 -1 = x + x2 z 2 . Using the same ~~ п п arguments as above one can check that 2 : L K is an 2 -derivation. By induction i we can find an an element ai = (1 + ai z ) ... (1 + a1 z )a such that ~ ai x ai ~~
-1 -1

-1

= (ax1 a-1 )(ax2 a-1 )

mod 2 = x1 x2 +(x11 x + x1 x21 )z 2

mod

2

=x+x

i+1 i+1

z

,
i+1

п and iп : L K is an i+1 -derivation. By lemma 0.12, iп is an inner +1 +1 a So there exists an element ai~ = (1 + ai+1 z i+1 )~i such that +1 ai~ xai~ +1 +1
-1

-derivation.

=x+x

i+2 i+2

z

for any x L. It is clear that the sequence {ai } converges in K . Since ai mod = a, ~ ~ п the limit of this sequence is a needed lifting. п Step 3. Now suppose a is a transcendental over K . Then by step 2 there exists a п lifting a O such that a commutes with every element in L. Then L[a] = 0 and the field of fractions L(a) can be embedded in O, which contradicts the maximality 18

п of L. So we can assume that K is algebraic over L. Suppose a is an algebraic and п п separable element over L. Using a generalisation of Hensel's lemma (see below) we can find a lifting a of a such that a commutes with elements of L and a is algebraic over L, which again leads to a contradiction. п пk п Finally, let a be purely inseparable over L, ap = x, x L. Let a be its lifting which п k commutes with every element of L. Then a p - x commutes with every element of L. k If (a p - x) = r = then similarly to the beginning of this proof we deduce that k k п the image of (a p - x)c(a p - x)-1 in K is equal to r (c), where c is an element from k lemma 0.14. Since r (c) = c, we get a contradiction. Therefore, a p = x and the field п п L(a ) can be embedded in O, which contradicts the maximality of L. Thus, L = K . The theorem is proved. 2 Prop osition 0.15 (Hensel's lemma) 1 Let O be a complete valuation ring in K , I be the valuation ideal, I n = 0, and let F be a subfield in O. Let A O be such that l F Al = lA. Let f (X ) F [X ], f (A) I and f (A) I . / ^ O such that Then there exists an element A ^ a) A commutes with A, ^ - A I, b) A ^ c) f (A) = 0 ^ ^ d) Al = lA l F ~ Pro of. If A commutes with A, then ~ ~ ~ f (A + A) = f (A)+ f (A)A + P A2 ~ ~ where P F [A, A]. We use Teilor's formula here. Put A = -(f (A))-1 f (A). It's clear ~ ~ ~ that A I and A commutes with A. Moreover, A commutes with every element in F . ~ ~/ ~ ~) = P A2 I 2 and f (A + A) = f (A)+ X A I , where X F [A, A]. ~ Thus, f (A + A ~ ~ ~ Similarly we can find the element A2 = -(f (A + A))-1 f (A + A) I 2 , which commutes ~ with A, A and with every element in F and such that ~~ f (A + A + A2 ) I
4

^ ~~ Continuing this line of reason we can find the element A = A + A + A2 + .... The sum is converge because of completeness of O. 2 Remark. If n = Id, then the theorem is not true (see an example in з3).
1

the idea of the proof of this lemma was offered by N.I.Dubrovin

19

Corollary 5 Proposition 0.10 is true for any two-dimensional local skew field with n = id for al l n N. Theorem 0.16 Let K, K п n = Id for al l n N, K , (i) K is isomorphic to п a K. be two-dimensional local skew fields such that n = Id, п are commutative fields. Then K п a two-dimensional local skew field K ((z )) where za = a z ,

п п (ii) K is isomorphic to K iff k k and there is an isomorphism f : K K such = -1 that = f f . Pro of. The proof follows from corollary 5 and from the known classification of onedimensional local fields (see for example [30]). 2 п п Definition 0.17 Let K be a one-dimensional local field with residue field k , charK = п chark , let be an automorphism of the field K . Put a1 = (u)u-1 mod k . Define i N as fol lows: i = 1 if a1 is not a root of unity in k else п i = ((n - Id)(u)), where n 1: an = 1, am = 1 m < n. 1 1 Lemma 0.18 Let k be a field of characteristic 0. Any k -automorphism of a field k ((u)) with (u) = u + a2 u2 + ..., where n = 1, n 1, m = 1 if m < n, is conjugate with an automorphism : (u) = u + xui + yu2i -1 , where x k , y k , x and y depend on . Moreover, i = i . Pro of. First we prove that = f f
-1

where
in+1

(u) = u + xu

+ yu

2in+1

for some natural i. Then we prove that i = i . Consider a set {i : i N} where i = fi i-1 fi-1 , fi (u) = u + xi ui for some xi k , 1 = . Write i (u) = u + a2,i u2 + a3,i u3 + ... One can check that a2,2 = x2 ( 2 - )+ a2,1 and hence there exists an element x2 k such that a2,2 = 0. Since aj,j +1 = aj,i , we have a2,j = 0 for all j 2. Further, a3,3 = x3 ( 3 - ) + a3,2 and hence there exists an element x3 k such that a3,3 = 0. Then a3,j = 0 for all j 3. Thus, any element ak,k can be made equal to zero if n |(k - 1) and so = f f -1 where ~ (u) = u +~ ~ a
in+1

u

in+1

+~ a

in+n+1

u

in+n+1

+ ...

20

for some i, aj k . Notice that ain+1 does not depend on xi . Put x = x() = a ~ ~ ~ Now we replace by . One can check that if n|(k - 1) then ~ a and a
k+in,k j,k

in+1

.

=a

j,k-1

for

2 j < k + in

= xk x(k - in - 1) + a

k+in

+ some polynomial which does not depend on x

k

From this fact it immediately follows that a2in+1,in+1 does not depend on xi and for all k = in +1 ak+in,k can be made equal to zero. Then y = y () = a2in+1,in+1 . Now we prove that i = i . Using the formula (u) = u + nx() -1 u
n in+1

+ ...

we get i = in +1. Since f -1 f = , f -1 (n - Id)f = n - Id. Therefore, (f -1 (n - п n 2 Id)f (u)) = (( - Id)(u)) = i . Suppose f (u) = u = f1 u + f2 u + ..., f1 = 0. Let п п us show that f -1 (n - Id)(u ) = i . It suffice to check that (n - Id)(u ) = i . We п have (n - Id)(u ) = [f1 (u + aп ui + ...)+ f2 (u + aп ui + ...)2 + ...] - [f1 u + f2 u2 + ...] = i i [(f1 u + f1 aп ui +. u i The lemma is proved. 2 Prop osition 0.19 Let barK be a one-dimensional local field with the residue field k п and charK = chark . Suppose k is algebraical ly closed and chark = 0. Let , be п automorphisms of the field K . п п = k ((u)) and = f -1 f (where f is an automorphism of K ) iff Then K (a1 ,i ,y ()) = (b1 ,i ,y ( )). Pro of. The "only if" part is clear. We prove the "if" part. It is easy to see that a1 = b1 if = f -1 f . If is not a root of unity, then by lemma 0.18 is conjugate with : (u) = u. Therefore, the "if" part is proved for the case i = i = 1. Suppose now i = i = 1 and a1 = b1 are roots of unity. Lemma 0.20 Let , be k -automorphisms of the field k ((u)): (u) = u + xu п п п п п yu2in+1 , (u) = u + xuin+1 + yu2in+1 , where x/x (k )in , y = (x/x)2 y . Then and are conjugate. 21
in+1 >i

)+(f2 u2 +. u

>i

)+(f3 u3 +. u
>i

>i

)+ ...]

-[f1 u + f2 u2 + ...] = f1 aп ui +. u i

+

Pro of. Put x0 = (x/x)( п Then we have

in)

-1

. Let f be an automorphism such that f (u) = x0 u.
in+1

п f (u) = x0 u + x(x0 u)in+1 + y (x0 u)2in+1 = x0 u + x0 xu 2

+ x0 yu п

2in+1

= f (u)

From this and previous lemmas we get the proof of the proposition. 2 Corollary 6 In the conditions of the proposition suppose k is not algebraical ly closed field. Suppose n = Id. Then there exists a parameter u in k ((u)) such that (u ) = a1 u . Pro of. The proof follows from lemma 0.18. From the proposition we get also the following result: Theorem 0.21 Let K, K be two-dimensional local skew fields with the last residue fields k and k and with canonical automorphisms , . Suppose charK = chark , charK = chark , n = Id, n = Id for al l n N, the fields k, k are algebraical ly closed of characteristic 0. K is isomorphic to K iff k k and (a1 ,i ,y ()) = (a1 ,i ,y ( )). = Now let us study skew fields with canonical automorphisms of finite order.

0.3

Classification of two-dimensional lo cal splittable skew fields of characteristic 0.

In this part we assume that a two-dimensional local skew field K splits, п k K , k K , k Z (K ), char(K ) = char(k ) = 0, n = id for some n 1, п for any convergent sequence (aj ) in K the sequence (zaj z -1 ) converges in K (i.e. the maps i , i 1 are continuous, see corol lary 3). We note that the continuity of the maps i , i 1 does not depend on the choice of parameters, as it follows from lemma 0.11 and corollary 3.

22

0.3.1

The case = Id.
i = ((z - 1)(u)) r = [((z - 1)(u))z п
-i

Definition 0.22 Define N ] i Z/iZ

mod

mod

where u, z are arbitrary local parameters of K , z : K K , z (a) = ad(z )(a). Prop osition 0.23 i and r do not depend on the choice of parameters u and z . Pro of. We fix some parameters u, z : K k ((u))((z )). Let u ,z be other parameters. = Then u = (x0 u + x1 u2 + ...)+ c1 z + c2 z 2 + ... z = a0 z + a1 z 2 + ..., where xi k, ci k ((u)), , x0 = 0;

ai k ((u)),

a0 = 0

Put z = a-1 z . It's clear that ((z - 1)(u)) = ((z - 1)(u)). From the other hand 0 by corollary 4, ((z - 1)(u)) = ((z - 1)(u)). So, i does not depend on the choice of parameter z . Now we prove that ((z - 1)(u )) = ((z - 1)(u)). One can obtain this property from the following lemma. Lemma 0.24 Suppose the fol lowing relation in K holds: zuz
-1

= u + uj z j + ...,

where 1 = ... = j -1 = 0, j = 0. Then (i) for u = u + bz q we have zu z where u
q -1

= u + u 1 z + ...u

q

-1

z

q -1

+ u q z q + ...,

= uq + b - /u(u )b.

(ii) Suppose (u) = u, k , n = 1 for some natural n. Then for u = u + bz q , n|q we have zu z
-1

= u + ... +(uq + b - b)z q + ... + u

q

+j -1

z

q + j -1

+u

q

+j

z

q +j

+ ...,

where u q+j = uq+j + bj - /u(uj )b (iii) If = id, then for u = x0 u + x1 u2 + ..., where xq k , x0 = 0, we have zu z
-1

= u +(u

j

u )z j + ... u

23

Pro of. (i) We have zu z
-1

= z (u + bz q )z

-1

= u + u1 z + ... +(b + b1 z + ...)z q =
1

u + u1 z + ... +(uq + b )z q + ... = u + u

+ ... +(uq + b - /u(u )b)z q + ...,

because u 1 = (u + bz q )1 = x0 (u + bz q )+ x1 (u + bz q )2 + ... = u1 + /u(u1 )bz q + ... if u1 = x0 u + x1 u2 + .... (ii) We have zu z
-1

= z (u + bz q )z

-1

= u + uj z j + ... +(b + bj z j + ...)z q =
+1

u + uj z j + ... +(uq + b )z q + u u + ... +(uq + b - b)z q + u (iii) We have zu z 2
-1 q

q

z

q +1

+ ... + u
q

q

+j -1

z

q +j -1

+(u
q
+j

q

+1

z

q +1

+ ... + u

+j -1

z

q + j -1

+(u

+ bj )z q+j + ... = + bj - (uj )b)z q+j u
+j

= x0 (u + uj z j + ...)+ x1 (u + uj z j + ...)2 + ... = u +(u

j

u )z j + ... u

Remark. Note that this lemma works also in characteristic p > 0. So, i does not depend on the choice of parameters u and z . Now we prove it for r. Recall that in our proposition = id (because i and r were defined only for = id). By lemma 0.24 for any parameter u we have zu z
-1

= u +(u

i

u )z i + ... u

п п Therefore, [((z - 1)(u ))z -i ] = (ui ) = [((z - 1)(u ))z -i ] п If we change z by z we get z uz Hence [((z - 1)(u))z п [((z - 1)(u))z п 2 Definition 0.25 Define a = resu u
2i -
i+1 2 i 2

-1

= zuz

-1

mod

i

-i

mod ] = [((z - 1)(u))z п

-i -i

mod ] = mod ] mod i

-i

mod ]+ (a-i ) = [((z - 1)(u))z п0 п

(ui )2 24

du

k

Prop osition 0.26 a ui+1 ,...,u2i-1 .

=

a(u

i+1

,...,u

2

i-1

), i.e. a depends only on the maps

Pro of. We comment on the statement first. The maps j are uniquely defined by parameters u, z and they depend on the choice of these parameters. So it suffice to show that a does not depend on the on the choice of parameters which preserve the maps i+1 ,...,2i-1 . We can assume that i+1 = 0,...,2i-1 = 0, because we can change the parameters to make this maps to be equal to zero (see lemma 0.11). First we show that any change of the type u u = u + c1 z + ... + ci z i is equivalent to a change of parameters as follows: z z = z + a1 z 2 + ..., u u = u + ci z i + ..., i.e. we get the same maps j in both cases. The proof is by induction. One can decompose the change u u = u + c1 z + ... + ci z i in a finite number of changes u u1 = u + ci z i , u1 u2 = u1 + ci-1 z i-1 ,...,ui-1 ui = ui-1 + c1 z . So it suffice to prove our assertion for any change of the type uj uj +1 = uj + ci-j z i-j . For j = 1 the assertion is trivial. Consider an arbitrary case. By lemma 0.24, 2i-j is the first map which is not invariant under this change. By lemma 0.11,(ii) there exists a parameter z = z + ai-j z i-j +1 such that the compositum of uj uj +1 = uj + ci-j z i-j and z z = z + ai-j z i-j +1 does not change this map. To use the induction hypothesis and complete the proof we have to show that there exists a parameter u = u + bz i such that the compositum of uj uj +1 = uj + ci-j z i-j , z z = z + ai-j z i-j +1 and u u = u + bz i does not change the map 2i . Denote by 2i the map which is given by the compositum uj uj +1 = uj + ci-j z i-j , z z = z + ai-j z i-j +1 . By lemma 0.24 there exists such a parameter u iff resu
j +1

(2i - 2i )(u (ui )2 j +1

j +1

)

duj

+1

= 0.

We have uj +1 = ci-j z i-j + ... + ci z i . One can decompose the change u uj +1 in two changes: u u = u + ci-j z i-j and u u = u + ci-j +1 z i-j +1 + ... + ci z i . The second change does not change the map 2i-j , so by the induction hypothesis it suffice to prove that the residue is equal to zero for the compositum of u u = u + ci-j z i-j and z z = z + ai-j z i-j +1 . Using lemma 0.24, we can calculate ai-j : ai-j = u ((jui )-1 ci-j )ui . Note that if (ci-j ) = r is big enough then the residue is equal to zero. One can show it with help п of lemmas 0.11 and 0.24. We denote by r the minimal positive integer which satisfy this property. Let ci-j = r =N xh uh + r+1 xh uh . Then we can decompose the change u h h= u = u + ci-j z i-j in finite number of changes u u1 = u + xN uN z i-j , ...,ur-N -1 ur-N = ur-N -1 + r+1 xh uh z i-j . It is clear that it suffice to prove our assertion for h= each change. So we have to prove it for an arbitrary change u u + xuh z j , x k . We have to check that the compositum of u u + xuh z j and z z - (j - i)-1 (h - 25

r)xuh-1 z j +1 does not change the map 2i , i.e. the residue above is equal to zero. Put -(j - i)-1 (h - r)xuh-1 = b, xuh = b . Let us show that such a compositum change only the maps i+qj , q N. Moreover, we claim that i+qj (u) = const § ur+q(h-1) . Indeed, if u = u + b z j we have zu z
-1

= u + ui z i + u2i z 2i + ... +(b + b

i

+b

2

i

z 2i + ...)z j =

u + ui z i + b i z i+j + u2i z 2i + ... = i 1 2 i 2 i+2j 1 3 i 3 (u )b z - (u )b z (u )b )z i+j - u + u i z i +( (b )ui - u u 2! u2 3! u3 1 e i e 2i (u )b )z +(u2i - e! ue where ej = i if j |i. If j | i, u2i does not change. Therefore, i (b )ui - (u )b , u i+j = u u and (u i+j ) = r +(h - 1). п Then 1 2 i 2 (u )b u i+2j = - (u i+j )b - u 2! u2 and (u i+2j ) = r +2(h - 1), п u and (u п
i+
qj

i+3j

- ...

=-

(u u

i+(

q -1)j

)b -

1 2 (u 2! u2

i+(

q -2)j

)b 2 - ... -

1 q (u i )b q ! uq

q

i+

qj

) = r + q (h - 1).
j +1

If z z = z + bz

we have
j +1

z u = (z + bz +u2i z
2i+1

)u = uz + ui z
i+j +1 i+j +1

i+1

+u

i+

j

z

i+j +1 2

+ ... +m. withz
>2i+1

+ ... + buz

j +1

+(j +1)bui z
i+1

+ ... +(j +1)bu + ... + u2i z
j +1 i+j +1 2i+1

i-j +1

=

uz + ui z u(z + bz Hence,
j +1

+u

i+

j

z
j

+ ... =
j +1 2i+1

)+ ui (z + bz

j +1 i+1

)

+u
i+
j

i+

(z + bz
i+
j

)

+ ... + u2i (z + bz

)

+ ...

u and (u п
i+
j

=u

+ b(j - i)u

i

) = r +(h - 1), u
i+2
j

=u

i+2

j

- Ci2+1 b2 ui - Ci1+ 26

j +1

bu

i+

j

+(j +1)bu

i+

j

and (u п u
i+
qj

i+2

j

) = r +2(h - 1),
qj

=u

i+
qj

- - Ciq+1 bq ui - Ciq+j1+1 b

q -1 i+

u

j

- ... - Ci1+(

q -1)j +1

bu

i+(

q -1)j

+(j +1)bu

i+(

q -1)j

and (u п

i+

) = r + q (h - 1).

So if j | i, u2i does not change. If j |i but e(h - 1) - r = -1, then the residue is equal to zero ( note that e(h - 1) - r = -1 if (r - 1,i) = 1). At last, if e(h - 1) - r = -1, then one can check the assertion by direct calculations. So we have shown that the change u u = u + c1 z + ... + ci z i is equivalent to the change z z = z + a1 z 2 + ..., u u = u + ci z i + .... By lemma 0.24 the change u u = u + ci z i + ... does not change a. By lemma 0.11 the change z z = z + a1 z 2 + ... does not change i+1 ,...,2i-1 only if a1 = a2 = ... = ai-1 = 0. But then it does not change also a. Therefore, any change of the type z z = z + a1 z 2 + ..., u u = u + c1 z + ..., which does not change i+1 ,...,2i-1 , does not change also a. To complete the proof we only have to show that changes of the type u u = x0 u + x1 u2 + ..., xj k and z z = a0 z , a0 = 0 k ((u)) don't change a. It is clear for the first change. For the second change we have u
2
i

=a

-2i 0

[u

2

i

+ ia0 (a-1 )i ui - a-i (a 0 0 a
-2i 0 (i )

i-1 i 0 a0

+ ... + a0 (a

i-1 i 0)

)] =

[u
2

2

i

+ i(i +1)/2a-1 ai ui ] 0 0
2 -2i i 0u

u Therefore, u
2
i

=a

- ia

-2i-1 i i a0 u 0
2

- (i +1)/2u(i ) u2i - (i +1)/2ui = =a (ui )2 (ui )2 The proposition is proved. 2 Remark. If a two-dimensional skew field K does not split, then the numbers i, r, a can not be defined as the following example shows (cf. also the remark after theorem 0.13). Example.2 Let "((u)) < x1 ,x2 > be a free associative algebra over "((u)) with generators x1 ,x2 . Let I =< [[x1 ,x2 ],x1 ], [[x1 ,x2 ],x2 ] >. It is easy to see that the quotient S = "((u)) < x1 ,x2 > /I

2

is a "-algebra which has no non-trivial zero divisors, and in which z = [x1 ,x2 ]+ I is a central element. The elements z , ui = xi + I (i = 1, 2) are algebraically independent. Any element of S can be uniquely represented in the form f0 + f1 z + f2 z 2 + ... + fm z
2

m

I thank N.I. Dubrovin for showing me this example

27

where f0 ,...fm are polynomials in the variables u1 ,u2 : a + bu1 + cu2 + d1 u2 + d2 u1 u2 + d3 u2 + ... 1 2 S is an Ore domain (see [4]), and the quotient skew field K has a discrete valuation such that (ui ) = 0, ("((u))) = 0, (z ) = 1. The completion of K with respect to is a two-dimensional local skew field which does not split as the following lemma shows. Lemma 0.27 Suppose there exist elements u1 ,u2 in the valuation ring O of a twodimensional local skew field K such that the element z = u1 u2 - u2 u1 is a parameter and for any m z O \ z 2 O the elements [ui ,m] = ui m - mui (i = 1, 2) belong to z 2 O. Then K does not split. п Pro of. Assume the converse. Let : K K be an embedding. Consider elements f (u1 ), g (u2 ). Then m1 = f - u1 , m2 = g - u2 z O and 0 = [u1 + m1 ,u2 + m2 ] = [u1 ,u2 ]+[m1 ,u2 ]+[u1 ,m2 ]+[m1 ,m2 ] = z +[m1 ,u2 ]+[u1 ,m2 ]+[m1 ,m2 ] Note that the second and the third summands belong to z 2 O and [m1 ,m2 ] z 2 O, because m1 m2 ,m2 m1 z 2 O. But then = (0) = (z +[m1 ,u2 ]+[u1 ,m2 ]+[m1 ,m2 ]) = (z ) = 1, a contradiction. 2 In this skew field we have i = , and r, a are not defined. From the other hand, if we consider the change z u1 z , then i become equal to 1, r = 0, a = 0. So these numbers depend on the choice of parameters in this unsplittable skew field. Prop osition 0.28 Let K be a skew of this paragraph. Let char k = 0, field k ((u))((z )) with zuz -1 = u + ui e = (r - 1,i); 2i (u) = (a(0,..., 0) + field which satisfy the conditions in the beginning = 1 and i 1. Then K is isomorphic to a skew z i + u2i z 2i , where i (u) = cur , c k /(k )e , where r(i +1)/2)u-1 (i (u))2 and j (u) = 0 for j = i, 2i.

Pro of. Consider the change z z = a0 z . By lemma 0.11, (iii) we have ui = a-i ui . 0 So, ui can be made to be equal to ui = c0 u
( u i ) п

mod

i

,

where c0 k /(k )i . By lemmas 0.24 and 0.11, c0 depend only on changes of the type z z = a0 z , u u = x0 u, where a0 ,x0 k . So, c0 can be made to be equal to - c = c0 a-i x0 r+1 . Therefore, c k /(k )e , where e = (r - 1,i). 0 28

Let us show that there exists a change z z = z + a1 z 2 + ... such that j (u) = 0 for 2i > j > i. Indeed, it can be done by a sequence of changes of the type z z = z + bz j +1 . By corollary 4, for any such j there exists b such that j (u) = 0. Let us show that there exists a change such that it changes the map 2i as follows: 2i (u ) = (a + r(i +1)/2)u -1 (u i )2 . To show it we use lemma 0.24, (ii). By this lemma it suffice to show that there exists an element b such that u
2
i

- (a + r(i +1)/2)u-1 (ui )2 + bi - (ui ) b = 0

, where the prime ' denote the derivation by u. By corollary 2, i is a derivation. Therefore, we can rewrite the equation above as follows u2i - (a + r(i +1)/2)u-1 (ui )2 + b ui - (ui ) b = 0 b One can find a solution of this equation in the form b = uj ~. The equation has a ~ + u2i (ui )-2 - (a + r(i +1)/2)u-1 = 0. The last equation holds, because solution if b 2 (u ) resu (ii(u))2 du = r. Using now the same arguments as in the previous paragraph, we can complete the proof. 2

0.3.2

The general case.

Consider now the case n = Id for some natural n > 1. Lemma 0.29 Suppose the canonical automorphism of a local skew field K satisfy the property n = 1 for some natural n > 1. Then there exists a parameter z = z +a1 z 2 +... such that z uz -1 = u + un z n + u2n z 2n + ... Here j = 0 if n |j . Pro of. Let zuz
-1

= u + u1 z + u2 z 2 + ...

By corollary 2, 1 is a (2 ,)-derivation. Since n > 1, 2 = . Therefore, by lemma 0.12, 1 is an inner derivation and 2 1 (u) = du - u d. Put z = z - dz 2 . By lemma 0.11, (i) we have z uz
-1

= u + u2 z 2 + ...

By corollary 2, 2 is a (3 ,)-derivation. If n = 2 then it is inner and we can apply lemma 0.11. By induction we get that there exists a parameter z such that z uz
-1

= u + un z n + u 29

n+1

z

n+1

+ ...

where n is a (n+1 ,) = (, )-derivation, i.e. n -1 is a derivation. Note that n+1 is a (2 ,)-derivation. Indeed, by proposition 0.7 we have
n+1

n+1 (ab) =
k=0

n+1-k (a) (S

k n+1

)(b),

п a, b K

But j = 0 if j < n. Therefore, n+1 (ab) = n+1 (a)
n+2 n

(b)+n (a)(
k=0

k 1

n-k

)(b)+(a)n+1 (b) = n+1 (a)2 (b)+(a)n+1 (b)

and by lemma 0.12, n+1 is an inner derivation. Using lemma 0.11 with z z = z + bz n+2 for an appropriate b, we have z uz
-1

= u + un z

n

+u

n+2

z

n+2

+ ...

with n+1 = 0. By induction we can assume that there exists a parameter z such that z uz Then if n |k +1, k
-1

= u + un z n + u is a (
k+2

2

n

z

2n

+ ... + u

k

+1

z

k+1

+ ...

+1

,)-derivation. Indeed,
k+1

k+1 (ab) =
l=0 k+2 x

(k

+1-l

l )(a) (Sk+1 )(b) =

k+1 (a)

(b)+
m=1

mn (a) (S

k+1-mn k+1

)(b)+ (a)k+1 (b)

where x N : xn k +1, (x +1)n > k + 1, because j = 0 if j < k +1 and n |j . k+1 Every monomial (Sk+1 -mn ) contain an element j with j < k + 1 and n | j . l It follows from the definition of Sk+1 together with n |k + 1 - mn. Therefore, k+1 (Sk+1 -mn )(b) = 0 m and k+1 is a (k+2 ,)-derivation. If n|k + 1, we can apply the same arguments and conclude that k+2 is a (k+2 ,)derivation. Therefore, by lemma 0.11 there exists a parameter z = z +bz k+2 (z +bz k+3 if n|k + 1) such that z uz
-1

= u + un z u
k
+1

n

+u +u

2

n

z z

2n

+ ... + u + ...

k

+2

z

k+2

+ ...

(or

z

k+1

k

+3

k+3

if n|k +1)

Since zl+1 = (1 + zll )zl , the sequence of the lemma. 2

{zl } l=1

converges in K . Therefore, we get the proof

30

Lemma 0.30 There exists a parameter u K such that (u) = u, where n = 1, and for al l j jn (u) = u( k yjk unk ) uk ((un )), where yjk k , and for k not divisible by n k = 0. Pro of We can assume that the relation from lemma 0.29 holds. We will do changes of the form u u = u + bjn z jn . We have seen in the proof of lemma 0.24 that the maps k , n | k don't change under such substitutions. By lemma 0.24, (i) we can see that u jn = ujn + b - /u(u )b. By corollary 6 we can assume (u) = u, where n = 1. Therefore, u jn = ujn + b - b and we can find an element b to satisfy the conditions of the lemma. 2 As in the case = Id we can define in ,rn and an . Definition 0.31 Put in = (( rn = [((zn - 1)(u))z п an = resu
z
n

- 1)(u)) mod
2
in in +1 2 in 2

N ] mod du

i
n

-in

Z/in Z

u

-

(u

in )2

k

where u, z are arbitrary local parameters in K , z : K K , z (a) = ad(z )(a). From the previous two lemmas we can derive that if z is a local parameter from lemma 0.29 then in nN and rn = 1 mod n. It is easy to see that the number in is the number of the first non-zero map in in lemma 0.30. In the same way as in the proof of proposition 0.26 we can get the following result: Prop osition 0.32 We have i an (uin +1 ,...,u2in -1 ). Prop osition 0.33 Let K be a ditions in the beginning of this Let chark = 0 and n = Id Then K is isomorphic to a zuz
n

=

in (uj ,j

/

nN), r

n

=

rn (in ), a

n

=

two-dimensional local skew field which satisfy the conparagraph. for some natural n. skew field k ((u))((z )) with the relation = u + u
in in

-1

z +u

2

in

z

2in

,

where n = 1, in = in (0,..., 0), in (u) = curn , c k /(k )e , e = (rn - 1,i), 2in (u) = (an (0,..., 0) + rn (in +1)/2)u-1 (in (u))2 .

31

Pro of. We can assume that the conditions of lemma 0.30 hold. Then, because of special choice of the element ujn it suffice to repeat the proof of proposition 0.28. 2 Combining all these results we get: Theorem 0.34 Let K and K be local skew fields of characteristic zero. Suppose they satisfy the conditions in the beginning of this chapter. Then K is isomorphic K iff k k and the sets (n, , in ,rn ,c,an ), (n , ,in ,rn ,c ,an ) coinside. = Remark. If n = 1 and in = , then K is a commutative two-dimensional local field k ((u))((z )). Let us now summarise all the classification results we have got above. Theorem 0.35 (I) Let K be a two-dimensional local skew field with a commutative residue skew field. It splits if the canonical automorphism satisfy the condition n = Id for al l n. If this condition does not hold, there are examples of non-splittable skew fields. (II) Let K, K be skew fields as in (I). Assume n = Id, (a) K is isomorphic to a two-dimensional local skew field п п a K and K is a one-dimensional local field with the residue = Id for al l n. Then п ((z )) where za = a z , K field k .
n

(b) K and K are isomorphic iff k k and there exists an isomorphism = п п f : K K such that = f -1 f . (c) If charK = chark , charK = chark and k, k are algebraical ly closed fields of characteristic 0, then K is isomorphic to K iff k k and = (a1 ,i ,y ()) = (a1 ,i ,y ( )). (III) Let K, K be two-dimensional splittable local skew fields of characteristic 0, k Z (K ), k Z (K ), and n = Id, n = Id for some natural n, n 1. Then (a) K is isomorphic to a two-dimensional local skew field k ((u))((z )) where zuz
-1

= u + u

in in

z +u

2

in

z

2in

,

where n = 1, in = in (0,..., 0), in (u) = curn , c k /(k )e , e = (rn - 1,i), 2in (u) = (an (0,..., 0) + rn (in +1)/2)u-1 (in (u))2 (in ,rn ,an were defined in 0.31). If n = 1, in = , then K is commutative. (b) K is isomorphic to K iff k k and the sets = (n, , in ,rn ,c,an ), (n , ,in ,rn ,c ,an ) coinside. 32

Corollary 7 Every two-dimensional local skew field K with the ordered set (n, , in ,rn ,c,an ) is a finite-dimensional extension of a skew field with the ordered set (1, 1, 1, 0, 1,a) if in < . Remark. It's easy to see from the corollary that skew fields in the theorem above are almost always infinite dimensional over the centre. Namely, the only finite dimensional skew fields are the skew fields with in = . In the case of skew fields of positive characteristic the situation is much more complicated.

0.4

Splittable skew fields of characteristic p > 0.

It is difficult to classify all splittable two-dimensional skew fields with the canonical automorphism of finite order in positive characteristic even if we consider only skew fields with = id, at least because there are infinitely many maps j which can not be removed by any change of parameters. Nevertheless, our methods give some useful tools for studying splittable skew fields finite dimensional over their centre. For splittable skew fields in positive characteristic one can define an invariant which is in some sense a replacement of the invariant an for skew fields of characteristic 0. Certainly, there are infinitely many of other invariants. Namely, if K is a splittable two-dimensional local skew field of positive characteristic п п п with K commutative, k = K K , k K , k Z (K ), of finite order we define dK = max (zuz
u,z (z ) -1

- (u) - in (u)z in ),

(z )

where in is a map defined by a parameter z , and in is a number defined in 0.31. In the case of a skew field of characteristic 0 we have dK = 2in (0,..., 0) or dK = , that is why it is in some sense a replacement of an : in characteristic 0 it reflects the property of an to be zero or not. In this section we will prove the following theorem: п Theorem 0.36 Suppose that a two-dimensional local skew field K splits, K is a field, п Z (K ), char(K ) = char(K ) = p > 2, = id, and d 2i = 2i or d = . п k=K K 1 K Then K is a finite dimensional vector space over its centre if and only if K is isomorphic to a two-dimensional local skew field k ((u))((z )), where z -1 uz = u + xz п with x K p , (i, p) = 1. 33
i

To prove this theorem we prove more general result about finite dimensional algebras, which generalises some known results of Jacob and Wadsworth in [9] and Saltman [28]. As a corollary we get the positive answer on the conjecture about exponent and index of a finite dimensional division algebra over a C2 -field for some types of C2 -fields. These results will be proved in the subsection below. Now we prove 2 п only the "if" part. Indeed, since x K p , we have i (u) = 0. Hence, by corollary 1 we j/i п have j = ci , c k if i|j , and j = 0 if i |j . But then zap z -1 = ap for any a K , so K is a finite dimensional skew field over its centre and the index indK = p. To prove the "only if" part we need results from the following subsection:

0.4.1

Wild division algebras over Laurent series fields

In this subsection we prove a decomposition theorem for some class of wild division algebras over a Laurent series field with arbitrary residue field of characteristic greater than two. Namely, we prove this theorem for wild division algebras which satisfy the folп п lowing condition: there exists a section D D of the residue homomorphism D D, where D is a central division algebra. This theorem is a generalisation of the decomposition theorems for tame division algebras given by Jacob and Wadsworth in [9]. An extensive analysis of the wild division algebras of degree p over a field F with complete п discrete rank 1 valuation with char(F ) = p was given by Saltman in [28] ( Tignol in [32] analysed more general case of the defectless division algebras of degree p over a field F with Henselian valuation). The main result of this subsection is Theorem 0.55; it is a corollary of Theorem 0.43 and propositions 0.51-0.54. Theorem 0.43 is a key tool in the proof of Theorem 0.55. As a corollary we get the positive answer on the following conjecture: the exponent of A is equal to its index for any division algebra A over a C2 -field F = F1 ((t2 )) (corollary 8) (see also [37], corollary 4, з8.3.2.), where F1 is a C1 -field. We note that the proof of the conjecture does not depend on the statement of theorem 0.55, but uses only several lemmas from it's proof. In fact, these lemmas are most important technical statements about the maps m and their generalisations. We change the notation in this subsection and use the notation of [9], because it is more convenient for applications in the valuation theory. We always denote here by D a division algebra finite dimensional over its centre F = Z (D). Recall that any Henselian valuation on F has a unique extension to a valuation on D. Given a valuation v on D, we denote by D its value group, by VD its valuation п ring, by MD its maximal ideal and by D = VD /MD its residue division ring. By [31], p.21 one has the fundamental inequality пп [D : F ] |D : E |§ [D : F ].

34

D is called defectless over F if equality holds and defective otherwise. It is known that D is defectless if it has a discrete valuation of rank 1. Jacob and Wadsworth in [9] introduced the basic homomorphism пп D : D /F Gal(Z (D)/F ) п induced by conjugation by elements of D. They showed that D is surjective and Z (D) п is the compositum of an Abelian Galois and a purely inseparable extension of F . п п We say D is tame division algebra if char(F ) = 0 or char(F ) = q = 0, D is defectless п ) is separable over F , and q ||ker(D )|. We say D is wild division algebra п over F , Z (D if it is non tame. п п We call a division algebra D inertial ly split if Z (D) is separable over F , the map D is an isomorphism, and D is defectless over F .

0.4.2

Cohen's theorem

There is a natural question if there exists a generalisation of Cohen's theorem, i.e. is any central division algebra splittable or not. It is not true if a division algebra is not finite dimensional over its centre, as Dubrovin's example shows. It is not true also for finite dimensional division algebras, as we will see in Wadsworth's example below. But it is true for tame division algebras over complete discrete valued fields. This easily follows from results of Jacob and Wadsworth [9]. Theorem 0.37 Let (F, v ) be a valued field which is complete with respect to a discrete п rank 1 valuation v . Suppose charF = charF . Let D be a tame division algebra with Z (D) = F and [D : F ] < . п п Then there exists a section D D of the residue homomorphism D D. Pro of. Since F is a complete field, F is a Henselian field and v extends uniquely to п a valuation w on D. Since D is tame, Z (D)/Z (D) is a cyclic Galois extension. There п exists an inertial lift Z of Z (D) over F , Z is Galois over F , and by classical Cohen's ~п theorem there exists a section Z (D) Z . п п Consider the centraliser C = CD (Z ) of Z in D. Then we have C = D. Indeed, by Double Centraliser Theorem we have [D : F ] = [C : F ][Z : F ] пп and [Z : F ] = |Gal(Z (D)/F )|. By [9], prop.1.7 a homomorphism D : D /F п )/F ) is surjective, so for any parameter z we have D (w(z )) = , where п Gal(Z (D пп < >= Gal(Z (D)/F ). It is clear that z C . Now let u1 ,...,u[C :F ] be a F -basis of / C . It is easy to see that the elements uj ,z uj ,...,z n-1 uj , j = 1,..., [C : F ], where n = ord( ), the order of , are linearly independent, so form a basis for D over F . Since w(F < z uj ,...,z n-1 uj ,j = 1,..., [C : F ] >) C = 0, 35

where F < z uj ,...,z n-1 uj ,j = 1,..., [C : F ] > denote a vector space in D generated by elements uj z i , this implies that for any element x D with w(x) can find elements r1 ,...r[C :F ] F such that x = r1 u1 + ... + r[C :F ] u[C :F ] mod п п Hence C = D. п п Fix an embedding F F and consider the algebra A = C F Z (C ). It is п п п see that A is an unramified division algebra with A = C = D. Therefore by [2], п A C ; so there exists a section D C . = The theorem is proved. 2

over F = 0 we MD . easy to Th.31,

Example (Wadsworth). Let p be any prime number, let k = Z/pZ , the field with p elements, and let r, s be independent indeterminates over k . Let F = k (r, s)((t)), the formal Laurent series field in t over the rational function field k (r, s). F has its complete п discrete t-adic valuation with residue field F = k (r, s). Let f = xp + tx - r in F [x]. The derivative test shows that f has no repeated roots. Let be a root of f , and let K = F (), which is separable over F . Let M be the separable closure of K over F . So, the Galois group Gal(M/F ) is isomorphic to a subgroup of the symmetric group Sp . Let L be the fixed field of a p-Sylow subgroup of Gal(M/F ), and let be a generator of Gal(M/L), a cyclic group of order p. The valuation on F extends uniquely to complete discrete valuations on L пп п and on M . Note that L doesn't contain r1/p , since [L : F ] divides [L : F ], which is п п п prime to p. (For the same reason, L doesn't contain a p-th root of s.) But M contains п : L] = [M : L] = p, and M = L(r1/p ), which is purely п п п , which is a p-th root of r. So, [M п п inseparable over L. Since acts trivially on M , the norm map from M to L induces п п the p-th power map from M to L. So, s is not in the image of the norm from M to L. Therefore, the cyclic algebra D = (M /L, , s) is nonsplit of degree p, so it is a division п algebra. With respect to the unique extension of the valuation on L to D, we have D 2 п : L] [L(r1/p ,s1/p )] = p2 . п п contains a pth root of r and also of s, so p = [D : L] [D п п п This shows that D is the field L(r1/p ,s1/p ), which is purely inseparable over L. Hence also, the ramification index of D over L must be 1.

0.4.3

Decomp osition theorem

In this part we prove a generalisation of Jacob-Wadsworth's decomposition theorem ([9], Th.6.3., lemma 6.2) for finite dimensional splittable division algebras over a Laurent series field k ((t)), chark > 2. So, in this section we consider only splittable division algebras. Moreover, we will need more strong condition: Definition 0.38 A division algebra D is cal led good splittable if there exists a section п п D D compatible with an embedding Z (D) Z (D), i.e. Z (D) Z (D) D. 36

It's easy to see that all tame division algebras are good splittable, because by Cohen's theorem any embedding Z (D) Z (D) can be uniquely extended to any separable extension of Z (D). We note that the skew field K from theorem 0.36 is good splittable if K is a finite dimensional division algebra over its center. Indeed, because of the condition of the theorem, we can assume k is an algebraically closed field. Then the center of K is a C2 -field by Tsen's theorem (see the definition and the properties of C2 -fields below, at the end of this subsection). Then it will be shown in corollary 8 that all division algebras over C2 -fields are good splittable. For division algebras of index p over a Laurent series field the condition to be splittable is equivalent to the condition to be good splittable, see the end of this subsection. Let D be a finite dimensional division algebra over a complete valued field F = k ((t)). Let w be a unique extension of the valuation v to D. We will denote by z any parameter of D, i.e. any element with < w(z ) >= D . п Prop osition 0.39 D is isomorphic to a local skew-field D((z )), where zaz
-1

= (a)+ 1 (a)z + 2 (a)z 2 + ...,

п a D;

п п п п here : D D is an automorphism and i : D D are linear maps such that the map i satisfy the identity
i

i (ab) =
k=0

(

i-k

)(a) (Sik )(b),

п a, b D

The proof is an easy combination of the proofs of propositions in section 1 and Cohen's theorem 0.37. Definition 0.40 Let us define maps z m az If m = 0, put
-m m i

п п : D D, m Z, i N as fol lows. п a D.

= m (a)+ m 1 (a)z + m 2 (a)z 2 + ...,

m i

= 0.

Note that if = id, then m i = 0 for m = pk , where k is sufficiently large, k depends on i. Moreover, m i = m+pk i for k sufficiently large. We will use also the following notation: ~ m i = -m i . Prop osition 0.41 (i) The maps
m i m i

satisfy the fol lowing identities:
i-1 m m i-k k=1

(ab) = m i (a)

i+m

(b)+ (a)m i (b)+ 37

(a)i-

k+m k

(b)

(ii) Suppose = id. Then the maps
m i

m i i-1

satisfy the fol lowing identities: (a)
(j1 ,...,jl )

(ab) = m i (a)b + am i (b)+
k=1

m i-k

Cil-

k+m j

1

...jl (b)

where the second sum is taken over al l the vectors (j1 ,...,jl ) such that 0 < l min{i - k k k jm = k ; Cj = 0 if j = 0, and Cj = Cj +pq for q >> 0 if j 0. k + m, k }, jm 1, п Pro of. For any a, b D we have m (ab)z m + m 1 (ab)z ()
m+1

+ m 2 (ab)z
m+1

m+2

+ ... = z m (ab) =
m+2

(m (a)z m + m 1 (a)z

+ m 2 (a)z

+ ...)b

If we represent the right-hand side of () as a series and then compare the corresponding coefficients on side, we get some formulas for m i (ab). We have to same as in our proposition. Let z i+m-k b = i+m-k (b)z i+m-k + ... and (m (a)z m + m 1 (a)z Then we have y
i+m m+1

with coefficients shifted to the left the left-hand side and right-hand prove that these formulas are the
i+m

+ xk z

+ ...

+ m 2 (a)z

m+2

+ ...)b = m (ab)z m + y
i-1

m+1

z

m+1

+y

m+2

z

m+2

+ ...

= (a)xi +
k=0

m

m i-k

(a)x

k

In the proof of prop. 0.7 we have shown that z
i+1-k

b=

i+1-k

(b)z

i+1-k

+ ... + (Sik )(b)z

i+1

+ ...
m i

Hence xk = (Sik+m-1 )(b) for k < i. It is easy to see that xi = and (Sik+m-1 ) = i+m-k k , which proves (i). For = id, by corollary 1, (S
k i+m-1

(b), x0 =

i+m

(b)

)(b) =
(j1 ,...,jl )

Cil-

k+m j

1

...jl (b),

where l, j1 ,...,jl were defined in our proposition. This proves (ii). The proposition is proved. 2 38

п Definition 0.42 Let D be a splittable division algebra. For any element a D define the numbers i(a) = max w(zaz -1 - (a)) N ,
j (a),z

п where j : D D, z -- parameter in D; dD (a) = max w(zaz
j (a),z -1

- (a) - ij

(z ) ij (a) (a ) z

) N ,

where ij (a) = w(zaz -1 - (a)) for a given embedding j . It does not depend on the choice of z as lemma 0.11 shows. The following theorem is a main technical result of this subsection. п Theorem 0.43 Let D be a good splittable division algebra. Let u Z (D) be a purely inseparable element over Z (D) and charF > 2. Then dD (u) = 2i(u)+ np, where n > 0, п and up F only if dD (u) = . п Pro of. By proposition 0.39 we can assume that u D D. Without loss of пп generality we can assume that Z (D)/F is a purely inseparable extension. Moreover, it п п can be assumed that Z (D) = F (u). Then by Scolem-Noether theorem we can choose a parameter z such that = id. Suppose zuz
-1

= u + i (u)z i + ...,
k

i.e. 1 |F (u) = ... = i-1 |F (u) = 0, i (u) = 0. Suppose up F . Without loss of generality it can be assumed that the following property holds: ) let j , j > i be the first map such that j = 0 if j is not divisible by i and j/i п j = cj/i i for some cj/i D otherwise; then j = 2i mod p. Indeed, let j be the first map such that j = 0 and i |j , j = 2i mod p. Then by lemma 0.11, (ii) there exists a parameter z such that j (u) = 0. Therefore by corollary 1, j |F (u) = 0. By Scolem-Noether theorem, j is an inner derivation. Therefore by lemma 0.11 (ii) there exists a parameter z such that j = 0. j/i If j is the first map such that i|j , j |F (u) = cj/i i |F (u) and j = 2i mod p,thenby j/i lemma 0.11, (ii) there exists a parameter z such that j (u) = cj/i i j/i (u) = cj/i i (u), where (i +1) ... (i(j/i - 1) + 1) (1) cj/i = (j/i)! One can easily show that j |F (u) = cj/i i j/i |F (u) . By Scolem-Noether theorem, (j - j/i cj/i j ) is an inner derivation. Therefore by lemma 0.11 (ii) there exists a parameter z j/i such that j = cj/i i . Let's divide our proof in two steps. 39

Step 1. First we prove that (i, p) = 1. In this Step we don't use the condition D is a good splittable algebra. We use only a condition that D is splittable. Lemma 0.44 Let j be the minimal positive integer such that j |F (upl ) = 0, l 0. Then the maps n m , kj m < (k +1)j , k {1,...,p - 1} satisfy the fol lowing properties: п i) there exist elements cm,k D such that (n m - c
l

m,1

- ... - c

m,k

k )|F

(up )
l

l

= 0,

п п where : D D is any F -linear map such that |
l

F (up )

is a derivation, (uj ) = 0 for

j pl N, (up ) = 1, ckj,k = c(j (up ))k , c Fp . / ii) ckj,k = 0 iff n, n + j,...,n +(k - 1)j = 0 mod p. Pro of. i) The proof is by induction on k . Let a, b F (up ). Put t = up . For k = 1, by proposition 0.41, (ii) we have
n m
l l

(ab) = n m (a)b + an m (b)
l

because all the maps q , q < j are equal to zero on F (up ). Hence, n m is a derivation l on F (up ) and cj,1 = n j (t) = nj (t). For arbitrary k , by proposition 0.41, (i) and by the induction hypothesis we have ()
q -2 l=0 q -2 l=0 n m

(t ) = q n m (t)t

q

q -1

+n j (t)(

(c1 +...+c

k -1

k -1

)(t

q -1-l

)t )+...+n s (t)(

l

(g1 )(tq

-1-l

)tl ),

п where cj ,gj D, s p- 2 j p-1-l l )t = l=0 cj (t and we only have to s Using (**) we can

> m - 2j . Therefore, n m (tp ) = 0, because k p - 1 and 0 for j p - 2. Hence, n m |F (t) = cm,1 + ... + cm,p-1 p-1 how that cm,q = 0 for q > k . calculate cm,j . We have c
m,1

= n m (t);

c

m,2

=

1 2 n m (t ) - 2c 2!
q -2 q -1

m,1

t=

1 n j (t)(c1 (t)) + ... + n s (t)(g1 (t)) 2 ...
q -2 q +1

c

m,q

=

1 (n j (t)( q!

c
l=0

q -1

(tq

-1-l

)tl )+ ... + n m- + ... + n m- 40

(t)(
l=0

g )

q -1

q -1

(tq

-1-l

)tl ))

1 = (n j (t)c q

q -1

q +1

(t)g

q -1

Hence, cm,k+1 = ... = cm,p-1 = 0 and ckj,k = cj (t)ck-1 = c(j (t))k , c, c Fp . Note that ~ ~ c(p-1)j,p = (j j (t))p-1 . ii) Suppose n = 0 mod p. For k = 1 we have n j |F (t) = nj |F (t) = 0. For arbitrary k we have
q -2 n kj kj -1 n+j (k-1)j r =0 q -2 n l

(t ) = q n kj (t)t

q

q -1

+n j (t)

(t

q -1-r

)t +...+
l=j +1

r

(t)
r =0

n+l kj -l

(tq

-1-r

)tr

Since n j (t) = 0 and m1 h |F (t) = c1 + ... + ck-2 k-2 for h < (k - 1)j , the same arguments as in i) show that ckj,k = 0. Suppose n + rj = 0 mod p, r < k - 1. The same arguments as above show that in this case ckj,k (n kj ) = 0 iff c(k-1)j,k-1 (n+j (k-1)j ) = 0. So, by induction, ckj,k (n kj ) = 0 iff c(k-r)j,k-r (n+rj (k-r)j ) = 0, which proves ii). The lemma is proved. 2 Lemma 0.45 If p|i, then there exists a map j such that j (up ) = 0. Pro of. We claim that pq i is the first map such that pq i |F (upq ) = 0. The proof is by q -1 induction on q . For q = 0, there is nothing to prove. For arbitrary q , put t = up . By proposition 0.41 we have
p- 2 pq i-1 1+p r =0
q -1 k

p- 2

pq i (t ) = p

p

q -1

i

(t)

i p

q -1

i(p-1)

(t

p-1-r

)t +
l =p
q -1

r

l (t)
i+1 r =0

1+l pq i-l

(tp

-1-r

)tr

By induction and lemma 0.44, 1+l pq i-l |F (t) = c1 + ... + cp-2 p-2 for l > pq-1 i. p- 2 p-1-r r )t = 0. By lemma 0.44, (ii), 1+pq-1 i pq-1 i(p-1) |F (t) = Therefore, r =0 1+l pq i-l (t c1 + ... + cp-1 p-1 with cp-1 = 0. Hence, pq i (tp ) = -cp-1 pq-1 i (t) = 0. The same arguments show that j (tp ) = 0 for j < pq i. So, pq i is the first non-zero q map on F (up ). 2 Step 2. п п From now on (i, p) = 1. Note that i (u) Z (D). Indeed, for any a D we have i (au) = i (a)u + ai (u) = i (ua) = ui (a)+ i (u)a Therefore, ai (u) = i (u)a. Since (i, p) = 1, there exists k1 N such that pk divides 1 + k1 i. Therefore by lemma 0.11 (iii) there exists a parameter z such that i (u) = п (i (u))1+k1 i Z (D). п So we can assume that i (u) Z (D) and ) holds. 41

Assume that dD (u) 2i(u). Then to prove our theorem it is sufficient to show that there exists a parameter z such that the maps j satisfy the following property: () If j is not divisible by i, then j |F (u) = 0. If j is divisible by i, then j |F (u) = j/i п cj/i i |F (u) with some cj/i D. To show it we prove that if property () does not hold, then there exists a map j k such that j (up ) = 0. Suppose () does not hold and 2i+mp is the first map which does not satisfy (). So, 2i+mp (u) = 0. Note that q (u) = 0 for i < q < 2i + mp. ~ ~ п Note that 2i+mp (u), 2i+mp (u) Z (D). Indeed, by proposition 0.41, i 2i+mp |F (u) is a ~ п derivation. Therefore, i 2i+mp (u) Z (D). Since i (u) Z (D) and q (u) = 0 for i < q < ~ ~ ~ п п 2i + mp, i 2i+mp (u) = i2i+mp (u) and 2i+mp (u) Z (D). Therefore, 2i+mp (u) Z (D). п First we prove that there exists a parameter z such that q = q for q 2i + mp п ~ (u) = 0 for q = 2i mod p, q > 2i + mp; here are maps given by п п and 2i+mp+(p-1)i q q the parameter z . Put j (1) = 2i + mp +(p - 1)i. п ~ Suppose j (1) q (u) = 0, q > 2i + mp and q = 2i mod p. By definition,
j (1) q

~ (u) = -j (1)q (u)+

k1 ...kl (u),

п п where ki < q . By lemma 0.11, (ii) for any a D there exists a parameter zq such that zq uzq 1 = u + i (u)пq + ... + q-1 (u)пq п п- zi zq п Therefore, there exists an element a D such that the sequence {zq } converges in D. So, z = limzq . п п п
-1

+ azq + ... пq

j (1) q

~ (u) = 0. It is easy to see that п

Lemma 0.46 Put = j (1) = 2i + mp + (p - 1)i. Then there exists a parameter z such that the fol lowing properties hold: ~ (i) 2i+
mp+(p-1)i

~ is the first map such that 2i+ ~ (up ) = 0 and r |F
(up )

mp+(p-1)i |F (up )

= 0.

~ (ii) 2i+

mp+(p-1)i+i+mp mp+(p-1)i

= 0 for j (1) < r < j (1) + i + mp. п (up ) Z (D).

~ (iii) 2i+

~ п (up ) Z (D), 2i+

mp+(p-1)i+i+mp

Pro of. i) Put w := 2i + mp +(p - 1)i. By proposition 0.41 we have
p- 2 p- 2 i- w-i q =0

~ ~ w (up ) = i (u)

(u

p-1-q

~ )u + 2i+
q

mp

(u)
q =0

2i+mp- (p-1)i

(u

p-1-q

)uq +

42

w -1 k=2i+mp+1

p- 2

~ k (u)
q =0

k- w-k

(u

p-1-q

)u

q

By lemma 0.44, k- w-k |F (up ) = c1 + ... + cp-2 p-2 for w - k < (p - 1)i and p- 1 with cp-1 = (i i (u))p-1 = 0. 2i+mp- (p-1)i |F (up ) = c1 + ... + cp-1 By proposition 0.41 we have
q -2 i- w-i

(u ) = q

q

i- w-i

(u)u

q -1

+

i- i

(u)
r =0

m1 w-2i

(u

q -1-r

)ur +

w-i-1 i- s s=2i+mp

q -2

(u)
r =0

s+i- w-i-s

(u

q -1-r

)u

r

By lemma 0.44, s+i- w-i-s |F (u) = c1 +...+cp-3 p-3 for w -i-s < (p-2)i.Since i- = 0 mod p, i- i (u) = 0 and i- 2i+mp (u) = 0. So, i- w-i |F (u) = c1 + ... + cp-2 p-2 . Hence, p- 1 ~p ~ ~ = -i 2i+mp (u)(i i (u))p-1 = 0 w (u ) = - 2i+mp (u)(i i (u)) ~ п and w (up ) Z (D). ~ ~ The same arguments show that w is the first map such that w |F (up ) = 0. ii) For j (1) < w 2i + mp +(p - 1)i + i + mp, by proposition 0.41 we have
p- 2 p- 2 i- w-i q =0 p- 2

~ ~ w (up ) = i (u)

(u

p-1-q

~ )u + 2i+
q

mp

(u)
q =0 w -1

2i+mp- w-2i-mp

(u

p-1-q

)uq +... +

p- 2

~ w

-(p-1)i

(u)
q =0

w-(p-1)i-

(

p-1)i

(u

p-1-q

)uq +
k=w-(p-1)i+1

~ k (u)
q =0

k- w-k

(u

p-1-q

)u

q

By lemma 0.44, k- w-k |F (u) = c1 + ... + cp-2 p-2 for w - k < (p - 1)i. Let us prove that 2i+mp- |F (u) = c1 + ... + cp-2 p-2 for (p - 1)i < < (p - 1)i + i + mp. If (p - 1)i < < 2i + mp, then it is clear that 2i+mp- |F (u) = 0. By proposition 0.41, for 2i + mp and q < p we have
2i+mp- q -2 2i+mp- i

(uq ) = q

2i+mp-

(u)u

q -1

+ (u
q -1-r

q -2 m -i

(u)
r =0

(u

q -1-r

)u +

r

2i+mp- 2i+mp

(u)
r =0 p- 3

m1 -2i-mp

)ur + ...

Since - 2i - mp < (p - 2)i,

m1 s |F (u)

= c1 + ... + c

p- 3

for s - 2i - mp.

43

To show that m -i |F (u) = c1 + ... + cp-3 p-3 we use induction on r, where i + mp - (p - 2 - r)i < 2i + mp + ri. For arbitrary r we can use the same calculations, so we only have to prove that m -(p-2-r)i |F (u) = 0 for some r 0. There exists r 0 such that i + mp - (p - 2 - r)i < 2i + mp. If 2i + mp > 2i, then k -(p-2-r)i = ci , k > p if i| and m -(p-2-r)i = 0 otherwise. So, m -(p-2-r)i |F (u) = 0. m If 2i + mp 2i, then (p - 1)i < < pi. So, i < - (p - 2)i < 2i + mp and m -(p-2)i = 0. Let us prove that 2i+mp- (p-2)i+2i+mp |F (u) = c1 + ... + cp-1 p-1 with cp-1 = 0. Note ~ that 2i+mp- 2i+mp+ri |F (u) = c1 + ... + cr+1 r+1 with cr+1 = 0 for any 0 r p - 2. Indeed, by proposition 0.41 we have
q -2 2i+mp-

~ 2i+

mp+ri

(u ) = q

q

2i+mp-

~ 2i+

mp+ri

(u)u

q -1

+

2i+mp-

~ i (u)
t=0

q -2 l C-i-mp j (j1 ,...,jl ),lp
1

...jl (u

q -1-t

)u +

t

2i+mp-

~ 2i+

mp

(u)
t=0

ri

(u

q -1-t

)ut + ...

By lemma 0.44, ri |F (u) = c1 + ... + cr r with cr = 0 and m s |F (u) = c1 + ... + cr-1 r-1 for s < ri. ^ If there exists jk 2i + mp, then j1 + ... + jk + ... + jl ri; so there exists jt < i and k j1 ...jl = 0. If there are no jk 2i + mp, then jk = ci , c " and j1 ...jl |F (u) = 0, because l p. ~ Hence by lemma 0.44, 2i+mp- 2i+mp+ri |F (u) = c1 + ... + cr+1 r+1 with cr+1 = 1 1~ ~ (u)(i i (u))r = r+1 i 2i+mp (u)(i i (u))r = 0. r +1 2i+mp- 2i+mp We have
2i+mp- (p-2)i+2i+mp

+

2i+mp- (p-2)i+2i+mp

~

+

2i+mp- i

§

2i+mp- (p-3)i+2i+mp

~

+ ... +

2i+mp- (p-2)i

§

2i+mp- 2i+mp

~

+

2i+mp- (p-3)i+2i+mp

§

2i+mp- i

~ =0

~ We have 2i+mp- ri § 2i+mp- (p-2-r)i+2i+mp |F (u) = c1 + ... + cp-1 p-1 with cp-1 = 1 ~ (u)(i i (u))p-2 . p-1-r i 2i+mp ~ Since (2i+mp- i )p |F (u) = 0, and using induction, we get 2i+mp- (p-2)i+2i+mp |F (u) = c1 + ... + cp-1 p-1 with c (1+...+
p- 1

= -(1 + ... +

1 ~ )i 2i+ p-1

mp

(u)(i i (u))p-2 -

w -(

1 ~ ~ ~ )i 2i+mp (u)(i i (u))p-2 -...-i 2i+mp (u)(i i (u))p-2 = -i 2i+mp (u)(i i (u))p-2 = 0 p-2 -2 ~ Note that w-(p-1)i (u) p=0 w-(p-1)i- (p-1)i (up-1-q )uq = 0 only if w = i mod p. q Indeed, suppose w - (p - 1)i - = i mod p. Therefore by lemma 0.44, (ii), p- 2 . p-1)i- (p-1)i |F (u) = c1 + ... + cp-2 44

Let us prove that
i- w-i q

p- 2 q =0 i- w-i

(u

p-1-q

)uq = 0. By proposition 0.41 we have
q -2 i- i

(u ) = q

i- w-i

(u)u

q -1

+

(u)
r =0

2i- w-2i

(u

q -1-r

)ur +

w-i-1 i- s s=2i+mp

q -2

(u)
r =0

s+i- w-i-s

(u

q -1-r

)u

r

Since i- = 0 mod p, i- s (u) = 0 for s < 2i+mp+(p-1)i.For s 2i+mp+(p-1)i k we have w - i - s mp and s+i- w-i-s = ci , c ". But m 0 by our assumption in the beginning of Step 2, so s+i- w-i-s = 0. ~ So, we have w (up ) = 0 only if w = i mod p or w = 2i + mp +(p - 1)i + i + mp.By lemma 0.11, (ii),(see the same arguments before this lemma, for example) there exists a ~ parameter z such that the map w (up ) becomes equal to zero on up if w = i mod p. Since 2i + mp +(p - 1)i + i + mp - w i by our assumption, the change from lemma ~ 0.11 does not change the map 2i+mp+(p-1i+i+mp . So, we get the proof of (ii). ~ ~ ~ Now we have 2i+mp+(p-1i+i+mp (up ) = - 2i+mp (u)(-i 2i+mp (u)(i i (u))p-2 ) п ), which proves (iii). Z (D The lemma is proved. 2 Consider the following two cases. ~ Case 1. i (2i+mp (u)) = 0 or i + mp < i. In this case we have shown that ~j (1) (up )) = 0 and i (i j (1)+i+mp (up )) = 0. ~ i (i Lemma 0.47 Let j (n+1) be the first map such that fol lowing conditions hold: ~ i') j (n) j (n+1)+i+ j (n +1)+ i + mp; ii') i (j ~
mp j (n+1) |F (u
pn+1

)

= 0. Suppose the

(u

p

n+1

)|F

(u

pn+1

)

= 0 and

j (n) r |F (u

~

pn+1

)

= 0 for j (n + 1) < r <

(n) j (n+1)+i+mp

(u

p

n+1

)) = 0 and i (j

(n) j (n+1)

~

(u

p

n+1

)) = 0.

Then there exists a parameter z such that the fol lowing conditions hold: i)
j (n+1) j (n+1) j (n+1)+i+mp+(p-1)j (n+1) (n+1)+i+mp+(p-1)j (n+1) |F (u

~

~ j

pn+2

is the first map such that ) = 0;

n+2 ~ ~ ii) j (n+1) j (n+1)+i+mp+(p-1)j (n+1)+i+mp (up ) = 0 and j (n+1) r |F (upn+2 ) = 0 for j (n +1)+ i + mp +(p - 1)j (n +1) < r < j (n +1)+ i + mp +(p - 1)j (n +1)+ i + mp;

45

n+2 ~ iii) i (j (n+1) j (n+2) (up )) = 0 and i (j 2) = j (n +1)+ i + mp +(p - 1)j (n +1).

(n+1) j (n+2)+i+mp

~

(u

p

n+2

)) = 0, where j (n + = for By By

п Pro of. First we prove that there exists a parameter z such that q |F (upn+1 ) п n+1 п q |F (upn+1 ) for q j (n +1)+ i + mp and j (n+1) ~q (up ) = 0 only if q = 2i mod p п п q > j (n +1)+ i + mp; here q are the maps given by the parameter z . n+1 ~ (up ) = 0, q > j (n + 1) + i + mp and q = 2i mod p. пq Suppose j (n+1) n+1 n+1 n+1 ~ definition, j (n+1) q (up ) = -j (n +1)q (up )+ k1 ...kl (up ), where ki < q . п lemma 0.11, (ii), for any a D there exists a parameter zq such that п zq u п
p
n+1

zq 1 = u п-

p

n+1

+ j

(n+1)

(u

p

n+1

)пq zj

(n+1)

+ ... + q-1 (u

p

n+1

)пq zq

-1

+ azq + ... пq

n+1 ~ п Therefore there exists an element a D such that j (n+1) q (up ) = 0. It is easy to see п п that the sequence {zq } converges in D. So, z = limzq . п ~j (n+1)+i+mp (upn+1 ) = 0 and j (n+1) r | pn+1 = 0 for j (n + ~ Now we prove that j (n+1) F (u ) n+1 ~ 1) < r < j (n +1)+ i + mp and i (j (n+1) j (n+1)+i+mp (up )) = 0. We have j (n +1) = j (n) mod p. Therefore,

z

-j (n+1) p

u

n+1

z

j (n+1)

=z

-pk

(z

-j (n) p

u

n+1

z

j (n)

)z

pk

=z

-pk

(u

p

n+1

+

j (n) j (n+1)

~

(u

p

n+1

)z

j (n+1)

+

j (n) j (n+1)+i+mp

~

(u

p

n+1

)z

j (n+1)+i+mp

+ ...)z
n+1

pk p

=
n+1

z

-pk p

u

n+1

z pk +z +

-pk

j (n) j (n+1)

~

(u

p

n+1

)z pk z

j (n+1)

+j

(n) j (n+1)+i+mp

~

(u

)z

j (n+1)+i+mp

+... =

up

n+1

j (n) j (n+1)

~

(u

p

n+1

)z

j (n+1)

+

j (n) j (n+1)+i+mp

~

(u

p

)z

j (n+1)+i+mp

+ ...,

because ii') provide i (r (up )) = 0 for j (n + 1) < r < j (n + 1) + i + mp. So, n+1 n+1 j (n+1)+2i+mp z -pk up z pk = up mod MD . n+1 i) Put w = j (n +1)+ i + mp +(p - 1)j (n +1), t = up . By proposition 0.41 we have
p- 2 q =0 w -1 j (n+1) k=j (n+1)+i+mp+1 p- 2 j (n+1)

n+1

~ w (tp ) =

j (n+1)

~ j

(n+1)+i+mp

(t)

i+mp (p-1)j (n+1)

(tp

-1-q

)tq +

~ k
q =0

k-j (n+1) w-k

(tp

-1-q

)tq

By lemma 0.44, k-j (n+1) w-k |F (t) = c1 + ... + cp-2 p-2 for w - k < (p - 1)j (n + 1) ~ and i+mp (p-1)j (n+1) |F (t) = c1 + ... + cp-1 p-1 with cp-1 = 0. Therefore, j (n+1) w (tp ) = ~ -j (n+1) j (n+1)+i+mp (t)cp-1 = 0. ~ ~ The same arguments show that j (n+1) w is the first map such that j (n+1) w (tp ) = 0. n+1 п ii) Put t = up . Using the same arguments as above, we can find a parameter z п п = q for q j (n + 1) + i + mp + (p - 1)j (n + 1) and j (n+1) ~q (tp ) = 0 such that q 46

п for q > j (n + 1) + i + mp + (p - 1)j (n + 1), q = 2i mod p. Since j (n+1) ~r |F (tp ) , j (n +1)+ i + mp +(p - 1)j (n +1) < r < j (n +1)+ i + mp +(p - 1)j (n +1)+ i + mp are п derivations (see lemma 0.44), j (n+1) ~r |F (tp ) = 0 for j (n +1)+ i + mp +(p - 1)j (n +1) < r < j (n +1)+ i + mp +(p - 1)j (n +1)+ i + mp. For j (n +1)+ i + mp +(p - 1)j (n +1) < w j (n +1)+ i + mp +(p - 1)j (n +1)+ i + mp, w = 2i mod p, by proposition 0.41 we have
p- 2 j (n+1)

~ w (tp ) =

j (n+1)

~ j

(n+1)+i+mp

(t)
q =0

i+mp (p-1)j (n+1)+i+mp

(tp

-1-q

)tq + ... +

p- 2 j (n+1)

~ w

-(p-1)j (n+1) w -1

(t)
q =0

w-(p-1)j (n+1)-j (n+1) (p-1)j (n+1) p- 2 j (n+1)

(tp

-1-q

)tq +

~ k (t)
q =0

k-j (n+1) w-k

(tp

-1-q

)tq

k=w-(p-1)j (n+1)+1

By lemma 0.44, k-j (n+1) w-k |F (t) = c1 + ... + cp-2 p-2 if w - k < (p - 1)j (n +1). -2 Therefore, p=0 k-j (n+1) w-k (tp-1-q )tq = 0. q -2 ~ Note that j (n+1) w-(p-1)j (n+1) (t) p=0 w-(p-1)j (n+1)-j (n+1) (p-1)j (n+1) (tp-1-q )tq = 0. q Indeed, w - (p - 1)j (n +1) - j (n +1) = 2i mod p. Therefore by lemma 0.44 (ii), p- 2 . w-(p-1)j (n+1)-j (n+1) (p-1)j (n+1) |F (t) = c1 + ... + cp-2 Let us prove that k-j (n+1) |F (t) = c1 + ... + cr r for (r + 1)j (n + 1) < < (r +1)j (n +1)+ i + mp, r p - 2. The proof is by induction on r. By ii') and i'), s (t) = 0 for j (n + 1) < s < j (n +1)+ i + mp. Therefore for r = 0, k-j (n+1) |F (t) = 0. For arbitrary r we have
q -2 k-j (n+1)

(tq ) = q

k-j (n+1)

(t)+

k-j (n+1)

j

(n+1)

(t)
r =0

k -j (n+1)

(tq

-1-r

)tr +

-1 k-j (n+1) s s=j (n+1)+i+mp

q -2

(t)
r =0

s+(k-j (n+1)) -s

(tq

-1-r

)tr

By lemma 0.44, s+(k-j (n+1)) -s |F (t) = c1 + ... + cr-1 r-1 for - s < rj (n +1). For any m m -j (n+1) |F (t) = 0 if r = 1, and m -j (n+1) |F (t) = c1 + ... + cr-1 r induction and lemma 0.44, because
q -2 m -j (n+1)

-1

by

(t ) = q m

q

-j (n+1)

(t)t

q -1

+ m j

(n+1)

(t)
l=0

m1 -2j (n+1)

(tq

-1-l

)tl +

47

-j (n+1)-1 m s=j (n+1)+i+mp

q -2

s
l=0

ms -j (n+1)-s

(tq

-1-l

)tl ,

and by lemma 0.44, ms -j (n+1)-s |F (t) = c1 + ... + cr-2 r-2 for s j (n +1)+ i + mp. ~ The same arguments show that k-j (n+1) |F (t) = c1 + ... + cr r for (r +1)j (n +1) < < (r +1)j (n +1)+ i + mp. Let us show that i+mp (p-1)j (n+1)+i+mp |F (t) = c1 + ... + cp-1 p-1 , cp-1 = 0. Put = (p - 1)j (n +1)+ i + mp. We have
w -k -1 i+mp

+

i+mp

~ +
s=1

i+mp -s

§

i+mp s

~ =0

~ First we prove that i+mp rj (n+1)+i+mp |F (t) = c1 + ... + cr r with ~ cr = 1 i+mp j (n+1)+i+mp (t)(j (n+1) j (n+1) (t))r-1 = 0.Weuse thesamearguments as above. r ~ The proof is by induction on r. For r = 0, since i + mp < j (n +1), i+mp i+mp |F (t) = 0. Put w = rj (n +1)+ i + mp. For arbitrary r we have
q -2 i+mp

~ w (tq ) = q

i+mp

~ w (t)tq
w -1

-1

+

i+mp

~ j

(n+1)

(t)
r =0

j (n+1)-i-mp w-j (n+1)

(tq

-1-r

)tr +

q -2 i+mp

~ s (t)
r =0

s-i-mp w-s

(tq

-1-r

)tr

s=j (n+1)+i+mp

By lemma 0.44, s-i-mp w-s |F (t) = c1 + ... + cr-2 r-2 for w - s < (r - 1)j (n +1) and r -1 with cr-1 = (j (n+1) j (n+1) (t))r-1 = 0. j (n+1) (r -1)j (n+1) |F (t) = c1 + ... + cr -1 By proposition 0.41 we have
j (n+1)-i-mp w-j (n+1)

(tq ) = q
q -2

j (n+1)-i-mp w-j (n+1)

(t)tq-1 +

j (n+1)-i-mp j (n+1) w-j (n+1)-1 j (n+1)-i-mp s s=j (n+1)+i+mp

(t)
r =0 q -2

m1 w-2j (n+1)

(tq

-1-r

)tr +

(t)
r =0

s+j (n+1)-i-mp w-j (n+1)-s

(tq

-1-r

)tr

By lemma 0.44, s+j (n+1)-i-mp w-j (n+1)-s |F (t) = c1 + ... + cr-3 r-3 for w - j (n +1) - s < (r - 2)j (n + 1). Since j (n +1) - i - mp = 0 mod p, j (n+1)-i-mp j (n+1) (t) = 0 and j (n+1)-i-mp j (n+1)+i+mp (t) = 0 (here we use also i') and ii')). Therefore, r -2 and lemma 0.44 completes the proof. j (n+1)-i-mp w-j (n+1) |F (t) = c1 + ... + cr -2

48

~ The same arguments show that i+mp rj (n+1) |F (t) = c1 + ... + cr-1 r-1 . Therefore, p- 1 ~ only if s = j (n + 1) or s > j (n + 1) i+mp -s § i+mp s |F (t) = c1 + ... + cp-1 ~ and s = rj (n + 1) + i + mp. Note that cp-1 = 1 i j (n+1)+i+mp (t)(i j (n+1) (t))p-2 if s = r rj (n +1)+ i + mp. Using the same arguments for the maps i+mp -rj (n+1) , we get i+mp |F (t) = c1 + ... + cp-1 p-1 , where c (1 + ... +
p- 1

= -(1 +

1 1 ~ + ... + )i j 2 p-1 (t)(i j
(n+1)

(n+1)+i+mp

(t)(i j

(n+1)

(t))p-2 - (t)(i j
(n+1)

1 ~ )i j p-2

(n+1)+i+mp

(t))p

-2

~ - ... - i j (t))p
-2

(n+1)+i+mp

(t))p

-2

=

~ -i j

(n+1)+i+mp

(t)(i j

(n+1)

= 0.

This completes the proof of ii) and iii). The lemma is proved. 2 ~ Case 2. i (2i+
mp

(u)) = 0.

Lemma 0.48 Suppose i + mp i. Put t = up . Then there exists a parameter z such that the fol lowing properties hold: i)
j (1) j (1)+i+mp+(p-1)j (1)

~

is the first map such that

j (1) j (1)+i+mp+(p-1)j (1) |F (tp ) j (1) j (2)+i

~

= 0.

~ ii) j (1) r |F (tp ) = 0 for j (2) < r < j (2) + i and j (2) = j (1) + i + mp +(p - 1)j (1). iii)
j (1) j (2)+i

~

(tp ) = 0, where

~

п (tp ) Z (D),

j (1) j (2)

~

п (tp ) Z (D).

Pro of. To prove i) one can repeat the proof of i) in lemma 0.47. Note that ~j (2) (tp ) = -j (1) j (1)+i (i j (1) (t))p-2 Z (D). ~ п j (1) п ii) As in the case 1 we can find a parameter z such that q |F (t) = q |F (t) for q п п j (1) + i + mp, ~q (tp ) = 0 for q = 2i mod p, q > j (2). For r = 2i mod p, by proposition 0.41 we have
p- 2 j (1)

~ r (tp ) =

j (1)

~ j

(1)+i+mp

(t)
q =0

i+mp r -j (1)-i-mp

(tp

-1-q

)tq +

r -1 j (1) k=j (1)+i+mp+1

p- 2

~ k (t)
q =0

k-j (1) r -k

(tp

-1-q

)tq

49

c1 for

For i+

By lemma 0.44, k-j (1) r-k |F (t) = c1 + ... + cp-2 p-2 if r - k < (p - 1)j (1). -2 ~ Note that j (1) r-(p-1)j (1) (t) p=0 r-pj (1) (p-1)j (1) (tp-1-q )tq = 0. q Indeed, r - pj (1) = 2i mod p. Therefore by lemma 0.44 (ii), r-pj (1) (p-1)j (1) |F (t) = + ... + cp-2 p-2 . The same arguments as in lemma 0.47 (ii) show that m s |F (t) = c1 + ... + cp-2 p-2 (p - 1)j (1) < s < (p - 1)j (1) + i. Let us prove that i+mp (p-1)j (1)+i |F (t) = c1 + ... + cp-1 p-1 with cp-1 = 0. ~ First let us show that i+mp |F (t) = c1 + ... + cr r for = (r +1)j (1) + i, r p - 2. ~ ~ r = 0, i+mp |F (t) is a derivation. Since j (1) (t) = 0 for = j (1) + i and j (1) = ~ ~ mp mod p and p > 2, we have i+mp (t) = 0 and i+mp |F (t) = 0. For arbitrary r we have
q -2 i+mp

~ (tq ) = q

i+mp

~ (t)tq

-1

+

i+mp

~ j (1) (t)
r =0

j (1)-i-mp -j (1)

(tq

-1-r

)tr +

-1 i+mp k=j (1)+i+1 k-i-mp -k |F (t)

q -2

~ k (t)
r =0

k-i-mp -k

(tq

-1-r

)tr

By lemma 0.44, j (1) + i +1. By definition,

= c1 + ... + c

r -1

r -1

for - k < rj (1), i.e. k

j (1)-i-mp -j (1)

=
(j1 ,...,jl ),lp

l Cj

(1)-i-mp j

1

...jl ,

because j (1) - i - mp = 0 mod p. Since l p, there exist jk ,jk1 such that jk < j (1) ^ and jk1 < j (1). Thus, j1 + ... + jk + ... + j^1 + ... + jl < rj (1) and j1 ...jl |F (t) = k r -1 ~ c1 + ... + cr-1 . Hence by lemma 0.44, i+mp |F (t) = c1 + ... + cr r . Now we have
(p-1)j (1) i+mp (p-1)j (1)+i

+

i+mp

~ (

p-1)j (1)+i

+
k =i

i+mp k

§

i+mp (p-1)j (1)+i-k

~

=0

with i+mp k § i+ Therefore,

mp (p-1)j (1)+i-k |F (t)

~

= c1 + ... + c
p- 2

p- 2

p- 2

for k = (p - 2)j (1) + i. §
i+mp j (1) |F (t)

i+mp (p-1)j (1)+i |F (t)

= c1 + ... + c

p- 2

-

i+mp (p-2)j (1)+i

~

So by induction,
i+mp (p-1)j (1)+i |F (t)

= c1 +...+~p-2 ~ c

p- 2

+

i+mp i (i+mp j (1)

)p-1 |F

( t)

= c1 +...+c

p- 2

p- 2

+

50

i+mp i

((i j (1) (t))p-1 )

p- 1 i+mp i (i j (1) -2

~ Since i j (1) (t) = -j (1) j (1) (t) = which completes the proof of ii). ~ Finally, j (1) j (2)+i (tp ) = -j (1) because i+mp i is a derivation and i 2

i 2i+mp

~

(u)(i i (u))p-1 , we have

(t)) = 0,

~ j (1)+i+mp (t)i+mp i (i j (1) (t))(i j (1) (t))p п j (1) (t) Z (D) . This proves iii).

п Z (D),

The following lemma completes the proof of Case 2 and of Theorem. Lemma 0.49 Suppose the fol lowing conditions hold: i') ii')
j (1) j (n)+i |F (u j (1) j (n)

~

pn

)

= 0 and ~

j (1) r |F (u
n

~

pn

)

= 0 for j (n) < r < j (n)+ i, n 1;

~

(up ),

n

j (1) j (n)+i

п (up ) Z (D).

Then there exists a parameter z such that i)
j (1) j (n)+i+(p-1)j (n)

~

is the first map such that
(u
pn+1

j (1) j (n)+i+(p-1)j (n) |F (u

~

pn+1

)

= 0;

n+1 ~ ~ ii) j (1) j (n+1)+i (up ) = 0 and j (1) r |F where j (n +1) = j (n)+ i +(p - 1)j (n);

)

= 0 for j (n + 1) < r < j (n + 1) + i,

iii)

j (1) j (n+1)

~

(u

p

n+1

),

j (1) j (n+1)+i

~

(u

p

n+1

п ) Z (D).

n ~ Pro of. By induction, j (n) = i mod p. Put a = j (1) j (n) (up ). We have akj (n)+1 = l ap Z (D) for some k Z. Put z = (a-kj (1) z j (1) )1/j (1) . We claim that i'), ii') hold for ~ ~ ~ j (1) j (n)+i , i.e. j (1) j (n)+i |F (upn ) = 0, and j (1) r |F (upn ) = 0 for j (n) < r < j (n)+ i, and pn pn pn ~ ~ ~ п j (1) j (n)+i (u ), j (1) j (n) (u ) Z (D ). Moreover, j (1) j (n) (u ) Z (D ). п Note that = id, because a Z (D). We have

z

-j (1) p

uz

n

j (1)

=z

-j (1) p

uz

n

j (1)

=u

p

n

+ az

j (n)

+

j (1) j (n)+i

~

(up )z

n

j (n)+i

+ ...

Let us show that z We have z
j (n)

=a
j (n)

-kj (n) j (n)

z

mod a z

j (n)+i+1

=z

j (n)-j (1) -kj (1) j (1)

.

It is easy to see that z = a-k z + xz z j (n)-j (1) = (a-k z )j (n)-j (1) mod (a-k z )j
(n)-j (1) -kj (1)

i+1 j (n)

п + ..., x D. Since j (n) - j (1) = 0 mod p, -j (1)+i+1 . Now we have z + xz
j (n)-j (1)+i

a

=a

-kj (n) j (n)-j (1)

+ ...,

51

where x = [-kj (1) - k (j (1) + 1) - ... - k (j (n) - 1)]a- z
-j (1) p

kj (n)-1

i (a) = 0. Therefore, z
j (n)+i

uz

n

j (1)

=u

p

n

+ ap z

l

j (n)

+

j (1) j (n)+i

~

(up )a

n

k(j (n)+i)

+ ...

n n n l ~ ~ ~ п and j (1) j (n)+i (up ) = j (1) j (n)+i (up )ak(j (n)+i) Z (D), j (1) j (n) (up ) = ap Z (D). So i'), ii') hold. Now to prove i) one can repeat the proof of i) in lemma 0.47. Note that ~j (n+1) (tp ) = -j (1) j (n)+i (i j (n) (t))p-2 Z (D). ~ п j (1) n ii) We use the same arguments as in ii) of lemma 0.48. Put t = up . For r = 2i mod p, by proposition 0.41 we have

p- 2 j (1) r

p- 2 j (n)-j (1) r -j (n)

~ (t ) =
p

j (1) j (n)

~

(t)
q =0

(t

p-1-q

~ )t +j (1) j
q

(n)+i

(t)
q =0

j (n)-j (1)+i r -j (n)-i

(tp

-1-q

)tq +

r -1 j (1) k=j (n)+i+1

p- 2

~ k (t)
q =0

k-j (1) r -k

(tp

-1-q

)tq

By lemma 0.44, k-j (1) r-k |F (t) = c1 + ... + cp-2 p-2 if r - k < (p - 1)j (n). Note that j (n)-j (1) r-j (n) |F (t) = c1 + ... + cp-2 p-2 . Indeed, by proposition 0.41 we have
q -2 j (n)-j (1) r -j (n)

(t ) = q

q

j (n)-j (1) r -j (n)

(t)t

q -1

+j

(n)-j (1) j (n)

(t)
s=0

2j (n)-j (1) r -2j (n)

(tq

-1-s

)ts +

q -2 j (n)-j (1) j (n)+i

(t)
s=0

2j (n)-j (1)+i r -2j (n)-i q -2

(tq

-1-s

)ts + ... +

j (n)-j (1) j (n)+2i

(t)
s=0

2j (n)-j (1)+2i r -2j (n)-2i

(tq

-1-s

)ts + ...

Recall that r pj (n) + 2i. By lemma 0.44, m s |F (t) = c1 + ... + cp-3 p-3 if ~ s < (p - 2)j (n). Since j (1) j (n) (t) Z (D), we have j (n) (t) Z (D). Since j (n) - j (1) = 0 mod p and j (n) (t) Z (D) and charF > 2, we have j (n)-j (1) j (n)+ei (t) = 0 for e p - 1, which completes the proof. -2 ~ Note that j (1) r-(p-1)j (n) (t) p=0 r-(p-1)j (n)-j (1) (p-1)j (n) (tp-1-q )tq = 0. q Indeed, r - (p - 1)j (n) - j (1) = 2i mod p. Therefore by lemma 0.44 (ii), p- 2 . r -(p-1)j (n)-j (1) (p-1)j (n) |F (t) = c1 + ... + cp-2 The same arguments as in lemma 0.47 (ii) show that m s |F (t) = c1 + ... + cp-2 p-2 for (p - 1)j (n) < s < (p - 1)j (n)+ i. Let us prove that j (n)-j (1)+i (p-1)j (n)+i |F (t) = c1 + ... + cp-1 p-1 with cp-1 = 0. ~ Note that j (n)-j (1)+i rj (n)+i |F (t) = c1 + ... + cr r with cr = 0 for any 1 r p - 1. 52

Indeed, by proposition 0.41 we have
q -2 j (n)-j (1)+i

~ rj

(n)+i

(t ) = q

q

j (n)-j (1)+i

~ rj

(n)+i

(t)t

q -1

+

j (n)-j (1)+i

~ j

(n)

(t)
s=0

q -2 j (1)-i

(

r -1)j (n)+i

(tq

-1-s

)ts +

j (n)-j (1)+i

~ j

(n)+i

(t)
s=0

j (1) (r -1)j (n)

(tq

-1-s

)ts + ...

By lemma 0.44, j (1) (r-1)j (n) |F (t) = c1 + ... + cr-1 r-1 with cr-1 = 0 and m s |F (t) = c1 + ... + cr-2 r-2 for s < (r - 1)j (n). Let us prove that j (1)-i (r-1)j (n)+i |F (t) = c1 + ... + cr-2 r-2 . By proposition 0.41 we have
q -2 j (1)-i (r -1)j (n)+i

(t ) = q

q

j (1)-i (r -1)j (n)+i

(t)t

q -1

+j

(1)-i j (n)

(t)
s=0

j (n)+j (1)-i (r -2)j (n)+i

(tq

-1-s

)ts +

q -2 j (1)-i j (n)+i

(t)
s=0

j (n)+j (1) (r -2)j (n)

(tq

-1-s

)ts + ...

~ Since j (1) j (n) (t) Z (D), we have j (n) (t) Z (D). Since j (1) - i = 0 mod p, r -3 j (1)-i j (n) (t) = 0 and j (1)-i j (n)+i (t) = 0. By lemma 0.44, m s |F (t) = c1 + ... + cr -3 for s < (r - 2)j (n). So, j (1)-i (r-1)j (n)+i |F (t) = c1 + ... + cr-2 r-2 . ~ Hence by lemma 0.44, j (n)-j (1)+i rj (n)+i |F (t) = c1 + ... + cr r with ~ ~ cr = 1 j (n)-j (1)+i j (n)+i (t)(i j (n) (t))r-1 = 1 i j (n)+i (t)(i j (n) (t))r-1 = 0. r r ~ The same arguments show that j (n)-j (1)+i |F (t) = c1 + ... + cr-1 r-1 for < rj (n)+ i. Therefore we have
j (n)-j (1)+i (p-1)j (n)+i (p-1)j (n)+i-1 j (n)-j (1)+i w w=1

+ §

j (n)-j (1)+i (p-1)j (n)+i

~

+ = 0,

j (n)-j (1)+i (p-1)j (n)+i-w

~

~ where j (n)-j (1)+i w § j (n)-j (1)+i (p-1)j (n)+i-w |F (t) = c1 + ... + cp-1 p-1 only if w = rj (n). ~ In this case cp-1 = p-1-r i j (n)+i (t)(i j (n) (t))p-2 . 1 ~ ~ Since j (n)-j (1)+i j (n) (t) Z (D), we have j (n)-j (1)+i i (j (n)-j (1)+i j (n) )p-1 = 0. So using p- 1 induction, we get j (n)-j (1)+i (p-1)j (n)+i |F (t) = c1 + ... + cp-1 with c
p- 1

= -(1 + ... +

1 ~ )i j p-1 53

(n)+i

(t)(i j

(n)

(t))p-2 -

(1 + ... +

1 ~ )i j p-2

(n)+i

(t)(i j

(n)

(t))p

-2

~ - ... - i j
-2

(n)+i

(t)(i j

(n)

(t))p

-2

=

~ -i j

(n)+i

(t)(i j
(n)+i

(n)

(t))p
p- 1

= 0.

~ ~ Therefore, j (1) j (n+1)+i (tp ) = -j (1) j The lemma is proved. 2 The theorem is proved. 2

(t)c

п Z (D). This proves ii) and iii).

пп Lemma 0.50 Let D be a splittable division algebra. Let n = |Gal(Z (D)/F )|. There exists a parameter z such that zaz So, j = 0 if n |j . One can repeat the proof of lemma 0.29 to prove the lemma. пп Prop osition 0.51 Let D be a good splittable division algebra. Suppose Z (D)/F is not a separable extension. пп Then p does not divide |Gal(Z (D)/F )|. пп Pro of. Suppose p divides |Gal(Z (D)/F )|. By lemma 0.50 there exists a parameter z such that п zaz -1 = (a)+ n (a)z n + 2n (a)z 2n + ..., a D, пп where n = |Gal(Z (D)/F )|. пп Since Z (D)/F is a compositum of a purely inseparable extension and Abelian Galois п extension, there exists an element u Z (D) such that (u) = u, i.e. u is a purely inseparable element; so by theorem 0.43 up Z (D). In this case lemma 0.44 holds for l = 0 and we can repeat the arguments in the proof of lemma 0.45 to show that pi (up ) = 0, which is a contradiction. 2 Prop osition 0.52 Let D be a good splittable division algebra. Then we have D = D1 F D2 , where D1 ,D2 are division algebras such that D1 is an inertial ly split algebra, пп Z (D2 )/F is a purely inseparable extension and D2 is a good splittable algebra (D1 or D2 may be trivial). So, D A F B F D2 , where A is a cyclic division algebra and B is an unramified division algebra.
-1

= (a)+ n (a)z n + 2n (a)z

2n

+ ...,

п aD

54

Pro of. By [25], p.261, D D1 F ... F Dk , where [D : F ] = pr1 ...prk and = 1 k пп [Di : F ] = pri . Let p2 = p. By proposition 0.51, Z (D2 )/F is a purely inseparable i extension. Since Di are defectless over F , D1 ,D3 ,...Dk are inertially split. Therefore, by theorem 0.37 the algebra D1 D3 ... Dk is good splittable. п Let L be an inertial lift of a Galois part of the extension Z (D)/Z (D). Consider the centraliser D = CD (L). It's clear that D D2 L B , where B is a division algebra = similar to the algebra D1 D3 ... Dk L. The algebra B is inertial, because п=п п п п Z (B )/Z (B ) is trivial and B is inertially split. Since D D2 B and D D is a п п п good embedding, D contain a subalgebra B B L L B D . Now the centraliser = D2 F L and it is good splittable, so D2 is good splittable. CD (B ) = Decomposition theorems [9], Thm. 5.6-5.15 complete the proof. 2 Prop osition 0.53 Let D2 be a good splittable division algebra such that is a purely inseparable extension. Then D2 D3 Z (D2 ) D4 , where D3 is = division algebra and D4 is a good splittable division algebra such that п п D4 /Z (D2 ) is a purely inseparable extension, [D4 : Z (D2 )] = [D4 : Z (D2 п Z (D2 )/Z (D2 ) an unramified п D4 is a field, ) ].

п Pro of. For a good embedding there exists a subfield Z (D2 ) K Z (D2 ) such п that the extension Z (D2 )/K has degree p. Then by theorem 0.37 and 0.43 there exists п ~ ~ ~ a lift K of K in D2 , i.e. K = K , K = Z (D2 ) , K K . ~ ~ п п п Consider the centraliser C1 = CD2 (K ). We have C1 = D2 , Z (C1 )/Z (C1 ) is a purely п1 ) = Z (C1 )(u). Using similar arguments as inseparable extension of degree p, say Z (C in the proof of theorem 0.43 one can show that there exists a parameter z such that C1 C1 ((z )) as a vector space with the relation =п zaz
-1 2 = a + i (a)z i + c2i i (a)z 2i + ...,

cki F

p

p and zuz -1 = u + xz i , where x Z (C1 ). Therefore, i is a derivation trivial on the п centre Z (C1 ), hence by Scolem-Noether theorem it is an inner derivation. п We claim that z p Z (C1 ). To prove it, consider a subalgebra W = C1 ((z i )) C1 (note that Z (W ) = Z (C1 )). It exists because of the type of the relation in C1 . We have п z -i az i = a - ii (a)z i , a C1

in W . Therefore, z and

-pi

p az pi = a - ip i (a)z pi , 2

п a C1
2pi

z pi az
p where 1 = ip i . So,

-pi

= a + 1 (a)z pi + 1 (a)z

+ ...,

z p az

-p

1 12 = a + 1 (a)z pi + c2 2 1 (a)z i i 55

2pi

+ ...,

p p where ck are given by (1) in theorem 0.43. So, z p Z (C1 ) iff i = 0. Suppose i = 0. Consider an element Y Z (C1 ), w(Y ) > 0. Let

Y = a1 z p + ... First note that Y = a1 z p + a 2 z п + ..., ai C1 п Indeed, Y must satisfy [Y, u] = 0. Since u Z (C1 ), we then have [z ik ,u] = 0 for every k , where + a3 z
2p 3p

Y=
k=1

ak z

ik

Therefore, p|ik . п п Then, Y must satisfy Ya = aY for any a C1 . Therefore, a1 ,...ai Z (C1 ) and we must have aai+1 - ai+1 a = a1 1 (a)/i and aa
2i+1

-a

2i+1

a = ai 1 (a)+ a1 c2 1 (a).

2

Since (a) = aa2i+1 - a2i+1 a is an inner derivation, we get 1 2 = , where is a derivation, which is a contradiction. Therefore, 1 2 = = 0 and 1 = 0, and z p Z (C1 ). ~ п п п Consider the centraliser C2 = CC1 (K (z )). It's clear that [C2 : Z (C2 )] = [C2 : п п п п Z (C2 )] = indC1 and there exists a subalgebra C2 C2 , C2 D2 . Consider now п2 C3 , C3 C2 , because [Z (C3 ) : п=п п the centraliser C3 = CD2 (Z (D2 )(z )). We have C п Z (D2 )] = [C3 : Z (D2 ) ] = [Z (C2 ) : Z (D2 )] = [K : Z (D2 )]. By induction on dimension п п п п п п2 we get the existence of a subalgebra C4 D2 such that [C4 : Z (C4 )] = [C2 : of D п п Z (C2 )], Z (C4 ) = Z (D2 ). Therefore there exists an unramified subalgebra C4 D2 such п п that [C4 : Z (D2 )] = [C4 : Z (D2 )] = [C4 : Z (C4 )] = [D2 : Z (D2 )]. By Double Centraliser C4 Z (D ) D4 , where D4 is a division algebra with D4 = Z (D2 ). Since п п Theorem, D2 = 2 п п D2 D2 is a good embedding, [D4 : Z (D2 )] must be equal to [D4 : Z (D2 ) ]. It is easy to see that D4 is also a good splittable division algebra. The proposition is proved. 2 п Prop osition 0.54 Let D2 be a good splittable division algebra such that D2 is a field, п D2 /Z (D2 ) is a purely inseparable extension and dD2 (uk ) 2i(uk ) or dD2 (uk ) = for п al l generators uk of the extension D2 /Z (D2 ). Then D2 A1 Z (D2 ) ... Z (D2 ) Am , where Ai are cyclic division algebras of degree = пi : Z (D2 )] = [A : Z (D ) ]. p, [A 2 i

56

Pro of. The proof immediately follows from theorem 0.43 and [26], Thm.3, з2.8.( see [1] for the proof of this theorem). 2 So, we get the following decomposition theorem. Theorem 0.55 Let D be a finite dimensional good splittable central division algebra over a field F with a discrete complete rank 1 valuation, char(F ) = p > 2, such that п dD2 (uk ) 2i(uk ) or dD2 (uk ) = for al l generators uk of the extension Z (D)/Z (D). D1 F D2 F A1 F ... F Am , where Ai are cyclic division algebras Then D = п of degree p, [Ai : Z (D)] = [Ai : Z (D) ], D1 is an inertial ly split division algebra, (ind(D2 ),p) = 1, D2 is an unramified division algebra (D1 ,D2 ,Ai may be trivial). Recall that a field F is called a Ci -field if any homogeneous form f (x1 ,...,xn ) of degree d in n > di variables with coefficients in F has a non-trivial zero. Corollary 8 The fol lowing conjecture: the exponent of A is equal to its index for any division algebra A (here we don't demand that A is splittable) over a C2 -field F (see for example [26], 3.4.5.) has the positive answer for F = F1 ((t)), where F1 is a C1 -field. п п Pro of. 1) Let's prove that A is splittable. Since F is a C1 -field, A is a field. We пп п п can assume A/F is a purely inseparable extension. We claim that A = F (u) for some п, so by classical Cohen's theorem, A is splittable. uA п1 п п Indeed, suppose A = F (u1 ,...,ur ). Consider the field K = F (up ,...,up ). By Tsen's r p p p p- 1 п theorem, K and A are C1 -fields. So, the form x1 + x2 u1 + ... + xp u1 + xp+1 u2 has a nonp п trivial zero in A. But xp K and elements 1,u1 ,...,up-1 ,u2 are linearly independent 1 i over K , a contradiction. 2) Assume the corollary is known in the prime exponent case. We deduce the corollary by ascending induction on e = expA.If e is not a prime number, then write e = lm. By assumption Am can be split by a field extension F F of degree l. This implies that AF has exponent dividing m. Note that F is also a Laurent series field. By the induction hypothesis applied to the pair (F ,AF ), there exists a field extension F L of degree dividing m splitting AF . Therefore A is split by the extension F L of degree dividing lm and we conclude the corollary. 3) So, let expA = l be a prime number. By the basic properties of the exponent and the index (see, e.g. [26]) we have then indA = lk for some natural k . Suppose (l, p = charF ) = 1. It is known that the conjecture is true for all division algebras of index indA = 2a 3b (see, e.g. [26]), so we can assume charF = 2, 3. Then we can assume F contains the group Іl of l-roots of unity, because [F (Іl ) : F ] < l and we can reduce the problem to the algebra A F F (Іl ). Then by the Merkuriev-Suslin theorem A is similar to the tensor product of symbol-algebras of index l. 57

Every symbol-algebra of index l over F is good splittable and cyclic and its residue п field is a cyclic Kummer extension of F . To conclude the statement of the corollary it is sufficient to prove that every two symbol algebras A1 ,A2 contain F -isomorphic maximal subfields. Since Ai , i = 1, 2 is cyclic, it contains an element zi , zil F . Since Ai is a good splittable algebra and by lemma 0.50 (which is true also if charF = 0), we can assume v (zil ) = 1 (v is the valuation on F ). To prove it we show that A1 contains any l-root of elements u in F with v (u) = 0. Since for any element 1 + b, v (b) > 0 there exists an element (1 + b)1/l F , it is п sufficient to prove that A1 contains any l-root of elements ct, c F , where we fix some п embedding i : F F . Indeed, since A1 is a good splittable algebra and by lemma 0.50 (which is true also if charF = 0) we can assume there exists an element z such that v (z l ) = 1, z l = ct, п п п c F , ad(z ) acts on A1 , where A1 is embedded in A1 by a good embedding with п respect to i. Note that for any element b A1 we have (bz )l = NA1 /F (b)z l . But the пп п is a C1 -field (see, e.g. [26], 3.4.2), so for any norm map NA1 /F is surjective because F пп c there exists b such that (bz )l = ct. 4) Suppose now expA = p. Then indA = pk . 1/p 1/p By Albert's theorem (in [1]) there exists a field F = F (u1 ,...,uk ) which splits A. Using the same arguments as in 1) one can show that every such a field has maximum 1/p 1/p two generators, say F = F (u1 ,u2 ). Therefore, indA p2 . If indA = p, there is nothing to prove, so we assume indA = p2 and F is a maximal subfield in A. 5) Suppose F1 is a perfect field. By Albert's theorem, A A1 F A2 , where A1 ,A2 are cyclic algebras of degree = пппп p, A1 = (L1 /F, 1 ,u1 ), A2 = (L2 /F, 2 ,u2 ). Since F1 is perfect, A1 /F , A2 /F are Galois extensions. So, A1 ,A2 are good splittable. Let us show that A1 ,A2 have common splitting field of degree p over F . This leads to a contradiction. п By lemma 0.50 there exist parameters z1 A1 , z2 A2 such that they act on A1 , pp п A2 as Galois automorphisms. Note that then z1 ,z2 F . Let us show that F (z1 ) splits A2 . - Consider the centralizer D = CA (F (z1 )). Consider the element t1 = z2 z1 1 . We have p t1 F , w(t1 ) = 0, where w denote the unique extension of the valuation v on F . Since п D/Z (п ) is a Galois extension, there exists an element b1 F such that w(t1 - b1 ) > 0. D - Since (t1 - b1 )p F , there exists natural k1 such that w((t1 - b1 )z1 k1 ) = 0. Denote - t2 = (t1 - b1 )z1 k1 . We have again tp F . Repeating this arguments and using the 2 completeness of D A we get k k z2 = t1 z1 = (t2 z1 1 + b1 )z1 = ... = b1 z1 + b2 z1 1 +1 + ..., so, z2 F (z1 ) = Z (D). 6) Suppose F1 is not perfect.

58

Since F is generated by two elements over F , it contains all p-roots of F . Then, / every two elements u, z F such that z 1/p F (u1/p ), where z 1/p ,u1/p F , also generate F over F . This follows from the same arguments as in 1), 4). p Now take u F1 \F1 , z = u + t. It's clear that p-roots of these elements generate F over F . Moreover, the fields F (u1/p ),F (z 1/p ) are "unramified" over F , i.e. [F (u1/p ) : п п F ] = p = [F (u1/p ) : F ], [F (z 1/p ) : F ] = p. Denote u1 = u1/p , u2 = z 1/p in F . Then A1 F A2 , where A1 ,A2 are cyclic algebras of degree p, by Albert's theorem, A = A1 = (L1 /F, 1 ,u), A2 = (L2 /F, 2 ,z ). Since the fields F (u1/p ) A1 , F (z 1/p ) A2 are "unramified" and purely inseparable of degree p over F , the algebras A1 ,A2 are good splittable. Moreover, there exist п п п п embeddings A1 A1 , A2 A2 such that u1 A1 , u2 A2 . Then by theorem pp 0.43 there exist parameters z1 A1 , z2 A2 such that z1 ,z2 F and
- i z2 u2 z2 1 = u2 + cz2 ,

where c F , v (c) = 0. So, for the element u2 = c-1 u2 we have
- i z2 u2 z2 1 = u2 + z2 ,

/ and u2 F , u p F . 2 п п п Since F is a C1 -field, we have A1 = A2 and therefore there exist an element b F (u1 ) A1 A such that w(u2 - b) > 0, where w is the unique extension of v on A. Since b commutes with u2 , we have (u2 - b)p F . Therefore w(u2 - b) 1/pZ (we assume the value groups of w and v lie in a common divisible hull v Z "). Hence -pw(u2 -b) -pw(u2 -b) ) = 0. Put u3 = (u2 - b)z2 . w((u2 - b)z2 Note that - i z2 (u2 - b)z2 1 = (u2 - b)+ z2 ,
i - z2 u3 z2 1 = u3 + z21 ,

i1 < i and is a A1 . and

So, the elements (u2 - b),z2 generate a division algebra C of degree p over F u3 C . Then, up commutes with z2 if i1 > 0. Therefore, in this case up F and C 3 3 good splittable division algebra. Note that u1 CA (C ), so A A1 F C with u1 = Using the same arguments we get that there exists an element u4 with w(u4 ) = 0
i - z2 u4 z2 1 = u4 + z22 ,

i2 0

So, i2 must be equal to 0 and therefore u4 ,z2 generate a division algebra C of degree пп p over F such that C /F is a Galois extension and u1 CA (C ). So, A D F C with = u1 D. Therefore, A contains the maximal subfield F (u1 )F (u4 ), which is a compositum of a purely inseparable and Galois extension. Moreover, this field is "unramified" over F , so it is good splittable field and A is a good splittable algebra with p dividing 59

пп |Gal(A/F )|. But this is a contradiction with proposition 0.51. 2 Corollary 9 Let A be a central division p-algebra over a C2 -field F = F1 ((t)), F1 is a C1 -field. Then A contains a maximal purely inseparable over F subfield, i.e. A is a cyclic algebra. Moreover, A is a good splittable algebra. Pro of. The proof of the first statement is by induction on degree of A. If indA = p, then by Tignol's theorem in [32] A is cyclic, so it contains such a maximal subfield. If indA = pk , k > 1, then by assumption a division algebra similar to Ap has the exponent and index pk-1 and so can be split by a field extension F F of degree pk-1 . By corollary 8, the exponent and the index of AF is p, so there exists an extension L/F of degree pk such that L splits A. To prove the second statement note that it is sufficient to prove it only for algebras пп A with A/F -- purely inseparable. Now to prove the assertion we use lemma 0.24. Note that, using a similar induction, it is sufficient to prove the statement for algebras A of degree p. Let z be a purely inseparable element in A, indA = p. If F (z ) is an "unramified" over F , there is nothing to prove. So, we may assume F (z ) is totally ramified over F and z is a parameter of A. п п п Choose an element a A such that a generates A over F . Suppose a A for some п п embedding A A. Suppose zaz Then we have z az
p -p -1

= a + i (a)z i + i+1 (a)z

i+1

+ ...

=a+
k=pi (i1 ,...ip )

i1 ...ip (a)z k = a,

(2)

where ij = k and the second sum is taken over all such nonrepeating sets (i1 ,...,ip ). p п Therefore, i (a) must be equal to zero. Since i is a derivation, it is trivial on F (ap ). п Every element in F (a) can be written as a polynomial c1 + c2 a + ...cp ap-1 , where p p p п ci F (a ). Therefore, we can write i = i (a)/ (a). So, i (a) = i (a)/ (a)(i -1 (a)). p p п Hence / (a)(i -1 (a)) = 0 and i -1 (a) F (ap ). p- 1 j j If i (a) = 0, then let j be the maximal natural such that i (a) = 0, i (a) j j j п п п F (ap ). Now put a1 = i -1 (a)(i (a))-1 . Note that a1 generates A over F Since i (a) = j -1 / (a)(i (a))i (a), we have i (a1 ) = 1. So we can put a := a1 and assume i = / (a). Now the proof is by induction on k in the formula 2. For k = ip +1 we have i1 ...ip (a) = 0
(i1 ,...,ip )

60

l By lemma 0.44, i+1 = i + ci , so i1 ...ip (a) = 0 if ip = i. Therefore, we have

i ...i i+1 (a) = 0 п Therefore, there exists an element b A such that i (b) = i+1 (a) and by lemma 0.24 there exists an element a2 = a + b2 z such that za2 z
-1

= a2 + z i + i+2 z

i+2

+ ...

Note that here the coeffitients on the right hand side belong to another embedding п п п п п of A given by element a2 . Since A is a C1 -field, A is generated by a2 over F . So, the п p-basis of A consists of 1 element. So, by classical Cohen's theorem, any lifting of this п element gives an embedding of A. Now using induction and completeness of A we get that there exists an element a3 such that za3 z
-1

= a3 + z

i

п п and a3 generates A over F . Therefore, ap commutes with z , from here follows that a п 3 is a purely inseparable element and F (a3 ) is an "unramified" extension. The corollary is proved. 2 This corollary concludes the proof of theorem 0.36.

3

0.5

Classes of conjugate elements

Let K be a splittable local skew field of characteristic 0 whose first residue skew field is commutative and whose last residue skew field k is contained in its centre. We have classified these skew fields in the preceding section. In this section we give necessary and sufficient conditions for two elements of K to be conjugate. We fix a representation of K in the form k ((u))((z )). Definition 0.56 Let = Id. A residue resi,r on K is defined to be a map res k ((u))((z )) k xi resi,r (X ) = res i du u where X = l xl z l . Prop osition 0.57 Let = Id. Let L, M K , (L) = (M ) = -1, M = b-1 z -1 + b0 + b1 z + ..., L = a-1 z -1 + a0 + a1 z + .... The fol lowing assumptions are equivalent:
i,r

:

61

(i) there is an S K , (S ) = 0, such that M = S (ii) a
-1

-1

LS

= b-1 , a0 = b0 , ..., a res a
i-1 - bi-1 ui a-1

i-2

=b

i-2

; u a
i-1 - bi-1 ui a-1

du Z and

k [[u]] L
j -1 Sj

~ resi,r (M j ) = resi,r (Lj ) for al l j 1, where Lj = Sj j ~ Sj (M, Lj -1 ).

-1

~ , L0 := L, Sj = ~

Pro of K has the form k ((u))((z )) with the relation zuz we have: SM = s0 b-1 z
-1 i-2

-1

= u + ui z i + .... Thus
i-1

+(s0 b0 + s1 b-1 )+ ... +(
j =-1 i-2

bj s

i-2-j

)z

i-2

+(
j = -1

bj s
i-1

i-1-j

)z

i-1

+ ...

LS = s0 a-1 z +(s0 a0 +s1 a-1 )+...+(
j = -1

-1

aj s

i-2-j

)z

i-2

+(-a

i -1 s 0

+
j =-1

aj s

i-1-j

)z

i-1

+...

It follows that the condition a-1 = b-1 , a0 = b0 , ..., ai-2 = bi-2 is necessary for M and L to be conjugate. Another necessary condition is given by the following equation for s0 : si ai-1 - bi-1 0 = s0 a- 1 Since i is a differentiation, we have
s u 0 i-1 - bi-1 ui a-1

s

0

=

a

Thus we obtain the second necessary condition: res a
i-1 - bi-1 ui a-1

du Z and

u

a

i-1 - bi-1 ui a-1

k [[u]]

Conversely, if these two conditions hold, then there is an s0 k ((u)) such that the first i + 1 summands in L1 = s-1 Ls0 are the same as those in M . It is clear that L 0 and M are conjugate if and only if L1 and M are conjugate. The conjugating element ~ ~ S has the form 1 + §§§ (S can be written as (1 + s1 z )(1 + s2 z 2 ) ....) Note that for every -1 x-1 z + x0 + x1 z + ... K holds: (1 + sj z j )-1 (x-1 z
-1

+ x0 + x1 z + ...)(1 + sj z j ) = x-1 z (x
i+j -1

-1

+ x0 + x1 z + ... + x + ...

i+j -2

z

i+j -2

+

+ jxi1 sj + x-1 si )z - j 62

i+j -1

since the proof of lemma 0.11, (ii) implies that (1 + sj z j )-1 (x
-1

+ x0 z + x1 z 2 + ...)(1 + sj z j ) = x (x
i+j -1

-1

+ x0 z + ... + x

i+j -2

z

i+j -1

+

+ jxi1 sj )z -

i+j

+ ...),

and (1+sj z j )-1 z -1 (1+sj z j ) = (1+sj z j )-1 (z -1 +sj z It follows that (s1 a-1 )i = bi - a
i j -1

-si z j

i+j -1

+...) = z -1 -si z j

i+j -1

+...

(j = 1),

~ ~ if M = S -1 L1 S , where ai is the coefficient of L1 . This equation is soluble if and only if res b i - ai du = 0, ui

that is, resi,r (M ) = resi,r (L1 ). Conversely, if the residues are equal then there is an s1 k ((u)) such that the first i + 2 summands in L2 = (1 + s1 z )-1 L1 (1 + s1 z ) are the same as those in M . п п Proceeding by induction, we obtain at the k th step that if M = S -1 Lk S , then ksk a
i -1

+ a-1 si = b k

i+k-1

-a

i+k-1

.

To solve this equation, we substitute sk = a-k s into it and obtain the equation -1 s =a which is solvable if and only if res - coefficient of z i in M k has the form
a k-1 bi+k-1 -1

-a ui

i+k-1

, . On the other hand, the

k -1 1 ai+k-1 u i

= res

u

-r a k -1 b -1 i+k-1 u i

ka

k -1 -1 bi+k-1

+ fM

where fM is a polynomial in bi+k-2 ,...,b-1 and the values of j at these points. The corresponding coefficient in Lk has the form k ka
k -1 -1 ai+k-1 u i k -1 -1 bi+k-1 u i

k -1 -1 ai+k-1

+ fLk

and fLk = fM , since aj = bj for j i + k - 2. It follows that resi,r Lk = resi,r M k if and k only if res 2
a

= res

a

. which completes the proof of the proposition.

63

Definition 0.58 Let = Id. We say that the residue res ofX = zero, if x0 im( - Id)

l

xl z l is equal to

We say that two elements have the same residue if the residue of their difference is equal to zero. We define : k ((u)) k ((u)), (x) = x

-1

/x.

Prop osition 0.59 Let = Id. Let L, M K , (L) = (M ) = -1, M = b-1 z -1 + b0 + b1 z + ..., L = a-1 z -1 + a0 + a1 z + .... The fol lowing conditions are equivalent: (i) there exists an S K , (S ) = 0, such that M = S (ii) b-1 /a-1 im; ~ res (M j ) = res (Lj ) for al l j 1, where Lj = Sj j ~ Sj (M, Lj -1 ).
-1 -1

LS

L

j -1 Sj

~ , L0 := L, Sj = ~

Pro of is similar to that of the preceding proposition. We have SM = s0 b-1 z LS = a-1 s
-1

-1

+(s0 b0 + s1 b 1 )+ ... - +(a0 s0 + a-1 s
1
-1

0

-1

z

-1

)+ ...

Therefore, s0 b-1 = a-1 s , that is b-1 /a-1 im. If this condition holds, then we put 0 L1 = s-1 Ls0 . The first coefficients in L1 and M are equal. 0 Now we observe that (1 + sj )-1 (x-1 z +x
j -2 -1

+ x0 + x1 z + ...)(1 + sj z j ) = x-1 z
j -1

-1

+ ...

z

j -2

+(x

+ sj x

j -1

- x-1 s

j

-1

)z

j -1

+ ...

for any x-1 z -1 + x0 + x1 z + ... K , which follows from the calculation in the proof of Lemma 0.11, (i). The arguments used in the proof of the preceding proposition yield at the first step the following condition that is necessary for conjugacy: s1 a
-1

- a- 1 s

1

-1

= (s

1

-1

a-1 ) - (s

1

-1

a- 1 ) = b 0 - a

0

This equation is soluble if and only if (b0 - a0 ) im( - Id). which is equivalent to the equality res M = res L1 . 64

At the j th step we have the condition sj a Hence, (a 1 a -
2 -1 j -1

- a- 1 s

j

-1

=a

j -1

-b

j -1

...a

j - -1

1

)(a

j -1

-b

j -1

) = (a 1 a -
1

2 -1

...a 1 )sj - (a -
-1

j

-1
-1

...a

j - -1

1

)s

j

-1

=

((a-1 ...a This equation is is equivalent to in Lj are equal power of z in M

j - -1

)s

j

-1

) - (a

...a

j - -1

1

)s

j

soluble if and only if (a-1 ...a 1 )(aj -1 - bj -1 ) im( - Id), which - the equality res (M j ) = res (Lj ), since the first (j - 1) coefficients j to the corresponding coefficients in M , and the coefficient of the 0th j is a
-1

j -1

...a

- -1

j +2

b

-j +1 j -1

+b

j - 1 a- 1

...a -

j -1

1

+

a sum of monomials with indices < j - 1 The corresponding coefficient in Lj is j a
-1

...a

- -1

j +2

a

-j +1 j -1

+a

j - 1 a- 1

...a

j - -1

1

+

a sum of monomials with indices < j - 1 Hence, (a
-1 - -1
j +2

...a

b

-j +1 j -1

-a
-1

-1

...a
- -1

- -1

j +2

a

-j +1 j -1

+b
-1

j - 1 a- 1

...a
j +2

j - -1

1

-a

j - 1 a- 1

...a

j - -1

1

)=

([a

...a

j +2

b

-j +1 j -1

-a

+ ...a

- -1

a

-j +1 j -1

]-

[...]+ [...] - 2 [...]+ 2 [...] ... + b
j - 1 a- 1

j -1

[...]+

...a
-1

-1

j -1

-a
1

j - 1 a- 1 j -1

...a
j -1

-1

j -1

)=

(2[a 2

...a

j - -1

(a

-b

)])

Remark It was shown in [18], that for the residue res1,0 in the skew field of pseudodifferential operators holds res1,0 [X, Y ] = 0, where [X, Y ] is the commutator of two pseudodifferential operators. The next statements provide other examples of skew fields with this property. Lemma 0.60 Let K be a skew field such that n = Id or n = Id, in = . Let X, Y K . Then res [X, Y ] = 0.

65

Pro of It is sufficient to prove the assertion for X = ul z k , Y = um z q . If k + q = 0, then res (XY ) = res (YX ) = 0. In the case k + q = 0 we have: XY - YX = ul (um ) - um (ul ) 2 In this case our propositions can be stated as follows: Corollary 10 Let K be this case K is the ring k (L) = (M ) = -1, M = b -1 z -1 + b0 + b1 z + L = a -1 z - 1 + a0 + a1 z + The fol lowing conditions a skew field such that = Id, i = 1, r = 0, a = 0 ((In ((u))(( -1 )) of pseudodifferential operators.) Let L, M K , ..., .... are equivalent:
-1
k -k

= k (um (ul ) ) - um (ul )

-k

-k

im( - Id)

(i) there is an S K , (S ) = 0, such that M = S (ii) a
-1

LS

= b-1 ; res a0 - b0 du Z and a- 1 u(a0 - b0 ) k [[u]] a- 1

res1,0 (M j ) = res1,0 (Lj ) for al l j 1. Corollary 11 Assume that n = Id for al l n N. Let L, M K , (L) = (M ) = -1, M = b-1 z -1 + b0 + b1 z + ..., L = a-1 z -1 + a0 + a1 z + .... The fol lowing conditions are equivalent: (i) there is an S K , (S ) = 0, such that M = S (ii) b-1 /a-1 im; res (M j ) = res (Lj ) for al l j 1. The following examples show that the identity res... ([X, Y ]) = 0 does not hold in other cases. Example (i) Let K be a skew field with = 1, a(0,..., 0) = 0, r = 1. We assume that K has the form specified in Theorem 0.35. Let M = z -1 , L = z -1 + z i k ((z )) K . If resi,r ([X, Y ]) = 0 holds, then M and L are conjugate by Proposition 0.57. Let S = 1 + s1 z + .... We have SM = z
-1 -1

LS

+ s1 + s2 z + ... = LS = (z 66

-1

+ z i )(1 + s1 z + ...) =

(z

-1

+ s1 + s2 z + ...)+(z i - si z i )+(s1 z 1

i+1

- si z 2

i+1

)+ ... +(si z 2i + s

2 i -2 1

i

z 2i - s

i 2i i+1 z

)+ ...

Hence, 1 - si = 0. Since r = 1, this equation is soluble, and s1 = (1 - r)-1 c-1 u1-r . 1 Solving the next equations, we obtain s2 ,s3 ,.... Each of these elements consists of a single monomial whose valuation is different from r - 1. 2 - i Further, we have si z 2i + s1i 2i z 2i - s+1 z 2i = 0. By Theorem 0.35, if a(0,..., 0) = 0, i then s1i 2i contains a monomial whose valuation is equal to r - 1. Therefore, the equation is insoluble with respect to si+1 , and M is not conjugate to L. This contradiction completes the proof of the assertion. (ii) Let K be a skew field with = 1, a(0,..., 0) = 0. In this case i > 1, since r = 0 for i = 1, and we obtain the ring of pseudodifferential operators. We assume that K has the form specified in Theorem 0.35. Then zuz -1 = u + cur z i + r(i +1)/2c2 u2r-1 z 2i . 2 Therefore, 2i = i . Then for any x k ((u)) holds: z -1 xz = x - xi z i +. z We put X = u
-r -1 -i >2i 2 -

z , Y = u2 . Then
1-r -i

XY = u

z

+ ... + Cur-1 z i + ...,

C ",C = 0

Hence resi,r ([X, Y ]) = 0. An example with a(0,..., 0) = 0, r = 1 can be obtained likewise. (iii) Let K be a skew field with n = 1, in = . We put X = u-rn z -in and Y = u. Then XY =
-in 1-r

u

n

z

-in

+ C + ...

where C = -in -in +1 c = 0. Hence, res ([X, Y ]) = 0. Remark These examples show that the Scolem-Noether theorem does not hold for skew fields defined here.
- Let K be the ring k ((u))((u 1 )) of pseudodifferential operators. We have shown that this is the only slew field such that res1,0 ([X, Y ]) = 0. Let us deduce a criterion for two elements of this skew field to be conjugate. - Let n N be a certain number. Consider the skew field K = k ((t))((t 1 )), where tn = u. Then t = ntn-1 u , and K K . m Lemma 0.61 Let L = l-m t + ... + l0 + l1 of K . L K if and only if li ti k ((tn )). -1 t

+ ... K -be an arbitrary element

Pro of Assume that L K . Then L = b Let j N. We have: - j - u = (n-1 t1-n t )j , u j = (t 1 ntn-1 )j . 67

m - m u

+ ..., where bi k ((u)) = k ((tn )).

We prove first the assertion of the lemma for l-i (i > 0). For i = 1 we have i u = n-1 t1-n t and b-1 u = l-1 n-1 t1-n t . The assertion of the lemma holds, since t1-n tk ((tn )). For an arbitrary i we have
i u =

t -1 1-n -1 1-n i-1 2 (n t )(n t t ) +(n-1 t1-n )2 t (n-1 t1-n t )i-2 = t t
2

2 (1 - n)(n-1 t-n )(n-1 t1-n t )i-1 +(n-1 t1-n )2 t (n-1 t1-n t )i-

Since the coefficients in the expression for L in K belong to k ((tn )), it is sufficient to i show that the lemma holds for u . We prove by induction that the assumption of the lemma holds for all the coefficients k in (n-1 t1-n t )i-1 . The same is true for (n-1 t1-n t )i-2 . Let (n-1 t1-n t )i-2 = i-2 l~ t k=0 k i (Let us note that there are no negative powers of t in the expansion of u , i > 0, and the minimal power oft is equal to 1). We have:
2 (n-1 t1-n )2 t ( i-2 k=0 k l~ t ) = (n-1 t1-n )2 ( k i-2

l~ k
k=0

k+2 t

i-2

+
k=0

l~ k

k+1 t

i-2

+
k=0

k l~ t ) k

Therefore, (n-1 t1-n )2 l~ tk+2 k ((tn )), (n-1 t1-n )2 l~ tk+1 k ((tn )), (n-1 t1-n )2 l~ k k k tk k ((tn )). For i = 0 we have l0 = b0 k ((tn )). Let us prove that the assertion of the lemma holds for -i , i > 0. For i = 1 we have: n-1 - - - u 1 = n k=0 (tn-1 )(k) t 1-k Ck 1 . - ~ ~- Assume that for k < i it is proved u k = lj t k-j , lj t-k-j k ((tn )). j =0
-i u

= (

-1 t

nt

n-1 i

n-1 k=0

) = (n

- Ck 1 (tn-1 )(k)

-1-k t

)(

-1 t

nt

n-1 i-1

)

=

n-1

(n
k=0

- Ck 1 (tn-1 )(k)

-1-k t

)(
j =0

~ lj

-i+1-j t

)

- - ~ ~ (p) - For every k {0,...,n - 1} t 1-k lj = Cp 1-k lj t 1-k-p . This yields the followp=0 ing conditions on the coefficients for fixed k and j : - at t 1-k-p-i+1-j , p 0, the coefficient belongs to t-1-k-i+1-j -p k ((tn )). Conversely, assume that the assumptions of the lemma on the coefficients hold. We have obtained that i i u = j 0 cj t -j , and cj ti-j k ((tn )) for any i Z.

68

Consider the highest monomial in L: l-m
m t m = l-m c-1 u - l-m ( 0 j 1 m We have l-m c-1 k ((tn )), l-m cj c-1 tm-j k ((tn )). Hence, L = l-m c-1 u + L1 , where 0 0 0 (L1 ) > (L), and the the assumptions of the lemma hold for the coefficients in L1 . We complete the proof by induction. 2

c j c -1 0

m-j t

)

Lemma 0.62 Let L, M K K and (L) = (M ) = -n. Let M = SLS S K . Then S K if and only if res l
(L)+1

-1

, where

-m l (L)

(M )+1

= 0 and

t

l

(L)+1

-m l (L)

(M )+1

k [[t]]

Pro of is similar to that of Proposition 0.57. 2 Theorem 0.63 Let L, M K = k ((u))(( - (M ) M = m (M ) t + ..., - (L) L = l (L) t + .... The fol lowing assumptions are equivalent:
-1 u

)), (L) = (M ) < 0,

(i) there is an S K , (S ) = 0, such that M = S (ii) (L) = (M ), m res res(M
j/(- (M )) (M )

-1

LS

= l

(L)

, t l
(L)+1

l

(L)+1

-m l (L)

(M )+1

= 0 and

-m l (L)

(M )+1

k [[t]]

) = res(L

j/(- (L))

) for al l j 1 in K .

Pro of follows immediately from Corollary 10, Lemmas 0.61, 0.62 and the fact that L (and M ) has precisely one nth root in K . 2
- Theorem 0.64 Assume that L, M K = k ((u))((u 1 )) and (L) = (M ) = 0. Then (i) If l0 = m0 = const and l1 = m1 , then M = SLS -1 . (ii) If l0 = m0 = const, then M = SLS -1 if and only if (M - m0 )-1 = S (L - l0 )-1 S -1 (see Theorem 0.63)

Pro of is obvious. 2 69

0.6

New equations of KP-typ e on skew fields

In this section we give an answer on a question given in [22]. Namely, the classical KP-hierarchy is constructed by means of the ring of pseudo-differential operators P = k ((x))(( -1 )). This ring is a skew field. The point is to consider other skew fields instead of this one. We will study if there exist some new non-trivial generalisations of the KPhierarchy for a list of two-dimensional skew fields. In particular, we give a number of new partial differential equations of the KP-type. For every two-dimensional skew field from the list of theorem 1.5 we can write down a decomposition K = K+ + K- , where K- = {L K : ord(L) < 0} and K+ consists of the operators containing only 0 powers of z , and a "KP-hierarchy" in the Lax form: L = [(Ln )+ ,L], tn where L z -1 + K- k [[...,tm ,...]]. Let L = z -1 + u1 z + u2 z 2 + ..., where um = um (u, t1 ,t2 ,...). Further we will denote /tn as n . One can check that if the canonical automorphism in the classification theorem 1.5 is not trivial, then our "KP-hierarchy" became trivial, i.e. it can be easily linearised and solvable. We omit calculations here. So, it can be assumed that = id. The same is true if i > 1, because [(Ln )+ ,L] = -[(Ln )- ,L] = 0 mod i in this case, where is a maximal ideal of the first valuation in K . So, our "KP-hierarchy" again is linear and easily solvable in this case. So, we assume i = 1, hence, r = 0 and c = 1, and there is only only one non-trivial parameter a. If a = 0, K is isomorphic to the ring P of pseudo-differential operators. Denote by u ,u ,... the subsequent derivatives by x. First for n = 1, we get 1 u1 = u1 This means that we can take t1 = x for u1 . Now we write down the first two equations for n = 2 and the first equation for n = 3. (3) 2 u1 = u 1 +2u2 2 u2 = 2u3 +2u1 u1 + u 3 u 1 = u
1 2

+2ax-1 u

2 1

(4) - x - 2 u1 ) (5)

+3u

2

+3u3 +6u1 u1 +3a(x-1 u

Let us introduce the new notation: u = u1 (x, y , t) with y = t2 , t = t3 . Also we use the standart notation ut ,uy ,uyy ,... for derivatives. We can eliminate u3 from equations 4 and 5 and then we get 3u2y - 2ut = -6uu - 3u
2

- 2u +6ax-1 u2 - 6ax-1 u +6ax-2 u

(6)

70

From 3 we find u
2

= 1/2(u

y

- u ),u

2y

= 1/2(u

yy

- u y)

Differentiating equation 6 by x and inserting these expressions we finally get new KPequation (4ut - u - 12uu ) = 3u
yy

+6a(5x-2 u - x-2 uy - 3x-1 u + x-1 uy - 4x-3 u )

One can see that if a = 0, we get the usual KP-equation (see also explicite calculations in [21]).

71

Chapter 1 Classification of automorphisms of a two-dimensional lo cal field.
1.1 Basic results.

In this chapter let K be a two-dimensional local field, K k ((u))((z )); Autk (K ) = be a group of continuous k -automorphisms of a field K with respect to the topology given by fixed parametrisation, i.e. by the parameters u and z (see [35] concerning the connection between a topology and a parametrisation). Introduce the following notation. By Greece letters , , we will denote automorphisms of a field K . An overline will denote the residue homomorphism. As before, п denote a valuation on the field K , denote a valuation on the field K , , are п п п valuation ideals of the valuations , , І(k ) is the group of roots of the unity, Autk (K ) п п is a group of continuous k -automorphisms of the field K . Recall some results from chapter 1, section 3. п Definition 1.1 Let K be a one-dimensional local field with the residue field k , п п charK = chark , Autk (K ). Put п -1 mod k and define i(п ) N by the fol lowing: (п ) = (u)u п i(п ) = 1 if (п ) І(k ), else / n i(п ) = (( -Id)(u)), where n 1, (п ) is a primitive root of degree n, ord( (п )) = n. пп Prop osition 1.2 Let k be an arbitrary field, chark = 0. Any automorphism п п Autk (k ((u))) with (u) = (п )u + a2 u2 + ... is conjugate with the automorphism : п п(u) = (п )u + xui(п ) + x2 yu2i(п )-1 , where x k /k (i(п )-1) , y k . пп Two automorphisms , are conjugate iff п п п п),i( ),x( ),y ( )) = ( ( ),i( ),x( ),y ( )). п п п п ( ( Corollary 12 1) i(п ) = 1 iff is an automorphism of infinite order and (п ) has п infinite order; 72

2) 1 < i(п ) < iff has infinite order and (п ) has finite order; п 3) i(п ) = iff has finite order. п Remark. i) In the notation of proposition we have n|(i(п ) - 1). ii) if k is a field of characteristic p > 0, then the following fact remains true: the п automorphism Autk (k ((u))) with (u) = (п )u + a2 u2 + ... is conjugate with the п п: (u) = (п )u + xui(п ) + ..., where x k /k (i(п )-1) . п automorphism пп п Lemma 1.3 Let a0 K , Autk (K ). The fol lowing property: if n = Id for some n, then dim(kerT ) = п n if = Id and chark = 0, then dim(kerT ) = п if n = Id and chark = p, then one of the fol п 1) dim(kerT ) = dim(cok erT ) = 0 or 2) dim(kerT ) = 0, dim(cok erT ) = or 3) dim(kerT ) = 1, dim(cok erT ) = . п п linear map T = - a0 : K K has the п dim(cok erT ) = d, where d = 0 or ; dim(cok erT ) = d, where d = 0 or 1; lowing cases holds:

Pro of. By proposition 1.2 we can assume (u) = u + xui(п ) + ... п primitive n-th root of unity. If (u) = u, n = 1, then the first claim of lemma is clear, so п (u) = u (note that we have proved the first claim in the case chark п by corollary 12, any automorphism of a finite order looks like this). Suppose the element a0 satisfy one of the following properties: (a0 ) = 0 or п (a0 ) = 0 but a0 = j for all j Z. Let's study values of the valuation п T (ul ) for different l. We have:

, where is a from now on = 0, because,

on elements п
l

(ul ) - a0 ul = (u + xu п

i(п )

+ ...)l - a0 ul = l ul (1 + -1 xu

i(п )-1

+ ...)l - a0 u

Therefore: п п (T (ul )) = l or l + (a0 ) if (a0 ) < 0. п So, we can solve any equation (y ) - a0 y = Y , and the map T is surjective. It is п п п injective, because the values (T (ul )) are finite and (T (ul )) = (T (ul1 )) if l = l1 . п j Suppose now a0 = . Since the injectivity and the pro jectivity of the map - a0 are п defined by the existence and the uniqueness of a solution of the equation (y ) - a0 y = Y п п for any Y K , we can replace y by yuj and assume that a0 = 1. Then a0 can be written as the product a0 = (1 + a0j uj ), a0j k . j =1 Put q = (a0 - 1). There are two possible case: q < i(п ) - 1 and q i(п ) - 1. п Let q < i(п ) - 1. Then we can assume n|q . To prove it we have to prove that a0 п can be written as the product a1 (x) for some x, where (a1 ) = q1 > q , n|q1 . п0 0x l) l) (1+cu п 1+c(u п l l Note that 1+cul = 1+cul = 1 + c(п (u ) - u )(1 + cul )-1 , where c is a constant. (ul ) - ul = (u + xu п
i(п )

+ ...)l - ul = l ul (1 + -1 xu 73

i(п )-1

+ ...)l - u

l

(1.1)

From this formula we get the following property: l п if n |l, then (п (ul ) - ul ) = l and (1+cu ) = 1 + ( l - 1)cul + .... Hence, a0 can be п 1+cul represented as the product above, because there exists a constant c such that the value l (a0 (1+cu ) ) increases. п п 1+cul So, let we now have: q < i(п ) - 1 and n|q . Let's study values of the valuation on п l elements T (u ) for different l. By formula (1.1) we have: (T (ul )) = l if n |l п (T (ul )) = l + q if (l, n) = 1. п Therefore, we can solve any equation (y ) - a0 y = Y , and the map T is surjective. п It is injective, because all the values (T (ul )) are finite and (T (ul )) = (T (ul1 )) if п п п l = l1 . Consider now the case q i(п ) - 1. As in the first case we can assume that n|q . We divide this case into three cases: q = i(п ) - 1, q > i(п ) - 1 and q is finite q is infinite, i.e. a0 = 1. Let q = i(п ) - 1. Let a0 = 1 + wuq + .... Note that
(u + xui(п ) + ...)ln (culn ) п = = 1 + nl -1 xu ln ln cu u i(п )-1

+ ...

Hence, if w = nl -1 x for all l, we can apply the same arguments as in the first case and get that T is injective and surjective, i.e. d = 0. Otherwise, we can write a0 = п0 a1 (uln )/uln , where (a1 - 1) > q , and reduce this case to the case q > i(п ) - 1. 0п Let chark = 0. We claim that the case q > i(п ) - 1 can be reduced to the case п q = . In this connection it is necessary to show that a0 = (A) . We know, that A (1 + cul ) п = 1 + c(п (ul ) - ul )(1 + cul )- 1+ cul
1

and (ul ) - ul has the valuation equal to l if (l, n) = 1, and to (i(п ) - 1) + l if n|l, п (l, chark ) = 1 and l = 0. From here we get the necessary result, because we can multiply a0 sequentially by suitable elements of the form 1 + cuj + ..., each of which can be got from a certain element of the form 1 + cj uj or 1 + cj uj -(i(п )-1) . It is clear that the product A = j -(i(п )-1) ) converges. j =q (1 + cj u Let now chark = 0 and q = , i.e. a0 = 1. Then we claim that d = 1. Let us first find the dimension of the kernel of the map T . To do that we investigate the values of the valuation of the elements T (ul ) by different l. We have: п l (T (u )) = l if n |l п (T (ul )) = l +(i(п ) - 1) if n|l and l = 0 п (T (1)) = if l = 0, i.e. T (1) = 0. п 74

From this follows that the kernel is one-dimensional and consists of the elements of п п the field k , because all (T (ul )) are finite if l = 0 and (T (ul )) = (T (ul1 )) if l = l1 . п On the other hand, we get also that the cokernel is one-dimensional, because we can get an element with any value of valuation except an element with the value (i(п ) - 1), and there exists a pullback of any convergent (to zero) sequence, which is also converge to zero. Now we must examine the cases, when chark = p. Let us first consider the case chark = p and q = . We prove that n = 1 if and only if dimk (ker(п - 1)) = . п п Let n = 1, n = pk m, (p, m) = 1. It is obvious that if exists an element x K , п x k such that ( - 1)(x) = 0, then dimk (ker(п - 1)) = . Suppose, that there is no / п such an element. Therefore: пп m (u) = u + a1 , a1 K , (a1 ) > 1, a1 = 0, п пп п 2m (u) = u +2a1 + a2 , a2 K , (a2 ) > (a1 ), a2 = 0, п ..., k пп п p m (u) = u + ... + apk , apk K , (apk ) > (apk -1 ), apk = 0, п and we get a contradiction. Conversely, let dimk (ker(п - 1)) = inf ty . Assume F = ker(п m - 1), m = ord( (п )). It's clear that F is a field. Let n N be a minimal positive value of the valuation on this field. п k k Then n = p , k Z. For, if n = p l, (l, p) = 1, then there exists an element x F п with such a value and, moreover, x = dl , d K . But then d F , because (п m ) = 1, a contradiction. п п So, K/F is a finite algebraic extension of degree pk , therefore m is an automorphism of a finite order. It is easy to see that the order is equal to n, i.e. is a generator of п п (kerT )). the cyclic Galois group Gal(K/ Remark. In particular, we have got a description of a subgroup of elements of finite order in the so-called "Nottingham" group. See [3], [12], [5] for further details п п about this group (i.e. the group Autk (K ), charK = p). Let be an automorphism of infinite order. Then kerT = k , dimk (kerT ) = 1. Let п (i(п ) - 1,p) = 1. We claim that for any integer N > 0 there exist numbers h(N ) N, h(N ) > h(N - 1) and x, h(N - 1) < x h(N ) such that the maximal value of the valuation on a preimage of arbitrary element with the value x less than -N (or the preimage is empty). From this follows that one can construct infinitely many elements, which are not in the image of the map T . For N = 1 it's clear -- h(1) = x = i(п ) - 1. For arbitrary N consider the vector space < T (ul ), -N l s(k ) >,where s(k ) = (i(п )-1)+p(i(п )-1)+...+pk (i(п )-1). l It's clear that dimk < T (u ), -N l s(k ) > (s(k )+ N ). From the other hand side, k (T (up (i(п )-1) )) = 2pk (i(п ) - 1) > s(k ), п k (i(п )-1)+pk-1 (i(п )-1) )) = 2pk-1 (i(п ) - 1) + pk (i(п ) - 1) > s(k ), (T (up п 75

... k (T (up (i(п )-1)+...+p(i(п )-1) )) > s(k ). п So, our property holds for k N . Indeed, assume the converse. Then < ul ,h(N - l 1) l s(k ) >< T (u ), -N l s(k ) > for all k ,and s(k ) - h(N - 1) s(k )+ N - k for all k , a contradiction. п In the case (i(п ) - 1,p) = 1 we have: (T (ul )) = (T (ul1 )) if l = l1 . Therefore, the п cokernel of the map T has infinite dimension. Let now chark = p, q > (i(п ) - 1) and q is finite. Note that T is not injective if and only if a0 = (A)/A. Therefore, if T is not injective, this case is equivalent to the п case q = . Let T be injective. Here two cases are possible: п п 1) there exists an integer i (a0 - 1) such that for some s1 ,s2 K a0 = s2 (s1 )/s1 , п п where (s2 - 1) = i and there are no elements s1 ,s2 such that a0 = s2 (s1 )/s1 , where п (s2 - 1) > i. п п 2) for any integer i such that i (a0 - 1) there exist s1 ,s2 K such that a0 = п s2 (s1 )/s1 , where (s2 - 1) i. п п For example, the first case takes the place when (T (a0 - 1)) < (i(п ) - 1), the п п second -- when (T (a0 - 1)) > (i(п ) - 1). If (T (a0 - 1)) = (i(п ) - 1), then may take п place either the first or the second case. Indeed, we have seen that (culn ) п = 1 + nl -1 xu culn
i(п )-1

+ ...
1

(1.2) (1.3)

(1 + cul ) п = 1 + c(п (ul ) - ul )(1 + cul )- l 1+ cu

п Hence, if (T (a0 - 1)) < (i(п ) - 1) or (T (a0 - 1)) = (i(п ) - 1), but a0 = 1 + wui(п )-1 + ... п and w = nl -1 x for all l, then 1) holds. If the rest inequalities hold, then one can see that 2) may take place. Let the case 1) holds. Let's show that T is surjective. п Indeed, this case is equivalent to the property (п (y ) - a0 y ) i +п(y ), y K . But п this means that < T (ul ),N l N1 >< ul ,N + min{i(п ) - 1,i} l N1 + max{i(п ) - 1,i} > for all integers N, N1 , N < N1 . From this follows that the cokernel of the map T cannot have infinite dimension. Suppose it has finite dimension, i.e. it is not equal to zero. Choose an element of the minimal value of the valuation in the cokernel and choose a number N1 : N1 + i < . Complete a basis of the vector space < T (ul ),N l N1 > with respect to the basis of the vector space < ul ,N + min{i(п ) - 1,i} l N1 + max{i(п ) - 1,i} >. Denote

76

~ these elements by ej , j {1,..., |i - i(п )+1|}. Then for any integer N1 > N1 we will have ~ < T (ul ),ej ,N l N1 ,j {1,..., |i - i(п ) + 1|} >=< ul ,N + min{i(п ) - 1,i} ~1 + max{i(п ) - 1,i} >, but this contradicts to the existence of elements of the lN cokernel. Let the case 2) holds. This case is a negation of the case 1); so, for any natural i there exists N1 N such that < T (ul ),N l N1 >< ul ,N + min{i(п ) - 1,i} l N1 + max{i(п ) - 1,i} >. Repeating the arguments of the case a0 = 1 we get that п there exists a converge sequence in K such that the maximal values of the valuation on preimages of elements of this sequence tend to -. From this we get that the cokernel has infinite dimension. The lemma is proved. 2 Corollary 13 In the notation of lemma let chark = 0. Then d = 1 if and only if п п п п has infinite order and a0 = (x)/x for some x K ; d = iff has finite order and a0 = j , j Z; d = 0 in the rest cases. п Let Autk (K ). Then the automorphism Autk (K ) and its invariants (п ) п i(п )-1 , y (п ) k are defined (see def. and prop. 1.2 ). Put k , i(п ), x(п ) k /(k ) -1 k ((u)). Note that the numb er (a ) do es not dep end on the choice of п0 a0 = (z )z the parameter z .

п п/ Theorem 1.4 (Theorem I) Let chark = 0. Let (a0 ) = 0 or (a0 ) = 0, but a0 п { (п )m ,m Z}. Then 1) The automorphism is conjugate with an automorphism given by the formula (u) = u + xu (z ) = u
(a 0 ) п i(п )

+ x2 yu
2n

2i(п )-1

a0 (1 + an un + a2n u п

+ ...a

i(п )-1

u

i(п )-1

)z

/ where = (п ), x = x(п ), y = y (п ), anq k , q {1,..., (i(п ) - 1)/n}, ai(п )-1 -1 n (п ) x(п )Z , Z = Z\{0}, n = ord( (п )). (i(п )-1) j 2) (a0 ), a0 , aj п п /x(п ) , (п ), x(п ), y (п ), i(п ) is the complete system of invariants with respect to the conjugation. Assume (u) = c0 + c1 z + c2 z 2 + ..., (z ) = a0 z + a1 z 2 + ..., ci k ((u)) ai k ((u))

Let us denote the additional notation: п i N {} -- such a minimal positive integer that ai = (Y )/Y for some Y k ((u)), 0 j = minq {iq : ciq = 0}, q 0, 77

i() = minq {iq +1 : aiq = 0}, a0 = 1 + an un + a2n u2n + ... + a ~ п f y2 -- ~ x0 = y1 -- ~ B1 = B2 = Aq = Note

i(п )-1

u

i(п )-1

,
)- 1 )- 2

п пп п Autk (K ) -- such an automorphism that f -1 f (u) = (п )u + xui(п ) + yu2i(п i(п )-1 2i(п such a solution of the equation (Y )/Y = (п )+i(п )xx0 п +(2i(п )-1)yx0 пy п f (u) that (~2 u- (~2 ) ) = 1, y пy such a solution of the equation (Y )/Y = ai that (~1 u- (~1 ) ) = 1, п ~0 y -(i()-1)/i y1 ~ a0 , ~ -2(i()-1)/i y1 ~ a0 , ~ -(j +q -1)/i п-1 f (п (~2 )). y1 ~ y that B1 ,B2 ,Aq are defined uniquely.

, ,

Theorem 1.5 (Theorem I I) Assume chark = 0 and let (a0 ) = 0 and a0 п п m { (п ) ,m Z} and be of infinite order. п Then is conjugate to , that is defined according to the next four possible cases: a) i() - 1 = j , i() < , so (u) = (п )u + xu
i(п )

+ x2 yu z

2i(п )-1

+ r1 A1 u

i(п )-1 j

z

(z ) = a0 z + s1 B1 u ~ where r1 k /(k )j , s2 k , s b) i() - 1 < j , i() <

i(п )-1 i()

+ s2 B2 u /r1

i(п )-1 2i()-1

z

j (i(п )-1) (i(п )-2)(i()-1) x 1

(i( bar)-1)(i()-1)

k.

(u) = (п )u + xui(п ) + x2 yu

2i(п )-1 i(п )-1 2i()-1

(z ) = a0 z + s1 B1 u ~
where s1 k /k (i()-1,i(п )-1) , s2 k . c) i() - 1 > j , i() < (u) = (п )u + xui(п ) + x2 yu 2i(п )-1

i(п )-1 i()

z

+ s2 B2 u

z

+ r1 A1 u

i(п )-1 j

z + ... + r + s2 B2 u

i()-1-j

Ai(

)-1-j

u

i(п )-1 i()-1

z

(z ) = a0 z + s1 B1 u ~ where r1 k /k j , s1 d) i() = (j ) x

i(п )-1 i()

z

i(п )-1 2i()-1

z

j (i(п )-1) (i(п )-2)(i()-1)

/r1

(i(п )-1)(i()-1)

k , rq ,s2 k , q = 1.
i(п )-1 j

(u) = (п )u + xui(п ) + x2 yu

2i(п )-1

+ r1 A1 u

z

(z ) = a0 z ~ where r1 k /k j .

78

We denote j () := j in the cases a), c), d), and j () := in the case b). Then (i(п )-1)/j (a0 ), aj п /x(п ), (п ), x(п ), y (п ), i(п ), i(), j (), i, and the elements rq ,s1 ,s2 with the relations defined in the items a)-d) are complete system of invariants with respect to the conjugation. Moreover, i|j (), i|(i() - 1) (we accept that in the case i = i|j means, that j = , i() = , i.e. there are no elements with z j , z i() ). i = if and only if a0 = 1 + ai(п )-1 ui(п )-1 , where ai(п )-1 = q -1 x, q ". ~ Let us introduce the additional notation: п qa := - (cj ) mod j , i.e. 0 qa > -j ; п qb := qa + min{ (aj ) - qa ; -1}; in the case of qb - qa = -1 we denote cb /ca := {resu (cj /aj ) if aj = 0, cb /ca := 1 otherwise }; in the case of cb /ca " we denote q1 := 1 if cb /ca Z, q1 is a denominator of the fraction cb /ca = p1 /q1 , where (p1 ,q1 ) = 1, q1 > 0 otherwise; in the last case let us denote by p1 Z the numerator of this fraction; the case q1 < j (), q1 |j () we denote by N a number that satisfies the properties n1 < q1 , q1 |(j () - n1 ); the case, when the equation (x +1)/j = p1 /q1 is solvable, we denote by N such a number that ib - qa + 1 is a solution of this equation. Consider the equations 0 = -(1 + w)n2 p1 + n1 jw - (q - 1)(2 + w)q1 n1 + q1 (-2+(1 + w)qa )+ 1 q1 [p1 (j (q - 1)(w +2)+ j - (q - 1)2 q1 +(q - 1)q1 +2q1 )+2j (qa - 1) + (q - 1)q1 ((1 + w)qa - 2)] (1.4)
2 (p2 (-1+ q - q1 (-1+ q - 2q 2 + qq1 )) - qq1 (qa - 1) + p1 q1 (1 - 3q - (q - 1)qq1 + qqa )) = 0 (1.5) 1

in n1 in ib

Theorem 1.6 (Theorem I I I) Let chark = 0 and let (a0 ) = 0 and a0 { (п )m ,m п п n Z} and = Id. п Then is conjugate to defined in one of the fol lowing ways, depending on the possible cases: O) i = . Then (u) = u (z ) = B0 z 79

where B0 k ((un )), that is, has a form of a canonical automorphism of onedimensional local field F ((z )), where F = k ((un )) in the appropriate case. O') i < . Then i = 1 and this case is divided into two ones: I) j > i() - 1, i.e. j () = (u) = u (z ) = z + B1 z
i() 2 + B1 B2 z 2i()-1

where B1 ,B2 k ((un )), B k ((un )) /k ((un ))(i()-1) , i.e. has the form of a canonical automorphism of one-dimensional local field F ((z )), where F = k ((un )). II) j i() - 1 (and in this case j = i() - 1). This case has two subsections: A) qb - qa < -1. Then (u) = u + ruqa z (z ) = z + s1 uqb z
j +1 j 2j +1

+ s2 B2 z
j (qa -1) 1

where B2 = uqb -1 if qa = 0, and 0 otherwise, the numbers r = s1 (cb /ca )-1 k /(k )(j,qa ) , s2 ,s

/r

(i()-1)qb

k.

B) qb - qa = -1. This case has two possibilities: 1) cb /ca ". Then / (u) = u + ruqa z (z ) = z + s1 u where numbers r k /(k )(
j,qa ) j

qa -1 j +1

z

,s

j (qa -1) 1

/r

(i()-1)qb

k.

2) cb /ca ". This case has three possibilities: a) q1 = j . Then (z ) = z + s1 B1 z
j +1

(u) = u + ruqa z + s2 u
-p1 -1+q
a

j

z

2j +1

+ s3 u

-p1 +2qa -2 3j +1

z

where B1 = uqa -1 if (x +1 - qa )/j = p1 /q1 for al l x N, and B1 = uqa -1 + rib uib , ib > qa - 1 otherwise, j (q -1) s2 ,s3 ,rib k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k .

80

b) q1 < j . Here we have two sub-cases: i) q1 |j . Then (z ) = z +s1 B1 z s
j/q
1

(u) = u + ruqa z
j +1

j
a

+(s2,1 u
-1 1

- p1 q

-1 1

(q1 -j )+2qa -2

+s2,2 u

-p1 -1+q
1

)z

j +1+q

1

+s3 u

-2p1 -1+q

a

z

j +1+2q

1

+

u

- p1 q

((j/q1 +1)q1 -j )+2qa -2 j +1+(1+j/q1 )q

z

+ sq2 Bq2 z

j +1+q1 (1+q2 )

where B1 = uqa -1 if (x +1 - qa )/j = p1 /q1 for al l x N and B1 = uqa -1 + rib uib , ib > qa - 1 otherwise, Bq2 = u-p1 (1+q2 )-1+qa if -qa +1 - (q2 - 1)p1 = 0, and 0 otherwise, j (q -1) s2 ,s3 ,sj/q1 ,sq2 k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k . ii) q1 |j . Then we again have two cases:
- i') n1 |q1 and -q1 n-1 (p1 q1 1 (n1 - j )+ qa ) = -p1 +1. Then 1

(u) = u + ruqa z (z ) = z + s1 u
qa -1 j +1

j j +1+lq
1

z

+ s2 u

-p1 q

-1 1

(n1 -j )+2qa -2 j +1+n1

z

+ sq Bq z

,

where Bq = u-p1 q-1+qa if (1.4) is fulfil led and 0 otherwise, l is a solution of an equation (1.4); j (q -1) s2 ,sq k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k .
- ii') n1 |q1 or -q1 n-1 (p1 q1 1 (n1 - j )+ qa ) = -p1 +1. Then 1

(u) = u + ruqa z (z ) = z + s1 u
qa -1 j +1

j -p1 -1+q
a

z

+ s2 u

-p1 q
a

-1 1

(n1 -j )+2qa -2 j +1+n1

z

+ s3 u
1

z

j +1+q

1

+

s4 u

-p1 -1+q

z

j +1+2q

1

+ sq Bq z

j +1+lq

where Bq = u-p1 q-1+qa if -qa +1 - (q - 1)p1 = 0 and 0 otherwise, l is a solution of the equation -qa +1 - (q - 1)p1 = 0, j (q -1) s2 ,sq ,s3 ,s4 k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k . c) q1 > j . There are two possibilities: i) j |q1 . Then (z ) = z + s1 u
qa -1 j +1

(u) = u + ruqa z z + s2 u
a

j
1

-p1 -1+q

a

z

j +1+q

+ s3 u

-p1 -2+2q
1

a

z

2j +1+q

1

+

s4 u

-2p1 -1+q

z

j +1+2q

1

+ sq Bq z

j +1+lq

81

where Bq = u-p1 q-1+qa if -qa +1 - (q - 1)p1 = 0, and 0 otherwise, l is a solution of the equation -qa +1 - (q - 1)p1 = 0, j (q -1) s2 ,sq ,s3 ,s4 k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k . ii) j |q1 . It also has two possible cases: i') -p1 + qa - 2 = q1 (qa - 1)/j . Then (u) = u + ruqa z (z ) = z +s1 u
qa -1 j +1 j
a

z

+s2 u

-p1 -1+q

a

z

j +1+q

1

+s3 u

-p1 -2+2q
1

z

2j +1+q

1

+s4 u

-2p1 -1+q

a

z

j +1+2q

1

+

sq Bq z

j +1+lq

where Bq = u-p1 q-1+qa if -qa +1 - (q - 1)p1 = 0, and 0 otherwise, l is a solution of an equation -qa +1 - (q - 1)p1 = 0, j (q -1) s2 ,sq ,s3 ,s4 k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k . ii') -p1 + qa - 2 = q1 (qa - 1)/j . Then (u) = u + ruqa z (z ) = z + s1 u
qa -1 j +1 j 2j +1+l1 q
1

z

+ s2 u Bq

-p1 -1+q ,1

a

z

j +1+q
1

1

+s

qn ,1

Bq

n

,1

z

+s
1

qn ,2

Bq

n

,2

z

2j +1+l2 q

1

+

s

qm ,1

m

z

j +1+l1 q

+ ... + s

qm ,w

Bq

m

,w

z

j +1+lw q

where Bqn ,i = u-p1 li -2+2qa if (1.5) is satisfied, and 0 otherwise, Bqm ,j = u-p1 lj -1+qa if lq are defined, and 0 otherwise, l1 ,l2 are the solutions of equation (1.5), l1 ,...,lw are solutions of some equation of degree w = q1 /j , j (q -1) s2 ,sqn ,i ,sqm ,j k , r k /(k )(j,qa ) , s1 a /r(i()-1)qb k . (a0 ), (п ), i(п ), i(), j (), i, qa , qb , q1 , n1 , ib , and those elements with relations п that were defined in al l items are the complete system of invariants with respect to the conjugation.

82

± Є¤ ¶ ‰

І

є ЅЅм јЄ· Є «Є·… Ћ є ЅЅ Ѕ ««« «««« ««« «««« ў
± Є¤ ¶ ‰‰ ° ‰

± Є¤ ¶ ‰‰‰ °Ѓ ‰‰

¬

Ѓ

Ѓ

Ѓ

Ѓ

Corollary 14 For given (a0 ), n, i(п ), i(), j (), i, qa , qb , q1 , n1 , ib , the set of conjuп gacy classes of the automorphism is parametrised by only finite number of parameters, except the cases Th.III O), O')(I), Th.I i(п ) = . Pro of of theorems (and of corollary) Recall that is an automorphism on п пп the field K , = mod . It is clear that if two automorphisms , are conjugate, then the automorphisms п are conjugate in the group Autk (K ). To prove the theorem we must prove the п , п existence of an automorphism f such that = f f -1 and is an automorphism, as defined in the formulation of the theorem. Thereto it would be also proved, that the 83

automorphism can be uniquely reconstructed by the automorphism and any -like automorphism gives its own conjugacy class. Assume f (u) = x0 + x1 z + x2 z 2 + ... f (z ) = y0 z + y1 z 2 + ... п п We choose the parameter x0 K in such a way that (x0 ) has a canonical form, that i(п ) 2i(п )-1 . Recall that x is a representative of a class k /k (i(п )-1) . is (x0 ) = x0 +xx0 +yx0 п

1.2

Pro of of the theorems I and I I

Let a0 fulfil the assumptions of the theorem. We prove, that there exists an automorphism f such that f (u) = f (u); f (z ) = f (z ), where is an automorphism, as defined in the theorem. To do that, we prove by induction that f (u) = f (u) mod m and f (z ) = f (z ) mod m+1 for all m N. From (1.2), (1.1), (1.3) (which remain true also in the case of finite order automorphism ) follows, that the set of representatives of classes of the elements п п п-1 (a0 a0 -1 (y0 )/y0 ) can be described as the set of the elements {u (a0 ) (1 + an un + f пп 2n i(п )-1 -1 ),anq k, ai(п )-1 = nl x, where is a primitive root from a2n u + ...ai(п )-1 u 1 of a degree n, l Z\{0}}. From the definition of the element a0 follows that the elements anq are uniquely defined by automorphism , that is, they don't depend on the choice of parameter z , and a0 is defined up to multiplication by an element m , п m Z. ~ пп Assume a0 = a0 u (a0 ) (1 + an un + a2n u2n + ...ai(п )-1 ui(п )-1 ). Then we have for m = 1 ~ that i(п ) f (u) = (x0 ) = (x0 ) = x0 + xx0 + ... = f (u); f (z ) = (y0 )(z ) = ~ ~ ~ ~ (y0 )a0 z = f (a0 )y0 z = f (a0 )f (z ) = f (z ) mod 2 . п -1 For an arbitrary m we replace by fm-2 fm-2 for a suitable automorphism fm-2 (that is, for any automorphism with suitable coefficients x0 ,...,xm-2 , y0 ,...,ym-2 ), ~ ~ and now can consider that c0 = u + xui(п ) + ..., c1 = ... = cm-2 = 0, a0 = a0 , a1 = ... = am-2 = 0, x0 = u, x1 = ... = xm-2 = 0, y0 = 1, y1 = ... = ym-2 = 0. Then f (u) = (u)+(xm-1 )(z f (u) = (u + x
m-1 m-1

) = u+xu
m-1

i(п )

+...+c

m-1

z

m-1

+п (xm-1 )~ a
i(п )

m-1 m-1 z 0 m-1

mod
m-1

m

z

m-1

)+ x(u + x i(п )xx

z

m-1 i(п )

)

+ ... = u + x + ... =
m-1

u + ... + x

z

+

m-1

u

i(п )-1 m-1

z

u + x

i(п )

u + ... + x

m-1

(

(п (u)))z (u)

mod

m

84

Hence, a (xm-1 )~ п And in the same way, ~ f (z ) = (z )+(ym-1 )(z m ) = a0 z +a
m-1 m-1 0

+c

m-1

=x

m-1

(

(п (u))). (u)

(1.6)

z m +((ym-1 )+...)(a0 z +...) п ~
m-1

m

mod

m+1

f (z ) = f (~0 )f (z ) = (a0 + a ~ a0 z + ~ Hence, a0 (ym-1 )~m + a п
m-1

(~0 )x a (u)

z

m-1

)(z + y

m-1

zm) =

(~0 )x a (u)

m-1

z m +~0 y a

m-1

z

m

mod

m+1

= a0 y ~

m-1

+

(~0 )x a (u)

m-1

.

(1.7)

By Corollary 12, if the conditions of the theorem are fulfilled, the equations (1.6), (1.7) have the unique solution with any cm-1 ,am-1 and with any m, whence follows the proof of the case 1). By Proposition 1.2 the proof of the case 2) is evident. Proof of the Theorem II If i = , we can apply entirely the same arguments as in the Theorem I, and get that is conjugate to the automorphism , where has the same form as in the Theorem I (i.e. this case corresponds to the case d), when j = ). In order that these iп arguments remain true, we must only show that the element a := + i()xx0-1 + ... in (1.6) can be represented in the form (y )/y . But it follows directly from the relations п (1.2), (1.3), (1.1). Let i < . We prove that there exists such an automorphism f that f (u) = f (u), f (z ) = f (z ), where automorphism is as defined in the theorem. The proof is the same as in the Theorem I. The case m = 1 coincides with the case m = 1 from the Theorem I. Applying the same arguments as there, we get equations of the form (1.6) and (1.7). By Corollary 2, these equations are solvable if i |(m - 1). They may be unsolvable if i|(m - 1). Since chark = 0, the kernel and the cokernel of the maps Tm-
1,1

= (xm-1 )~ п a

m-1 0

- ( + xu

i(п )-1

+ ...)x

m-1

,

Tm-

1,2

= (ym-1 )~m - a0 y п a0 ~

m-1

are one-dimensional if i|(m - 1). k/i k/i We put xk = y1 y2 1 xk , yk = y1 yk for k = iq , q N . Then ~ ~- ~ (xk )~ -( +xu п a
k 0 i(п )-1

+...)xk = п

-k/i (~1 y2 xk y ~

(~1 ) (~2 ) пy пy - ) k/i - xy ~ y2 k 1 ~ y1 ~

k/i

k/i

= (~2 )~1 пy y

-k/i

(п (xk )-xk ),

85

(yk )~ п a

k+1 0

- a0 y k = ~ п

-k/i (~1 yk y

)

(~1 ) пy
k/i y1 ~

k/i

a 0 - a0 y k y 1 ~ ~~

-k/i

= a0 y 1 ~~

-k/i

(п (yk ) - yk )

Now we can write down the kernel and cokernel of these maps in the explicit form. -k/i -k/i For Tk,1 the kernel is y1 y2 (xk )0 , where (xk )0 k , cokernel -- cui(п )-1 y1 (~2 ), ~ ~ ~ пy -k/i ~ c k ; in the same way, for Tk,2 the kernel is y1 (yk )0 , where (yk )0 k , cokernel -- i(п )-1 -k/i y 1 a0 . ~ ~п c1 u Step 1 We show that is conjugate to an automorphism , which has all the coefficients cq and aq , q 1 , satisfying the property: if i |q , then aq = cq = 0; if i|q , then (cq y1 (~2 1 )) i(п ) - 1, (aq y1 a-1 ) i(п ) - 1 п ~ п y- п ~ ~0 (1.8) We even show that in (1.8) we have either equalities or cq = 0 (aq = 0). In fact, let be such an automorphism that c0 = c0 = u + xui(п ) + ..., a0 = a0 . ~ Let us find the rest coefficients satisfying these properties. Applying induction on m, we have for arbitrary m that
q/i q/i

f (u) = (u)+ (xm-1 )(z (u)+ c1 z + ...c п
m-1

m-1

)=
m-1

z

m-2

+c

m-1

z

m-1

+((xm-1 )+ ...)(a0 z + ...) п ~
m-2

mod
m-1

m

f (u) = f (п (u)) + f (c1 )f (z )+ ... + f (c (u)+ п Hence, (п (u))xm-1 z u a (xm-1 )~ п
m-1 0 m-1

)f (z

m-2

)+ f (c
m-1

m-1

)f (z

)=

+ c1 z + ... + c =x (

m-1

z

mod

m

(п (u))) + cm-1 (1.9) u If i |(m - 1), then cm-1 = 0 and by Corollary 2 the solution of this equation exists and is unique. If i|(m - 1), then for solvability of this equation it is enough to select -(m-1)/i (m-1)/i ~ (~2 ), i.e. (cm-1 y1 пy п ~ (~2 1 )) i(п ) - 1. п y- cm-1 in the form cui(п )-1 y1 Further, f (z ) = (z )+ (ym-1 )(z m ) = +c
m-1 m-1

a0 z + a1 z 2 + ... + a ~

m-2

z

m-1

+a

m-1

z m +((ym-1 )+ ...)(am z m + ...) mod п ~0
m-1

m+1

f (z ) = f (~0 )f (z )+ f (a1 )f (z 2 )+ ... + f (a a a0 z + ~ +(a
m-1

)f (z m ) =

(~0 )xm-1 z m +~0 ym-1 z m +(a1 + a (a )xm-1 z m-1 )(z + ym-1 z m )2 + ... a u u 1 + (am-1 )xm-1 z m-1 )(z +ym-1 z m )m = a0 z + (~0 )xm-1 z m +~0 ym-1 z m +a1 z 2 +... ~ a a u u +am-1 z m mod m+1 86

Hence

(~0 )xm-1 + am-1 , a u and in the same to the previous case way we get the desired result. (ym-1 )~m + a п a0
m-1

= a0 y ~

m-1

+

(1.10)

Step 2 Here two cases are possible: 1) j i(); 2) j < i(). Case 1). We show that = f -1 f , where (u) = (u). п -1 To do that we find the sequential conjugations = fmi fmi , where fmi (u) = ~- ~ u + xmi z mi , fmi (z ) = z , m 1, xmi = y1 m y2 (xmi )0 . If m = (j - i()+1)/i, we have for the coefficients cq that: f (u) = (u)+ (xim )(z im ) = (u)+ cj z j + ... + п ~ xim )z j + ...)(a0 z + ai()-1 z i() + ...)im = u (u)+ cj z j + (xim )~im z im + (xim )~0 -1 ai()-1 z j mod п п a0 п aim п (п (xim )+ ( f (u) = f (п (u))+f (c
(m+1)i

j +1 (m+1)i

z

(m+1)i

)+... = (u)+xim п

(п (u))z im +...+f (c u

(m+1)i

z

)+...

+ f (cj z j ) mod Since x
mi

j +1

(1.11)

= y1 m y2 (xmi )0 , the equation at z ~- ~ (xmi )~ п a
im 0

mi

has the form (1.12)
q/i

- xmi (

(п (u))) = 0 u

п y п y- We show that all the coefficients cq in (1.11) can be chosen so that (~1 (~2 1 )cq ) > i(п ) - 1. In order to do that if q < j , we prove, applying induction on q/i, that all the coefficients at z in degrees higher than im in f (п (u)), f (cli z li ) satisfy this property, supposing that cli satisfies this property at l < q /i. For f (ul ), l > 1 we have by Newton's binomial formula that
l

f (u ) = u +
k=1

l

l

u

l -k k xim z imk

Clk ,

whence mk (ul-k xk y1 (~2 1 )) = l -k +(k +1) (~2 ) = l -k +(k -1)i(п ) = l -k +(k -1)(i(п )-1) > п п y- пy im ~ (i(п ) - 1) for k > 1, what proves our assertion for f (п (u)). For f (cli z li ) we have f (cli z li ) = f (cli )z li , and, using Newton's binomial formula again, we get 87

l > i(п ) - 1 - (~1 )+ (~2 ) = i(п ) - 1 - (~1 )+ i(п ), where from l - 1+(k - 1)(i(п ) - 1) > п yl п y п yl l i(п ) - 1 - (~1 ) for all k , what proves our assertion in this case also. пy At z j we have the equation a (xim )~ п
im-1 ai()-1 0

+ cj = cj ,

(1.13)

and we must only solve the equation ~ п y- y1 (~2 1 cj + y1 (~2 1 )a ~ п y-
j/i j/i

~im-1 i()-1 a0

(xim ) = 0 mod п

i(п п

)

(1.14)

in order to finish the induction step for the coefficients cq . We have: ~ п y- y1 (~2 1 )cj + y1 (~2 1 )a ~ п y-
j/i j/i j/i j/i

~im-1 i()-1 a0

(xim ) = п
i()-1

y1 (~2 1 cj + y1 (~2 1 )п (~1 m )п (~2 )(xmi )0 a ~ п y- ~ п y- y- y ~ y1 (~2 1 )cj + y1 ~ п y-
j/i (i()-1)/i (i()-1)/i

=
)

a

i()-1 j/i

(xmi )0

mod

i(п п

Since (~1 пy ai()-1 ) = i(п ) - 1 = (~1 (~2 1 cj ), there exists a unique constant п y п y- (xmi )0 , with which the equation (1.14) is solvable. Let us show that the coefficients cq , q > j , aq , q 1 satisfy the properties (1.8). As for coefficients cq , it is remained to prove, that the coefficients at z d in (xim )(z im ) for d > j satisfy (1.8) . It's clear that (1.8) remains true if i |q . But if i|q , then (z im ) = z im D, where D is a series with coefficients of the same behaviour as aq . It follows from the Newton's binomial formula. Applying the same arguments as for f (cli )z li , we get that (1.8) holds for (xim )z im , where from (1.8) also holds for the product (xim )z im D, because (1.8) holds for each series (xim )z im dq z q , where D = q0,i|q dq z q . For the coefficients aq we have f (z ) = (z ) f (z ) = f (~0 )z + f (a a
i()-1

)z

i()

+ ...

(1.15)

where from, using calculations for f (ul ), we get that (1.8) holds for aq . Therefore, q/i since a0 = 1 mod i(п )-1 , we have (aq y1 a-1 ) > i(п ) - 1 for all q < i() - 1. ~ п п ~ ~0 To complete the induction, let's show that = f -1 f , where the coefficients cq ,aq of the automorphism satisfy (1.8) and cq = 0, 1 q j , aq = 0, 1 q < i() - 1. The proof is again by induction on m (m ). Let's use the calculations before the formulas (1.9) and (1.10). The equation (1.9) is solvable with cm-1 = 0, (m-1)/i (cm-1 y1 п ~ (~2 1 )) > i(п ) - 1, m - 1 < i() - 1, and to complete the proof we п y- 88

only have to check that the properties of the coefficients cq ,aq remain true by the -1 conjugation fm-1 fm-1 . But this follows from the same arguments as in the case, (m-1)/i -1 п ~ y2 ) = 0, because we used ~ when fm-1 (u) = u + xm-1 z m-1 , fm-1 (z ) = z , (xm-1 y1 (m-1)/i -1 ~ y2 ) 0, which is true also in our case, because ~ only the inequation (xm-1 y1 п xm-1 is a solution of an equation of the type (1.9). In the same way one can prove this (m-1)/i п ~ ) 0. From the fact in the case fm-1 (u) = u, fm-1 (z ) = z + ym-1 z m , (ym-1 y1 other hand side, one can see from the equation (1.9) that the conjugation fm-1 (u) = u, fm-1 (z ) = z + ym-1 z m does not change the coefficient cm-1 ,soany conjugation fm-1 can be decomposed into composition of two conjugations fm-1 ,fm-1 such that fm-1 (z ) = z , fm-1 (u) = u. Thus, we have proved that is conjugate to , where j > j , i = i(). Since our arguments do not depend on j , we get the required assertion by induction. In the same way with Proposition 1.2 it is proved now that = f -1 f , where (i()-1)/i п ~ )= (u) = (u), (z ) = a0 z + ai()-1 z i() + a2(i()-1) z 2i()-1 , where (ai()-1 y1 п ~ 2(i()-1)/i i(п ) - 1, (a2(i()-1) y1 п ~ ) = i(п ) - 1 (i.e it is the case b) of the theorem). Case 2). This case is divided into two ones: a) j = i() - 1, b) j < i() - 1. Let us look first at the case a). We show that = f f -1 , where is defined in the case a) of the theorem. -1 To do that we make sequential substitutions = fmi fmi , fmi (u) = u + ~- ~ ~- xmi z mi , fmi (z ) = z + ymi z mi+1 , where xmi = y1 m y2 (xmi )0 , ymi = y1 m (ymi )0 . It is enough to show that for every m a corresponding automorphism has coefficients cq ,aq , which satisfy the property: q/i q/i пy (~1 (~2 1 )cq ) > i - 1, (~1 aq ) > i(п ) - 1, i|q , im q mi + j , q = j, 2j , п y п y- п cq = aq = 0, q < im, because then can be reduced to the case, when the appropriate coefficients cq ,aq are equal to zero. That is done using the same substitutions, as by deriving equations (1.9), (1.10), and with the help of result from the case 1). Since for every m the number of necessary conjugations is finite, the desired automorphism f : = f f -1 exists. Let us write down the calculations for an arbitrary m: f (z ) = (z )+(ymi )(z
mi+1

) = a0 z +B1 u ~ + ( п

i(п )-1 i()

z

+B2 u

i(п )-1 2i()-1

z

+a

mi+i()-1

z

mi+i()

+

(ymi )~ п a

mi+1 mi+1 z 0 i(п )-1

ymi )Au u
mi+i()

i(п )-1 im+1+j

z

+

(mi +1)ami B1 u ~0 (~0 )xmi z a u

(ymi )z п
im+1

mod

mi+i()+1

f (z ) = f (~0 )f (z )+ f (aim )f (z a a0 z + ~
mi+1

)+ ... + f (a +...+a

im+i()-1

f (z

im+i()

)=
i()-1

+~0 ymi z a

mi+1

+aim z

im+1

im+i()-1

z

im+i()

+i()ymi a

z

im+i()

+

89

)xmi z im+i() mod mi+i()+1 (1.16) (a u i()-1 Because of the special form of yim , (ymi )~im+1 = a0 ymi . The coefficients aq , q < п a0 ~ im + i() - 1 can be chosen so that they have the pointed properties, in the same way, as in the case 1). For q = im + i() - 1 it is necessary to show, that there exists (ymi )0 : y1 ~
m+(i()-1)/i

(a

mi+i()-1

+п (

(ymi ))Au u (a u
i()-1

i(п )-1

+(mi+1)ami B1 u ~0
)

i -1 п

(ymi )-i()ymi a п

i()-1

-

)xmi ) = 0 mod i(п п

(1.17)

Since m+(i()-1)/i m+(i()-1)/i (~1 пy A( u ymi )ui(п )-1 ) > i(п ) - 1, (~1 п пy ami+i()-1 ) i(п ) - 1, m+(i()-1)/i m+(i()-1)/i i(п )-1 (~1 пy B1 (ymi )u п ) = i(п ) - 1, ( u (ai()-1 )xmi y1 п ~ ) i(п ) - 1 - 1+ i(п ) > i(п ) - 1, the element (ymi )0 exists and is defined uniquely if (im +1) = i(), i.e. q = 2j . Further,
f (u) = (u)+ (xmi )(z mi ) = (u)+ Aui(п п )- 1 j

z +c

mi+i()-1

z

mi+j

+ (xmi )~mi z mi + п a0
mi+i()

( п

xmi )Au u

i(п )-1 mi+j

z

+ mia ~

mi-1 0

B1 u

i -1 п

(xmi )z п
im+j

mi+i()-1

mod

f (u) = f (п (u))+f (cim )f (z im )+...+f (c +c
im+j

im+j

)f (z

) = (u)+ п

(п (u))xim z im +cim z im +... u (1.18)

(cj )xmi z im+j + jymi cj z mi+j , u whence we get similarly that we must solve an equation over (xmi )0 : z
im+j

+

y1 ~

m+j/i -1 y2 ~

(c

mi+j

+ mia ~

mi-1 0

B1 u

i -1 п

(xmi )- п

п п (1.19) (cj )xmi - jymi cj + ( xmi )Aui(п )-1 ) = 0 mod i(п ) u u Since (ymi )0 was already defined (if mi = j , we can take (ymi )0 equal to a constant), m+j/i -1 (~1 пy y2 ( u xmi )Aui(п )-1 > i(п ) - 1, ~ п m+j/i -1 m+j/i (~1 пy y2 u (cj )xmi ) > i(п ) - 1, (~1 ~ пy y2 B1 ui(п )-1 (xmi )) = i(п ) - 1, ~ п so an element (xmi )0 does exist. Let us now examine the case b). Now by the similar arguments as in a), we get that is conjugate to , (u) = (u)+ A1 z j + A2 z j +1 + ... + Ai()-1-j z i()-1 , п q/i q/i (z ) = a0 z + B1 z i() + b2 z 2i()-1 , (Aq y1 (~2 1 )) = i(п ) - 1 or Aq = 0, (Bq a-1 y1 ) = ~ п ~ п y- п ~0 ~ i(п ) - 1 if i() is finite, and 90

(u) = (u)+ Az j , п (z ) = a0 z if i() = (see cases c) and d) correspondingly). ~ In fact, let us use formulas 1.16 and (1.18). Since m+j/i+q -1 п y ) > i(п ) - 1, the arguments from deriving the formula (1.17) (Aq ( u ymi )~1 п remain true for all coefficients aq , q i + i() - 1, q = 2i() - 1, and the property from m+j/i+q -1 -1 ~ y2 ) > i(п ) - 1, we can ~ the case a) is realized. Similarly, since (п ( u xmi )Aq y1 п apply formula (1.19) for the coefficients cq , q i + i() - 1 and get the desired result. ~ Remark. In the case b) of theorem, if a0 = 1 or a0 = 1 but y = 0, where y ~ is a second parameter of the canonical representation of , one can show by direct п ~ calculations that is conjugate with : (u) = (u) + Az j , (z ) = a0 z + Bz i() , п where A satisfies (1.8) and B does not. But, if a0 = 1 and y = 0, then for any k 1 ~ (B )~2 Im(п - Id), where B = cy2 1 u1+k(1-i(п )) , whence, by formulas (1.17) and y ~- u (1.19), one can derive that does not exist and the number of parameters can not be decreased. Remark. 1. In the case of characteristic p > 0 we have in general dim(kerT ) = dim(cok erT ), as it was shown in lemma 1.3. From this follows that automorphisms can not be parameterised by finite number of parameters in more cases than in the case of chark = 0. For example, can not be always redused to , where (u) = (u)+ A1 z j + ... + Ak uj +k : k may be equal to the infinity. п 2. The classification can be easily generalised to the case of n-dimensional local field, because we used only the property dim(kerT ) = dim(cok erT ) and arguments with valuations. In the case of multidimensional equal characteristics local fields of characteristic 0 all our arguments can be carried over to the case of higher dimension if we assume that the value group of is Z ... Z. п Now we only have to prove that the automorphisms , are conjugate if and only if = , where , are automorphisms from the formulation of theorem. It's clear п п ~ ~ п that if is conjugate with , then a0 = a 0 and (u) = (u) = (u) is a nesessary condition, whence is defined up to the change u x0 : (x0 ) has the canonical view п and z cz , c k . Then, and must have the same numbers j, j and i(),i . Indeed, if and are conjugate, then = f -1 f , and f can be decomposed in a composition of automorphisms f = f1 f2 ...fm , where fq (u) = u + xq z q , fq1 (z ) = z + yq1 z q+1 . Then from (1.9), (1.10) follows that for q < min{j, j } we have xq kerTq,1 , for q1 < min{i() - 1,i - 1} we have yq1 kerTq,2 . From the proof of the case a) follows that the conjugations fq with this numbers preserve properties (1.8) of the coefficients cq ,aq1 for q min{j, j }, q1 min{i() - 1,i - 1}. Therefore, if j = j or i() = i , then the first nonzero coefficient of (u) or (u) or (z ) or (z ) must lie in the kernel of the map Tj (j ),1 (Ti()(i ),2 ), but this contradicts to the choice of these coefficients. Therefore, j = j and i() = i . So, and are in the same class defined by the 91

pair (j, i()). In this case the equality follows from the special choice of coefficients of z j ,z i() and the proves of the corresponding cases.

1.3

Pro of of the theorem I I I

Let n = 1. Then, by Proposition 1.2, there exists x0 such that (x0 ) = x0 , where п п is a primitive root of 1. As in the theorem II, we consider that (u) = u + cj z j + ..., (z ) = a0 z + ai()-1 z i() + .... п At first we note that i = 1 or . Indeed, if ai = (y) , then, as it was shown in the 0 y theorem II, we may suppose a0 = 1 + cuqn + .... But in this case ai = 1 + icuqn + .... 0 Further, п2 п пn п ai (ai ) ... n-1 (ai ) = (y) ((yy)) ... -(y) ) = 1. So we have that: 0п 0 0 y п п n 1 (y ai (ai ) ... n-1 (ai ) = (1 + icuqn + ...) ... (1 + icuqn + ...) = 1 + nciuqn + ... = 1, where п 0п 0 0 from a0 = 1 or i = . Thus, (u) = u + cj uj + ..., (z ) = z + ai()-1 z i() + ... (in this case dim(kerTk,1 ) = = dim(kerTk,2 ) = dim(cok erTk,1,2 )). Further let us consider that cq uk ((un )), aq k ((un )), because by going over to conjugations as in (1.9) and (1.10), we can solve all the equations п (y ) - y = cq mod uk ((un )), (y ) - y = aq mod k ((un )) п (in theorem 2 we have reduced general case to a case aq ,cq cok erTq,1 ,Tq,2 in the same way). As in Proposition 1, it is proved that if i = , then takes place the case O) of the theorem. Let i = 1. The case j i() coincide with the case j i() of the theorem 2: by writing over the formula (1.11), we get that holds (1.12) and there from holds cq = 0, q < j , and the equation (1.13) always has a solution xm uk ((un )) , when aq k ((un )), cq ,cq uk ((un )). It holds from (1.15) that by conjugation the coefficients aq = 0, q < i() - 1. All other arguments from the theorem 2 should be applied here also, and in the same way as with the case O), we get the case O') I). Let now j +1 = i() and let f (z ) = y0 z , y0 k ((un )). Then we have: f (z ) = (y0 )(z ) = (y0 + (y0 )cj z j + ...)(z + a u
i()-1

z

i()

+ ...) (1.20)

f (z ) = f (z )+ f (a1 z 2 )+ ... = y0 z + j п So we get from here that aq = 0, q < j , f (aj )y0+1 equation u (y0 )cj + y0 ai()-1 = 0 isn't solvable, then п from (ai()-1 /cj ) < -1 or (ai()-1 /cj ) = -1, but п 92

2 f (a1 )y0 z 2 + ...

= u (y0 )cj + y0 ai()-1 . If the dlog (y0 ) = -ai()-1 /cj , where res(-ai()-1 /cj ) Z. If it is /

solvable, then we can consider that (ai()-1 /cj ) = -1, res(ai()-1 /cj ) ", setting п y 0 = u. The case j +1 < i() is reduced to a case j +1 = i(): indeed, by setting y0 = u, we get that aj = 0 , and it's the first non equal to zero coefficient in the decomposition п (z ), while (aj /cj ) = -1, res(aj /cj ) ". The next part of the proof is following the proof of the theorem 2: in order to prove the rest items of the theorem, we shall go over to conjugations, and as a result get formulas of the kind (1.17) and (1.19). However, we cannot completely repeat the arguments from the previous case, because the kernel and cokernel of corresponding maps Tk,1 and Tk,2 are infinite-dimensional in our case. The formulas (1.17) and (1.19) are now written down as (ym )A +(m +1 - i())Bym - (B )xm = a u u mB xm - (1.21)

(1.22) (A)xm - jym A + (xm )A = b u u where A = cj , B = ai()-1 , a and b -- arbitrary elements from uk ((un )) and k ((un )) correspondingly, xm uk ((un )), ym k ((un )). It turns out that the solutions of this system strongly depend on the properties of the numbers, defined before the formulation of the theorem 3. From now on in the proof we are going to investigate the solvability of this system in dependence from the behaviour of these numbers. First of all we note that we can put A = c1 uk , k 0, |k | < j , c1 k . Indeed, we write down the conjugation f -1 f , f (z ) = y0 z , y0 = uq , where q 0 is a minimal п positive integer such that qi (A), f (u) = x0 , (x0 ) = 1. Then п f (u) = (x0 ) = x0 + (x0 )cj z j + ... u

j п f (u) = f (u)+ f (cj z j )+ ... = x0 + f (cj )y0 z j + ...,

where from we get y
j (-k + 1)c1 y0 /cj = solvable. And here

п j f (c j ) k 0 cj = u (x0 ). We consider cj = A = c1 u , k (x-k+1 ). We can choose c1 k so, that the 0 u - - п also aj = f -1 ( u (y0 )y0 j -1 cj + y0 j aj ).

= - (y0 /cj ). Then пj equation would be

We show that in all the cases (except the cases 2) a), 2) b) i) of the theorem) such a conjugation could be found, that it holds A = c1 uk , B = c2 uk1 . By that it appears, that the coefficients A1 ,B1 in all the cases of the theorem have the form, as mentioned above. Let (B/A) = -1, res(B/A) = p1 /q1 ", (p1 ,q1 ) = 1. п 93

We note that q1 doesn't depend on conjugation. Indeed, A and B change only by п conjugation f -1 f , f (u) = x0 , (x0 ) = 1, f (z ) = y0 z . But then from the (1.20) follows, that res(B /A ) - res(B/A) Z, hence we get that q1 doesn't depend on conjugations. Let us now show that there is such a conjugation, that B/A = res(B/A)u-1 , if q1 |j or res(B/A) < 0. Therefore we look for a conjugation f , f (u) = x0 , f (z ) = y0 z , so that the automorphism = f -1 f would have A = c1 uk , k < 0, B = c2 uk-1 . For that, considering (1.20), we must solve a system j п f (A )y0 = (x0 )cj , u п f (B )y
j +1 0

=

(y0 )cj + y0 a u

j (y0 )+y0 B/A
u

п Dividing the first equation by the second, we get: f (B /A )y0 = from c2 x c1
-1 0

u

(x 0 )

, where

(x0 ) = u

u

(y0 ) + B /A, y0

j c1 y0 A-1 =

(x0 )x u

-k 0

= (1 - k )-

1

(x u

1-k 0

) (1.23)

We look for x0 ,y0 in a form of x0 = u(1 + 1 u + 2 u2 + ...), y0 = u (1 + 1 u + ...). Let c = c2 /c1 , B/A = cba u-1 + 0 + 1 u + .... Then from the first equation (1.23) we get: cu u
-1 -1

+ c(1 +22 u +33 u2 + ...)(1 + 1 u + 2 u2 + ...)
-1

-1

=

+(1 +22 u + ...)(1 + 1 u + ...)-1 +(cba u

+ 0 + 1 u + ...)

Suppose c = + cba = 0 (we can always find such 0). By comparing the coefficients in the left and right sides, we get linear equations of the form ci = i + i-1
i-1

+ i-1 i ,

where i -- certain polynomial from q , q , q < i (they are determined from the previous equations). From the second equation we get: c1 uj (1 + 1 u + ...)j c-1 u a
-k
a

=u

-k

+(2 - k )1 u

1-k

+(3 - k )(2 + ...)u

2-k

+ ...

(where A = ca uka ). Suppose c1 = ca , -k = j - ka . Then k 0. Because of cba = 0, we can put = 0. Hence k = ka > -j . Comparing the coefficients, we get linear equations of the form ~ ~ (1.24) ji = (i +1 - k )i + i = (i +1+ j - ka )i + i For every i the system has a solution, if (i +1 - ka )/j = cba = p1 /q1 , what holds true always under the condition that q1 |j or cba = res(B/A) < 0. If these conditions are not fulfilled, then B can have the form B = c2 uk-1 + cib uib , what is evident from the arguments, mentioned above. If res(B/A) ", then applying the same thoughts, we also get the same result. / 94

Let (B/A) < -1. п Then we look for B in the form B = c2 uk+п(B/A) . System (1.23) will now have the form (y0 ) (B/A) п j + B /A, c1 y0 A-1 = (1 - k )-1 (x1-k ) cx0 (x0 ) = u u y0 u 0 Hence cu u
-1 (B/A) п

+ cu

(B/A)+1 п

(1 +22 u + ...)(1 + 1 u + ...)-1 =
(B/A) п

+(1 +22 u + ...)(1 + 1 u + ...)-1 +(cba u
ba

+

(B/A)+1 п

u

(B/A)+1 п

+ ...)

whence c = c

and equation i looks like following: cii = i+п
(B/A)+1

(i + (B/A)+1) + п

i

where i+п(B/A)+1 = 0 if (i + (B/A) + 1) 0. Equations (1.24) are written over п without changes, where from we get that every system i is solvable, and our proposition is proved. Let us now go back to a system (1.21), (1.22). We show, that system of the equations (1.21), (1.22) is solvable, if m = j . It holds: ym = (mB /(jA) - Hence 2 (xm )+ (xm )((2m - j )B/A - (A)/A)+ xm ((m - j ) (B )B/(BA)- 2 u u u u 2 2 (2m - j ) (A)B/A - ( 2 (A) (A))/( (A)A)+ u u u u (A))2 /A2 +(m - j )mB 2 /A2 ) - (b/A) - (m - j )Bb/A2 - ja/A = 0 u q = (B/A). From this: q -1, if q = -1, then res(B/A) Z. п / us show that the equation (A)/(jA))xm + (xm )/j - b/(jA) u u (1.25)

u We set Let (

(1.26)

2 (xm )((2m - j )B/A - (A)/A)+ xm ((m - j ) (B )B/(BA)- (xm )+ 2 u u u u (2m - j ) ( 2 (A)B/A2 - ( 2 (A) (A))/( (A)A)+ u u u u
k

(A))2 /A2 +(m - j )mB 2 /A2 ) = cu u 95

mod k п

+1

(1.27)

is solvable, if q < -1 or q = -1, but res(B/A) ", for all k Z and every constant / c k . From here we immediately get the solvability of the equation (1.26) for all b and a, and also of a system (1.21), (1.22). From here will follow the proof of the items A), B) 1). If q < -1, then п ((2m - j )B/A - u (A)/A) = (B/A) (if 2m = j ) and -1 (if 2m = j ), п 2 ((m - j ) u (B )B/(BA) - (2m - j ) u (A)B/A2 - ( u2 (A) u (A))/( u (A)A) + п 2 2 2 2 2 2 п ( u (A)) /A +(m - j )mB /A ) = (B /A ), because m = j . Thus, ( п 2 (xm )((2m - j )B/A - (A)/A)+ xm ((m - j ) (B )B/(BA)- (xm )+ 2 u u u u

(2m - j )

2 (A)B/A2 - ( 2 (A) (A))/( (A)A)+( (A))2 /A2 +(m - j )mB 2 /A2 ) = u u u u u 2 ( 2 (xm )+ п п (xm )(2m - j )c1 uq + xm (m - j )mc2 u2q ) = (xm (m - j )mc2 u2q ) u u where from immediately follows solvability of the equation (1.27). If q = -1, п п п then we put qa = (A), qb = (B ), k = (xm ), x = res(B/A). And now for the solvability of the equation (1.27) is necessary to show that the equation k (k - 1) + k (2m - j )x - kqa +((m - j )qb - (2m - j )qa )x + qa +(m - j )mx2 = 0 (1.28)

doesn't have a solution. - a This quadratic equation has the critical points - k-qj , - km1 (and if m = j , then one m- - of the points is - km1 ), so if res(B/A) ", then our assertion is proved. Moreover, in the / case when q = -1, res(B/A) " we have proved the solvability of the equation (1.26), / and through that also of a system (1.21), (1.22) for all m, by this proving the case B) 1). If m = j , q < -1, qa = 0, then the equation (1.27) has the form 2 (xm )+ (xm )(mB /A - (A)/A)+ xm (-mB (A)/A2 - 2 u u u u 2 (A) (A))/( (A)A)+( (A))2 /A2 = cuk mod k+1 , п 2 u u u u is always solvable, because (xm B u (A)/A2 ) < (xm B/A) < (xm ). п п п qa = 0, then this equation isn't solvable with k = qa - 1 + q . Thus, < -1, is the conjugation to automorphism : (u) = u + Auj , п = z + Bz i() + cu (A)-1+q z 2i()-1 (see case A)). (

that If if q (z )

96

The case res(B/A) " should be studied precisely. Recall that in this case (B/A) = -1. п Let res(B/A) = p1 /q1 (= cb /ca ), (p1 ,q1 ) = 1. The following proof of the theorem would be divided into three cases (which do not coincide with the corresponding cases from the formulation of the theorem), in order to make the proof easier: a) q1 |j , q1 = j b) q1 |j c) q1 = j . a) (see the case B) 2) b) i)). - qa 1 2- Here - k11 -j = - km11 = p1 . Then there exist c1 ,c2 k such that the equation m q (1.26)+ c1 uk1 -2 + c2 uk2 -2 = 0 has solutions with m = m1 , what follows from the solvability of the equation (1.27) for all k , except of k = k1 - 2, k = k2 - 2, and m1 -- is the first index, when the system (1.21), (1.22) isn't solveable in a general case. Also in this case the space of solutions of the homogeneous equation (1.26) is generated by x1 and п п x2 , (x1 ) = k1 , (x2 ) = k2 . Thus, automorphism is conjugate to the automorphism , (u) = u + Az j + cj +1+2m1 z j +1+2m1 + ..., (z ) = z + Bz j +1 + B2 z j +1+m1 + ..., where B2 = c1 uk1 -2+qa + c2 uk2 -2+qa . Now let us investigate behaviour of the values k1,mq ,k2,mq for different mq , for which the equation (1.26) has no solutions, where k1,mq ,k2,mq are solutions of the equation (1.28). Obviously, mq = qq1 , q N. Note that (k1,mq - k2,mq ) doesn't depend on mq k ,m -q k -1 1 (q = j/q1 ). Indeed, 1mqq-j a = 2,mq = - p1 . Hence k2,mq = -p1 q + 1, k1,mq = mq q -p1 (q - j/q1 )+ qa , and (k1,mq - k2,mq ) = p1 j/q1 + qa - 1. We observe, that k2,mq = k2,mq-1 + k2,m1 - 1, k1,mq = k1,mq-1 + k2,m1 - 1 = k2,mq-1 + k1,m1 - 1. We write down the formula (13) for the case, when ym1 = 1 ym1 ,1 + 2 ym1 ,2 , xm1 = 1 xm1 ,1 + 2 xm1 ,2 , where 1 ,2 k , xm1 ,ym1 are solutions of the homogeneous system (1.21), (1.22) for m = m1 . Because of (xm1 ,1 ) = k1,m1 , (xm1 ,2 ) = k2,m1 , we have (ym1 ,1 (xm1 ,1 )) = п п п п п п k1,m1 -1, (ym1 ,2 (xm1 ,2 )) = k2,m1 -1. Indeed, from the formula (1.25), (ym ) = (xm )-1, п if mp1 /q1 - qa + k = 0, where k = (xm ). Let be mp1 /q1 - qa + k = 0. Then p1 /q1 = k1,m1 -qa - m1 , whence j = 0. It's a contradiction. Analogously for k2,m1 qa = 1, but qa 0, also a contradiction. So we have: f (z ) = z + Bz (y u
m1 i()

+ B2 z

i()+m1

+a z

2m1 +i()-1

z

2m1 +i()

+ ... + y
m1

m1

z

m1 +1

+

)Az

m1 +i()

+(m1 +1)By + Bz
i()

m1

m1 +i()

+(m1 +1)B2 y
m2 +i()

z

2m1 +i()

+ ...
m1 +i()

f (z ) = z + y

m1

z

m1 +1

+ B2 z

i()+m1

+ B3 z

+

(B )x u

m1

z

+

97

i()y

m1

Bz

m1 +i()

+2-

1

2 (B )x u2 +

2 2m1 +i() m1 z

+Ci2() y
m1 ym1

2 m1

Bz

2m1 +i()

+

(B2 )x u z

m1

z

i()+2m1

+

(i()+ m1 )y

m1

B2 z

2m1 +i()

(B )i()x u z
2m1 +j

z

2m1 +i()

+a +

2m1 +i()-1

2m1 +i()

+ ...

f (u) = u + Az j + c

2m1 +j m1

+ ... + x
m1

m1

z

m1

m1 Bx f (u) = u + xm1 z
m1

z

m1 +j

+ m1 B2 x

z

2m1 +j

(x u + ...

m1

)Az

m1 +j

+

2 (A)xm1 z m1 +j + jym1 Az m1 +j +2-1 2 (A)x2 1 z 2m1 +j + m u u 22 Cj ym1 Az 2m1 +j + (A)jxm1 ym1 z 2m1 +j + c2m1 +j z 2m1 +j + ... u As m1 < j , in the expression for f (z ), f (u) there is the only term at z 2m1 +i() . For ymq = 1 ymq ,1 + 2 ymq ,2 , xmq = 1 xmq ,1 + 2 xmq ,2 the formula 1.16 should have the form + Az j + f (z ) = z +Bz (y u
mq i()

+B2 z

i()+m1

+B3 z

i()+m2

+a

mq +m1 +i()-1

z

mq +m1 +i()

+...+y

mQ

z

mq +1

+

)Az

mq +i()

+(mq +1)By

mq

z

mq +i()

+(mq +1)B2 y + ...

mq

z

mq +m1 +i()

+

(mq +1)B3 y f (z ) = z + y
mq

mq

z

mq +m2 +i()

z

mq +1

+ Bz

i()

+ B2 z

i()+m1

+ B3 z +

i()+2m1

+

(B )x u (B2 )x u
mq

mq

z

mq +i()

+ i()y B2 z

mq

Bz

mq +i()

z

i()+m1 +mq

+(i()+ m1 )y z
i()+m2 +mq

mq

m1 +mq +i()

+a

mq +m1 +i()-1

z

m1 +mq +i()

+

(B3 )x u

mq

+(i()+ m2 )y
mq +j +m1

mq

B3 z
mq

m2 +mq +i()

+ ...
mq

(1.29)
mq +j

f (u) = u + Az j + c

mq +j +m1 mq

z

+ ... + x
mq

z

mq

+

(x u

)Az

+

mq Bx f (u) = u + xmq z Whence follows: 2-1 ( 2-1 ( 2 (B )x u2
2 m1 mq

z

mq +j

+ mq B2 x
mq

z

mq +m1 +j

+ ... +c
mq +m1 +j

+ Az j +

(A)x u

z

mq +j

+ jy

mq

Az

mq +j

z

mq +m1 +j

+ ...

2 ) = Cqa -1 (cb u qa - 2

qa - 3

2 2 + ...)(1 x2 +21 2 x1 x2 + 2 x2 ) 1 2

2 (A)x u2

2 m1

2 ) = Cqa (ca u

2 2 + ...)(1 x2 +21 2 x1 x2 + 2 x2 ) 1 2

(1.30)

98

y

2 m1

B = (cb u

qa - 1

22 22 + ...)((1 y1 +21 2 y1 y2 + 2 y2 )

y

2 m1

22 22 A = (ca uqa + ...)((1 y1 +21 2 y1 y2 + 2 y2 )

2 2 (B )i()xm1 ym1 = (qa - 1)i()(cb uqa -2 + ...)(1 x1 y1 + 2 x2 y2 + 1 2 (x1 y2 + x2 y1 )) u 2 2 (A)jxm1 ym1 = qa j (ca uqa -1 + ...)(1 x1 y1 + 2 x2 y2 + 1 2 (x1 y2 + x2 y1 )) (1.31) u y
mq mq

B2 = 1 c1 y

mq ,1

u

k

1,mq

-2+q

a

+1 c2 y

mq ,1

u

k

2,mq

-2+q

a

+c1 2 y

mq ,2

u

k

1,mq

-2+q

a

+c2 2 y

mq ,2

u

k

2,mq

-2+q

a a

x (

B2 = 1 c1 x
mq

mq ,1

u

k

1,mq

-2+q

a

+1 c2 x

mq ,1

u

k

2,mq

-2+q

a

+c1 2 x

mq ,2

u

k

1,mq

-2+q

a

+c2 2 x u
k
2,mq

mq ,2

uk 2 (1.32)
a

,mq

-2+q

(B2 ))x u

= (k1,mq -2+qa )1 c1 x
mq ,2

mq ,1
a

u

k

1,mq

-3+q

a

+(k2,mq -2+qa )1 c2 x
mq ,2

mq ,1

-3+q

+

(k1,mq - 2+ qa )c1 2 x

u

k

1,mq

-3+q

+(k2,mq - 2+ qa )c2 2 x

u

k

2,mq

-3+q

a

(1.33)

Let (k1,mq - k2,mq ) < 0. We shall show, that in formulas (1.30)- (1.33) monomials with valuation (k1,mq + k1,m1 + qa - 3), belong to the image of the map (1.27), it means that the equation (1.27) with the right side in a form of these monomials is solvable. Indeed, in a case of (k1,mq - k2,mq ) < 0wehave p1 /q1 < (-qa +1)/j < (-qa +1+ i)/j , if i 1. But then A = c1 uqa , B = c2 uqb = c2 uqa -1 , and ymq = 1 uk1,mq -1 + 2 uk2,mq -1 , xmq = 1 uk1,mq + 2 uk2,mq , because all the coefficients of the homogeneous system (1.21), (1.22) have the monomial form. Since (k1,mq + k1,m1 - 1) < k1,mq + k2,mq - 1 = k1,mq+1 < k2,mq+1 , the equation (1.27) has monomial solutions of a form, mentioned above. If (k1,mq - k2,mq ) > 0, so (k1,mq + k1,m1 - 1) > k1,mq + k2,m1 - 1 = k1,mq+1 > k2,mq+1 , where from follows the same result. If (k1,mq - k2,mq ) = 0, then ymq and xmq consist of the only monomial, i.e. 1 = 0 and expressions in (1.30)-(1.33) are simplified to the one monomial, which is in the general case not in the image of the map (1.27). Now we show that for all q except q = 1, q = j/q1 , q = (1 - qa )/p1 +1 there exist the coefficients 1 ,2 (coefficient 2 , if k1,mq = k2,mq ) are such that (c )/j - cm1 +mq +j u (A)/(jA) + (mq+1 - amq +m1 +i()-1 + u mq +m1 +j j )p1 u-1 cmq +m1 +j /(jq1 )+(1 - i())B2 ymq - u (B2 )xmq belongs to the image of the map (1.27), i.e. the equation (1.27) with the right side in the form of these expressions is solvable. According to (1.32), (1.33), we need to show that (1-i()+m
q -1

)c2 2 y

mq ,2

u

k2 -2+q

a

-(k2,m1 +qa -2)c2 2 x

mq ,2

u

k2 -3+q

a

+

(b)/j +b (A)/(jA) u u

+(m

q +1

- j )p1 u-1 b/(jq1 ) = 0 mod k п

2,mq +1

+ qa - 1

99

where b = mq c2 2 x (1-i()+m
q -1

mq ,2

u

k

2,m1

-2+q

a

, only if (q - 1)q1 (1 - qa - (q - 1)p1 ) = 0 and
mq ,1

)1 c2 y

mq ,1

u

k2 -2+q

a

-(k2,m1 -2+qa )c2 1 x

u

k2 -3+q

a

+

(b)/j +b (A)/(jA) u u

+(m

q +1

- j )p1 u-1 b/(jq1 ) = 0 mod k п

1,mq +1

+ qa - 1

where b = mq c2 1 xmq ,1 uk2,m1 -2+qa only if -(q - 1)2 q1 p1 + (q - 1)(qa - 1)q1 + (1 - j/q1 )(jp1 + q1 (qa - 1)) = 0 (we remark that 2 does not depend on 1 ). p -q +k Since ymq ,1,2 = 1 a j 1,2,mq xmq ,1,2 u-1 + ..., it is necessary to show that -(1 - i()+ m
q -1

)

mq (k2,mq+1 - 1+ qa ) p1 - qa + k2,mq - (k2,m1 - 2+ qa )+ - j j mq qa (m + j
q +1

- j )p1 q =0 j

if (q - 1)q1 (1 - qa - (q - 1)p1 ) = 0, -(1 - i()+ m
q -1

)

mq (k1,mq+1 - 1+ qa ) p1 - qa + k1,mq - (k2,m1 - 2+ qa )+ - j j mq qa (m + j
q +1

- j )p1 q =0 j

if -(q - 1)2 q1 p1 +(q - 1)(qa - 1)q1 +(1 - j/q1 )(jp1 + q1 (qa - 1)) = 0. But -(1-i()+m
q -1

)

mq (k2,mq+1 - 1+ qa ) mq qa (m p1 - qa + k2,mq -(k2,m1 -2+qa )+ - + j j j = (q - 1)q1 (1 - qa - (q - 1)p1 ) , j

q +1

- j )p1 q j

-(1-i()+m =

q -1

)

mq (k1,mq+1 - 1+ qa ) mq qa (m p1 - qa + k1,mq -(k2,m1 -1+qa )+ - + j j j

q +1

- j )p1 q j

-(q - 1)2 q1 p1 +(q - 1)(qa - 1)q1 +(1 - j/q1 )(jp1 + q1 (qa - 1)) j

We observe here, that -(q - 1)2 q1 p1 +(q - 1)(qa - 1)q1 +(1 - j/q1 )(jp1 + q1 (qa - 1)) = 0. In fact, if -(q - 1)2 q1 p1 +(q - 1)(qa - 1)q1 +(1 - j/q1 )(jp1 + q1 (qa - 1)) has solutions 2 in integers, then its discriminant must be equal to q1 l2 , where l Z. But D = (qa - 2 1)2 q1 +4(qa - 1)q1 p1 (q1 - j )+4p2 j (q1 - j ), whence follows, that j (q1 - j ) = (q1 - j )2 , 1 what is wrong. So, we have shown that is conjugated to : (u) = u + Az j , 100

(z ) = z + Bz i() + B2 z i()+q1 + B3 z i()+2q1 + Bj/q1 z i()+(1+j/q1 )q1 + Bq2 z i()+q1 (1+q2 ) , k -2+qa where Bj/q1 = cu 1,mj/q1 +1 , Bq2 = cuk2,m1+q2 -2+qa if q1 (1 - qa - (q - 1)p1 ) = 0 and ~ Bq2 = 0 otherwise. Let's show now, that B3 can be taken as cb3 uk2,m2 -2+qa . In order to do that, we 2 exhibit that in formulas (1.30)-(1.33) monomials with 2 belong to the image of the map (1.27). Then the case q = 1 is equivalent to a general case, and since q = 1 is one of the solutions of (q - 1)q1 (1 - qa - (q - 1)p1 ) = 0, B3 is defined in the same way as Bq . For that, according to (1.30)-(1.33), we must show that (b)/j - b (A)/(jA)+(m u u
q +1

- j )p1 u-1 b/(jq1 )+ a = 0

222 2 2 2 where b = Cj 2 y2,m1 cb q1 /p1 u2k2,m1 -2+qa + Cqa cb q1 /p1 2 u2k2,m1 -2+qa + qa jcb q1 /p1 2 (p1 - 2 qa + k2,m1 )/j , a = Cqa -1 + Ci2() (p1 - qa + k2,m1 )2 /j 2 +(qa - 1)i()(p1 - qa + k2,m1 )/j . In fact, 2 Cj

(p1 - qa + k2,m1 )2 q1 2 q1 (2k2,m1 - 2+ qa ) (2k2,m1 - 2+ qa )+ Cqa + j 3 p1 p1 j

qa j qa j

q1 (p1 - qa + k2,m1 ) qa 2 (p1 - qa + k2,m1 )2 q1 2 q1 (2k2,m1 - 2+ qa ) - (Cj + Cqa + 2 2p p1 j j j1 p1

q1 (p1 - qa + k2,m1 ) m2 - j 2 (p1 - qa + k2,m1 )2 p1 - qa + k2,m1 2 2 +Cqa +qa j )+ (Cj )+Cqa -1 + 2 p1 j j j j Ci2() (p1 - qa + k2,m1 )2 /j 2 +(qa - 1)i()(p1 - qa + k2,m1 )/j = 0,

and it proves our assumption. The case k1,mq = k2,mq is more simple, and all the arguments remain true. b) - 1 1 1 In this case the system - km-qja = p1 , - k2m 1 = p1 is incompatible, that is why for - q q all m the "cokernel" of the map (1.27) is one-dimensional, A = c1 uqa , B = c2 uqa -1 . 1 Let denote as k1,nq the solution of the equation - k1q-qja = p1 , and as k2,mq the n- q 1 2- solution of the equation - kmq 1 = p1 . It is clear that mq = qm1 = qq1 , as in the case a), q and nq+1 = nq +m1 only if nq +m1 = j . But in this case the next value of nq+1 is nq +2m1 , so we consider this recurrence relation to be true always. Further, (k1,nq - k2,mq ) doesn't depend on q , as in a), and k2,mq+1 = k2,mq + k2,m1 - 1, k1,nq+1 = k1,nq + k2,m1 - 1. The proof, that follows, would be divided into three cases: 1) q1 < j (see case B) 2) b) ii)), 2) q1 > j , j |q1 (see case B) 2) c) i)), 3) j |q1 (see case B) 2) c) ii)). We put B2 = c1 uk1,n1 -2+qa , B3 = c2 uk2,m1 -2+qa , if n1 < m1 , and B2 = c1 uk2,m1 -2+qa , B3 = c2 uk1,n1 -2+qa otherwise. 101

In the case 1) n1 ,m1 < j , n1 < m1 . The idea of the following proof is the following: we look for the sequentional conjugations fnq and fmq , where fnq (u) = u + xnq z nq , fnq (z ) = z + ynq z nq , and xnq ,ynq are solutions of the homogeneous system (1.21), (1.22) with m = nq ; fmq (u) = u + xmq z mq , fmq (z ) = z + ymq z mq , where xmq ,ymq are solutions of the homogeneous system (1.21), (1.22) with m = mq . We choose xnq ,ynq ,xmq ,ymq so that with these conjugations the equation (1.26) become solvable with m = nq+1 , m = mq+1 . For the automorphism fmq we can use the results from the case a), because xmq and ymq here have the form 2 xmq ,2 , 2 ymq ,2 , (ymq ,2 ) = k2,mq - 1, (xmq ,2 ) = k2,mq , п п and m1 = q1 < j , j = mq . We rewrite for fnq the formula (1.29): if q > 1, then fnq (u) = u + Az j + c nq Bxnq z
nq +j nq +j +m1

z

nq +j +m1

+ ... + xnq z + nq B3 xnq z

nq

+

(xn )Az u q + ...
nq +j

nq +j

+

+ nq B2 xnq z
nq

nq +n1 +j

nq +m1 +j

fnq (u) = u + xnq z c
nq +n1 +j

+ Az j +

z

nq +n1 +j

(A)xnq z nq +j + jynq Az u + cnq +m1 +j z nq +m1 +j + ...
nq +m1 +j

+

(since nq > m1 , there are no more terms with z fnq (z ) = z +Bz (yn )Az u q
nq +i() i()

), z
nq +m1 +i()

+B2 z

i()+n1

+B3 z

i()+m1

+a

nq +m1 +i()-1

+...+ynq z

nq +1

+

+(nq +1)Bynq z
nq +1

nq +i()

+(nq +1)B2 ynq z +B3 z

nq +n1 +i()

+(nq +1)B3 ynq z

nq +m1 +i()

fnq (z ) = z +ynq z (B2 )xnq z u

+Bz

i()

+B2 z

i()+m1

i()+n1

+

i()+n1 +nq

+(i()+ n1 )ynq B2 z +a
nq +n1 +i()-1

n1 +nq +i()

(B )xnq z nq +i() +i()ynq Bz u + (B3 )xnq z i()+m1 +nq + u
nq +m1 +i()-1

nq +i()

+

(i()+ m1 )ynq B3 z

m1 +nq +i()

z

nq +n1 +i()

+a

z

nq +m1 +i()

+ ...

The formula remains true for q = 1 also, as it is seen from the calculations, similar to q = 1 in case a). If n1 |m1 and k1,nq + (m1 /n1 - 1)k1,n1 = k2,mq , i.e. m1 /n1 k1,n1 = k2,m1 , then the coefficient anq +m1 +i()-1 (cnq +m1 +i()-1 ) depends on anq +n1 +i()-1 (cnq +n1 +i()-1 ) in a general case. In this situation for almost all q the conjugation fnq can be chosen so that the equation (1.26) is solvable for m = mq , and fmq so that equation (1.26) is solvable for m = nq+1 . Thus, the arguments, similar to the case a), tell us that is conjugated to : (u) = u + Az j , (z ) = z + Bz i() + B2 z i()+n1 + Bq z i()+mq + Bq2 z i()+nq2 , 102

where Bq = cu from

k

2,mq

-2+q

a

or equals to zero. It depends on that, if at least one expression

(b)/j - b (A)/(jA)+(mq+1 - j )p1 u-1 b/(jq1 )+ a u u equals to zero or not, with b = uk1,nq +wk1,n1 -2+qa nq , a = (k1,n1 -2+qa )uk1,nq +wk1,n1 -3+qa + (1 - i()+ mq-1 )uk1,nq +wk1,n1 -3+qa (p1 - qa + k1,nq )/j , w has the values from 1 to q1 /n1 - 1, in other words, if equals to zero at least one of expressions -(1 + w)n2 p1 + n1 jw - (q - 1)(2 + w)q1 n1 + q1 (-2+(1 + w)qa )+ 1

q1 [p1 (j (q - 1)(w +2)+ j - (q - 1)2 q1 +(q - 1)q1 +2q1 )+2j (qa - 1) + (q - 1)q1 ((1 + w)qa - 2)] (1.34) k1,nq2 -2+qa or zero, in accordance with equality to zero of the expresFurther, Bq2 = cu sion (b)/j - b (A)/(jA)+(nq2 - j )p1 u-1 b/(jq1 )+ a u u with b = nq2 -1 uk2,mq2 -1 +k1,n1 -2+qa , a = (k1,n1 - 2+ qa )u k +k -3+qa (p1 - qa + k2,mq2 -1 )/j , i.e. mq2 -2 )u 2,mq2 -1 1,n1
k
2,mq -1 2

+k

1,n1

-3+q

a

+(1 - i()+

2 - (q2 - 2)2 q1 p1 +(q2 - 2)q1 jp1 + p1 j (j - n1 )+(3j + n1 )q1 (qa - 1)

(1.35)

We note, that this equation doesn't have solutions in integers, i.e. Bq2 = 0. Really, 2 2 2 its discriminant must be equal to q1 p2 l2 , l Z. But D = q1 j 2 p2 + 4q1 p2 j (j - n1 ) + 1 1 1 3 22 2 2 4q1 p1 (qa - 1)(3j + n1 ), hence 0 < p1 q1 (j - n1 ) = q1 p1 (qa - 1)(3j + n1 ) < 0, a contradiction. Thus we have proved the case B) 2) b) ii) i'). If n1 |m1 or (m1 /n1 )k1,n1 = k2,m1 , then the solvability of the equation (1.26) for m = nq+1 , a = anq +m1 +i()-1 , b = cnq +m1 +i()-1 doesn't depend on coefficients anq +n1 +i()-1 (cnq +n1 +i()-1 ). In this case for almost all q fnq can be chosen so that the equation (1.26) is solvable for m = nq+1 , and fmq so that equation (1.26) is solvable for m = mq+1 . Not very complicated modification of all arguments, mentioned before, leads us to the conclusion, that is conjugated to : (u) = u + Az j , (z ) = z + Bz i() + B2 z i()+n1 + B3 z i()+m1 + B4 z i()+2m1 + Bq z i()+m1 + Bq2 z i()+nq2 , where Bq = cuk2,mq -2+qa or equals to zero in accordance with equality to zero of the expression q1 (1 - qa - (q - 1)p1 ), B4 = Bq for q = 1, Bq2 = cuk1,nq2 -2+qa or equals to zero in accordance with equality to zero of the expression (b)/j - b (A)/(jA)+(nq2 - j )p1 u-1 b/(jq1 )+ a = 0 u u 103

with b = nq2 -1 uk1,nq2 -1 +k2,m1 -2+qa , a = (k2,m1 - 2+ qa )u k +k -3+qa (p1 - qa + k1,nq2 -1 )/j , i.e. nq2 -2 )u 1,nq2 -1 2,m1

k

1,nq -1 2

+k

2,m1

-3+q

a

+(1 - i()+

-n2 p1 -q1 (p1 (2j +q1 (q2 -3)2 )-(j +q1 (q2 -2))(qa -1))+n1 q1 (-1-2p1 (q2 -3)+qa ) (1.36) 1 This equation has no solutions in integers by the same reasons as (1.35), where from Bq2 = 0 (see case B) 2) b) ii) ii') ). In the case 2) m1 < n1 , and since j |q1 , we can apply here the arguments from the case 1). Then the result would coincide with the result of the previous paragraph. In the case 3) m1 < n1 , but j |q1 , so we rewrite the formula (1.29) for the conjugation fmq , q 1 in the following way: fmq (u) = u + Az j + c mq Bx z
mq +j mq +m1 +j

z

mq +m1 +j

+ ... + x z

mq

z

mq

+

(x u

mq

)Az

mq +j

+ +

mq

+

1 2 (xmq )A2 z 2 u2 mq B2 xmq z mq

mq +2j +m1 +j mq

2 + Cmq B 2 x

mq

mq +2j

+ mq B + ...
mq +j

(x u

mq

)Az

mq +2j

+ mq B3 x
mq

mq

z

mq +n1 +j

fmq (u) = u + xmq z jy
mq

+ Az

+ Az j +
mq +m1 +

Az

mq +j

+c

nq +j

z

nq +j

+ ... + c +a

(A)x u mq +m1 jz z

z

mq +j

+

+ ... +...+y
+1 mq

fmq (z ) = z +Bz (y u
mq

i()

+B2 z

i()+m1

+B3 z z

i()+n1

mq +m1 +i()-1

mq +m1 +i()

z

mq +1

+ +

)Az

mq +i()

+(mq +1)By
mq

mq

mq +i()

1 2 + (y 2 u2

mq

)A2 z

mq +2i()-1

2 +Cmq

B2y

mq

z

mq +2i()-1

(mq +1)B

(y u

)Az

mq +2i()-1 mq +1

+ ... +(mq +1)B - 2y + Bz Bz
i()

mq

z

mq +m1 +i() i()+n1

+ ...

fmq (z ) = z + y (B )x u a
nq +i()-1 mq

mq

z

+ B2 z +

i()+m1

+ B3 z
mq

+ + + ...

z

mq +i()

+ i()y

mq

mq +i()

(B2 )x u

z

i()+m1 +mq

z

nq +i()

+ ... + a

mq +m1 +i()-1

z

mq +m1 +i()

+(i()+ m1 )y

mq

B2 z

m1 +mq +i()

Hence follows, that two cases a') the solvability of the coefficients cnq +j ,...,cmq +m1 , amq +m1 +i()-1 (and it is equal

are possible: equation (1.26) for m = mq + m1 doesn't depend on anq +j ,...,an1 +m1 , that is it depends only on cmq +m1 +j , to -p1 + qa - 2 = q1 (qa - 1)/j ). 104

b') the solvability of the equation (1.26) for m = mq + m1 depends on coefficients nq +j ,...,cmq +m1 , anq +j ,...,an1 +m1 , i.e. -p1 + qa - 2 = q1 (qa - 1)/j . In the case a'), repeating the proof as in the previous cases, we get the same result as in the case 2) (it corresponds to a case B) 2) c) ii) i')). In the case b') the conjugations fmq determine the solvability of the equation (1.26) for m = nq , and conjugations fnq -- for m = mq+1 . Here, in this case of m = nq , as seen from the formula above, can exist not more than two q -- solutions of the equation c (b)/j - b (A)/(jA)+(nq2 - j )p1 u-1 b/(jq1 )+ a = 0 u u
2 2 with b = 2-1 u2 (uk2,mq )u2qa + Cmq u2qa -2 p2 /q1 uk2,mq + mq p1 /q1 u2qa 1 2 2 2 a = 2-1 u2 (uk2,mq -1 )u2qa (p1 -qa +k2,mq )/j +Cmq +1 u2qa -2 p2 /q1 uk2,mq 1 (mq +1)p1 /q1 u2qa -1 u (uk2,mq -1 )(p1 - qa + k2,mq )/j , i.e.
2

-1 (uk2,mq ), u -1 (p1 -qa +k2,mq

)/j +

2 (p2 (-1+q -q1 (-1+q -2q 2 +qq1 ))-qq1 (qa -1)+p1 q1 (1-3q -(q -1)qq1 +qqa )) = 0 (1.37) 1

and in the case m = mq+1 not more than (m of the appropriate equation

q +1

- nq )/j +1 = q1 /j = w q -- solutions

(b)/j - b (A)/(jA)+(nq2 - j )p1 u-1 b/(jq1 )+ a = 0 u u

(1.38)

Thus, is conjugated to : (u) = u + Az j , (z ) = z + Bz i() + B2 z i()+m1 + Bqn ,1 z i()+nq1 + Bqn ,2 z i()+nq2 + Bqm ,1 z i()+mq1 + ... + Bqm ,w z i()+mqw , k -2+qa where Bqn ,i = ci uk1,nqi -2+qa , Bqm ,j = cj u 2,mqj or 0 depending on solvability of corresponding equations (it is the case B) 2) c) ii) ii')). In the case c), when q1 = j , we use arguments from both: the previous case and the case a), and get from there, that is conjugated to : (u) = u + Az j , (z ) = z + Bz i() + B2 z 2i()-1 + B3 z 3i()-2 + Bqm,1 ,1 z i()+q0 q1 + Bqm,1 ,2 z i()+q2 q1 + Bqm,1 ,3 z i()+q3 q1 + Bqm,2 ,1 z i()+q1 q1 + Bqm,2 ,2 z i()+q2 q1 + Bqm,2 ,3 z i()+q3 q1 , -1 where B2 = cb2 u-p1 -1+qa , B3 = cb3 u-p1 +2qa -2 , Bqm,1 ,i = cbqm ,1 ,i u-p1 q1 (qi q1 -j )+2qa -2 , Bqm,2 ,j = cbqm ,2 ,j u-p1 qj +qa -1 or 0 depending on solvability of corresponding equations (b)/j - b u (A)/(jA)+(nq2 - j )p1 u-1 b/(jq1 )+ a = 0. If we denote as b50 ,a50 b and u a in (1.37), then the appropriate b, a, b ,a for two equations are equal to b = b50 + c-2 mq u a
k
mq ,1

u

-p1 +1-2+q

a

105

a = a50 + c-2 (1 - i()+ mq - m1 )u a a = a50 + c-2 (1 - i()+ mq - m1 )u a

-p1 +qa -2+q k
mq ,2

a

u

k

mq ,1

-1

(p1 - qa + k1,mq )/j
a

(1.39)

b = b50 + c-2 mq u a

u

-p1 +1-2+q

-p1 +qa -2+q

a

u

k

mq ,2

-1

(p1 - qa + k2,mq )/j

(1.40)

where ca is a constant for the coefficient A, and the rest notations are taken from the case a). By direct calculations it is not difficult to show, that these equations are not solvable. The proof of the last statement in the theorem is similar to the proof of the statement in the theorem 2. The theorem is proved. 2

106

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