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The Classical and Adelic Point of View in Number Theory K. Conrad 1. a) Show JQ(i) = Q(i)в · (Cв в v= Z[i]в ) and Q(i)в (Cв в v {±1, ±i}. This is an analogue of the formula JQ = Qв · (Rв в Qв (Rв в p Zв ) = {±1}. p

10/13/2010 Problem Set 4
в v = Z[i]v ) = в p Zp ), where

b) Using the fact that Rв = {±1} в R>0 , the formula JQ = Qв · (Rв в p Zв ) p can be refined to JQ = Qв в (R>0 в p Zв ). Let's try to generalize this to p JQ(i) using the decomposition for JQ(i) in part a. Assuming there is a direct product decomposition Cв = {±1, ±i} в H for some subgroup H of Cв , show we can write JQ(i) = Q(i)в в G for some group G. Then show the assumption is wrong: Cв = {±1, ±i} в H for any subgroup H because Cв has no subgroup of index 4, and in fact Cв has no subgroup of any finite index greater than 1. 2. Let K be a number field. We want to show JK admits a direct product decomposition which is compatible with the idelic norm. Choose an archimedean absolute value v0 on K . Write down the indices (absolute values) of JK so the v0 -component comes first. Define an embedding f0 : R>0 JK , depending on whether the chosen v0 is real or complex, by the formula (t, 1, 1, 1, . . . ), if v is real, 0 f0 (t) := ( t, 1, 1, 1, . . . ), if v is complex. 0
1 Show the multiplication map M : R>0 в JK JK where M (t, x) = f0 (t)x is an isomorphism of topological groups and it fits into the commutative diagram below, where the maps along the top row are the standard inclusion and pro jection for subgroups of a direct product of groups and the maps along the bottom 1 row are the inclusion of JK into JK and the idelic norm on JK .

1

/

1 JK

/

R

>0 M

вJ


1 K

/

R
id.

>0

/

1
/

id.

1

/

1 JK



/ JK

||·||

/

R


>0

1


(Remark. Unless K has exactly one archimedean absolute value, i.e., unless 1 K is Q or an imaginary quadratic field, the isomorphism JK R>0 в JK is not = canonical since it depends on the choice of v0 .) 3. Let G be a locally compact Hausdorff group and µ be a Haar measure on G. a) If H is an open subgroup, show the restriction µ|H is a Haar measure on H . b) Give an example of G and a subgroup H which is not open such that µ|H is not a Haar measure on H . 4. A Borel measure µ on a topological space which satisfies the two conditions µ(A) = inf µ(U ), µ(A) = sup
cpt. K A

open U A

µ(K ),

for all Borel sets is called a regular measure. A Borel measure satisfying the first condition for all Borel sets A and the second condition when A is open or finite is called a -regular measure, so by definition a Haar measure is -regular. This exercise gives an example of a Haar measure which is not regular. Set G = S 1 в Rd , where Rd is the real line with the discrete topology. This is a direct product of two locally compact (abelian) groups, so G is locally compact with the product topology. Think about G as an infinitely long vertical cylinder whose horizontal slices are "topologically independent" circles. Let µ be a Haar measure on G and set A = {1} в [0, 1] (a vertical line segment of length 1). a) Show A is not open in G but it is closed. b) Show µ(K ) = 0 for all compact subsets K of A. c) Let U be any open set in G containing A. Therefore U (1 - y , 1 + y ) в {y },
0y 1

where 0 < y < 1 for all y . Show there's an > 0 such that y for infinitely many y . (Hint: Assume otherwise. There are uncountably many y in [0, 1].) d) By part c, U to show µ(U ) = regular. (Page 1 of measure is regular. (1 - , 1 + ) в {y } for infinitely many y [0, 1]. Use this and conclude from -regularity that µ(A) = so µ is not Rudin's book "Fourier Analysis on Groups" says any Haar This is wrong.)