Let be a prime. The ring of -adic integers is defined to be the pro – p -completion
Each element of is of the form satisfying the condition for all . We have
, with .
where and the second = is because .
So, by induction, we can write each element of in the form .
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Consider the continuous mapping multiplication by as follows
$latex \alpha_p : \mathbb{Z}_p \to \mathbb{Z}_p$ defined by
Then , and $latex p^n\mathbb{Z}_p=Ker(\varphi_n)$.
So we have .
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Subgroups of .
1. Open subgroups:
Assume is an open subgroup of oder , but from we have .
2. Closed subgroups
Let be a closed subgroup, then we have
since in profinite group with open in .
Since each $latexK+p^n\mathbb{Z}_p$ is open, then we have .
So either or .
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is an integer domain
In fact, for , we consider the mapping multiplication by defined by , then is an open subgroup of , so either 0 or . But in , then . So we have
which implies that is an integer domain.
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Ideals
Take a principal ideal which is a closed subgroup of . Then either or .
An ideal is a union of principal ideals, then we conclude that any ideal is principal.
is the unique maximal ideal, so is a local domain.
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Units
Every in is invertible iff iff .
Let be the group of units in . We are now going to find elements of finite order in .
1. Assume that with . Let
.
(a) is a adic integer, since
, with .
We have
(b) We have
.
So since and is a domain.
If .
Thus we find roots of 1. They are all the roots of 1.
2. (a) If is odd, then does not contain nontrivial roots of 1.
For with . Then
.
Assume , then
with .
We can write as where .
So
since is odd.
where , which is a contradiction.
(b) , then if has order a power of 2, then .
By contradiction, assume , so . We have
, where , So
which is a contradiction.
3. where with , then .
Assume .
For any n, we have
.
Since .
Let , then we have
, with .
So we have the following theorem
Theorem
Let be the group of units of finite order in . Then
- if p is odd.
- if p=2.
Proof.
, let be such that where .
We have
, then .
If then , so , so , which proves that .
If is odd, then we have
, so from 2(a) above, we have and . So they are all roots of unity, which proves that .