field is discrete and cocompact in its adèles

field is discrete and cocompact in its adèles

For brevity, we write

P

f

for the set of

finite places

of

K

, and

P

∞

for the set of infinite places. We also write

∏

′

for a

restricted direct product

. Then

𝔸

K

=

∏

′

v

∈

P

f

′

⁢

K

v

×

∏

v

∈

P

∞

K

v

Theorem 1

.

K

is

discrete

as a subgroup of

A

K

.

Proof.

Since

𝔸

K

is a

topological ring

, it suffices to show that there is a

neighborhood

in

𝔸

K

meeting

K

in only

0

.

Let

U

=

∏

v

∈

P

f

𝔬

v

×

∏

v

∈

P

∞

B

⁢

(

0

,

1

2

)

Since

𝔬

v

is open in

K

v

for

v

finite, this is an

open set

. (Note that

𝔬

v

=

𝒪

K

v

, the

ring of algebraic integers

of

K

v

).

Now consider an

element

x

∈

U

∩

K

⊂

𝔸

K

. If

x

=

(

x

v

)

, then for

v

finite,

x

v

∈

𝔬

v

, and for

v

infinite

,

x

v

∈

B

⁢

(

0

,

1

2

)

. Assume

x

≠

0

. Then

|

x

v

|

v

≤

{

1

v

⁢

finite

1

2

<

1

v

⁢

infinite

but then

∏

|

x

|

v

=

∏

|

x

v

|

v

<

1

in

contradiction

to the

product formula

. Thus

x

=

0

and we are done.

The above

theorem

is very sensitive to the fact that all places are included in

𝔸

K

. For example, it is clear that the image of

𝔸

K

in

∏

v

∈

P

∞

K

v

is

dense

, since

K

v

is characterized by an

embedding

K

↪

K

v

≅

ℝ

,

ℂ

,

ℚ

p

. Then by an

argument

familiar from Minkowski’s theorem,

𝒪

K

is a full-rank

lattice

in the image of

K

. But

K

is the

ℚ

-span of that lattice, so is dense in

K

v

.

Furthermore, the same is true for the finite places:

Proposition 2

.

The image of

K

in

∏

′

v

∈

P

f

⁢

o

v

is dense.

Proof.

Suppose

x

=

(

x

v

)

v

∈

∏

′

v

∈

P

f

⁢

𝔬

v

. We show that

x

can be approximated as closely as desired by an

element of

K

by showing that for any ideal

I

⊂

𝒪

K

, there is

y

∈

K

such that

y

-

x

v

∈

I

⁢

𝔬

v

for each

v

∈

P

f

.

First multiply through by some

z

so that everything is in

𝒪

K

: choose

z

≠

0

such that

z

⁢

x

v

∈

𝔬

v

for all

v

∈

P

f

. This is possible since all but finitely many

x

v

are already in

𝔬

v

. Thus

y

-

x

v

∈

I

⁢

𝔬

v

is

equivalent

to

z

⁢

y

-

z

⁢

x

v

∈

(

z

⁢

I

)

⁢

𝔬

v

⊂

I

⁢

𝔬

v

. So assume wlog that

x

v

∈

𝔬

v

for all

v

; in the end simply divide by

z

to recover the general case. But then the

existence

of

y

is guaranteed by the

Chinese Remainder Theorem

, since if

I

=

∏

𝔭

i

e

i

, then

I

⁢

𝔬

v

=

𝔭

i

e

i

for some

i

.

It is true, though somewhat harder to prove, that

K

is in fact dense in

𝔸

K

if even one place is missing from the

product

!

Theorem 3

.

𝔸

K

/

K

is

compact

.

Proof.

The set

U

=

∏

v

∈

P

f

𝔬

v

×

∏

v

∈

P

∞

K

v

is open in

𝔸

K

.

Claim first that

K

+

U

=

𝔸

K

. Choose

(

x

v

)

∈

𝔸

K

. There is a

finite set

S

of finite places

v

such that

x

v

∉

𝔬

v

for

v

∈

S

. Using an argument identical to the approximation argument above, choose

y

∈

K

such that

y

-

x

v

∈

𝔬

v

,

v

∈

S

and

y

∈

𝔬

v

,

v

∉

S

. Then

(

x

v

-

y

)

v

is in

𝔬

v

for

v

∈

S

, is in

𝔬

v

for

v

∉

S

but finite, and is in

K

v

for

v

infinite. Thus

(

x

v

-

y

)

v

∈

U

and we are done.

Claim next that

K

∩

U

=

𝒪

K

.

⊃

is obvious. To see

⊂

, note that an element of

K

∩

U

is an element of

K

that is

integral

at every finite place, so it is integral and is in

𝒪

K

.

Thus we get a natural map

U

↪

𝔸

K

↠

𝔸

K

/

K

. This map is

surjective

since

K

+

U

=

𝔸

K

, and its kernel is

K

∩

U

. So it suffices to show that

U

/

(

K

∩

U

)

=

U

/

𝒪

K

is compact. There is obviously an exact sequence induced by the projection

U

→

∏

v

∈

P

∞

K

v

,

∏

v

∈

P

f

𝔬

v

→

U

/

𝒪

K

→

∏

v

∈

P

∞

K

v

/

𝒪

K

→

0

The left-hand side is compact since each

𝔬

v

is, and the right-hand side is

ℝ

r

1

+

2

⁢

r

2

/

𝒪

K

which know is compact since

𝒪

K

forms a full-rank lattice in

ℝ

n

. Thus

U

/

𝒪

K

is also compact and we are done.

So we have shown that

𝔸

K

is a

locally compact

ring, and that

K

⊂

𝔸

K

is discrete and cocompact. This is analogous to two other situations with which we are familiar:

ℝ

⁢

is locally compact,

⁢

ℤ

⊂

ℝ

⁢

is discrete and cocompact

∏

v

∈

P

∞

K

v

⁢

is locally compact,

⁢

𝒪

K

⊂

∏

v

∈

P

∞

K

v

⁢

is discrete and cocompact

This is a useful

concept

because in such a situation one can do

Fourier analysis

. For example, if

f

:

ℝ

→

ℝ

is a

C

∞

function

with

exponential decay

(or at least integrable on all of

ℝ

), then we can define its

Fourier transform

f

^

, and the

Poisson summation formula

∑

n

∈

ℤ

f

⁢

(

n

)

=

∑

n

∈

ℤ

f

^

⁢

(

n

)

relates the two. The same theory thus exists for appropriately defined functions

f

:

𝔸

K

→

ℝ

, and the Poisson formula again holds with the sum over

K

rather than over

ℤ

. This can be used to show that the

L

-functions have analytic continuations, just as the real Poisson formula is used to show this for

ζ

.

Title

field is discrete and cocompact in its adèles

Canonical name

FieldIsDiscreteAndCocompactInItsAdeles

Date of creation

2013-03-22 18:00:05

Last modified on

2013-03-22 18:00:05

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11R56

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