Ring of p-adic integers

September 5, 2009 — dhdung1309

Let p be a prime. The ring of p-adic integers is defined to be the pro – p -completion

\mathbb{Z}_p=\underleftarrow{lim}(\mathbb{Z}/p^n\mathbb{Z})

Each element of \mathbb{Z}_p is of the form \{x_n+p^n\mathbb{Z}\}_{n\in\mathbb{N}} satisfying the condition x_{n+1}\equiv x_n~ mod~ p^n for all n. We have

x_1+p\mathbb{Z}=a_0+p\mathbb{Z}, with 0\leq a_0\leq p-1.

x_2+p^2\mathbb{Z}=b_0+a_1p+p^2\mathbb{Z}\equiv a_0+a_1p+p^2\mathbb{Z}^2 where 0\leq b_0,a_1\leq p-1 and the second = is because  x_2\equiv x_1 ~mod ~p.

So, by induction, we can write each element of \mathbb{Z}_p in the form \sum_{0\leq i}a_ip^i.

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Consider the continuous mapping multiplication  by p as follows

$latex  \alpha_p : \mathbb{Z}_p \to \mathbb{Z}_p$ defined by x\mapsto xp

Then Im(\alpha_p)=p\mathbb{Z}_p=Ker(\varphi_1), and $latex  p^n\mathbb{Z}_p=Ker(\varphi_n)$.

\mathbb{Z}_p=\underleftarrow{lim}(\mathbb{Z}/p^n\mathbb{Z})\xrightarrow{\varphi_n} \mathbb{Z}/p^n\mathbb{Z}

So we have \mathbb{Z}_p/p^n\mathbb{Z}_p=\mathbb{Z}/p^n\mathbb{Z}.

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Subgroups of \mathbb{Z}_p.

1. Open subgroups:

Assume H\leq\mathbb{Z}_p  is an open subgroup of oder p^n, but from\mathbb{Z}_p/p^n\mathbb{Z}_p=\mathbb{Z}/p^n\mathbb{Z} we have H=p^n\mathbb{Z}_p.

2. Closed subgroups

Let K be a closed subgroup, then we have

K=\bar K=\bigcap_{n}(K+p^n\mathbb{Z}_p) since in profinite group \bar X=\bigcap_{N\triangleleft G}XN with N open in G.

Since each $latexK+p^n\mathbb{Z}_p$ is open, then we have K\bigcap_{n}p^n\mathbb{Z}_p.

So either K=\{0\} or K=p^n\mathbb{Z}_p.

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\mathbb{Z}_p is an integer domain

In fact, for 0\ne x\in\mathbb{Z}_p, we consider the mapping multiplication by x \alpha_x:\mathbb{Z}_p\to\mathbb{Z}_p defined by z\mapsto xz, then Ker(\alpha_x is an open subgroup of \mathbb{Z}_p, so either 0 or p^n\mathbb{Z}_p. But x\ne 0 in \mathbb{Z}_p, then Ker(\alpha_x)=\{0\}. So we have

Ker(\alpha_x)=Ann(x)=\{z\in\mathbb{Z}_p : xz=0\}=\{0\}

which implies that \mathbb{Z}_p is an integer domain.

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Ideals

Take a principal ideal x\mathbb{Z}_p  which is a closed subgroup of \mathbb{Z}_p. Then either x\mathbb{Z}_o=\{0\} or z\mathbb{Z}_p=p^n\mathbb{Z}_p.

An ideal is a union of principal ideals, then we conclude that any ideal is principal.

p\mathbb{Z}_p is the unique maximal ideal, so \mathbb{Z}_p is a local domain.

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Units

Every x=a_0+a_1p+\cdots in \mathbb{Z}_p is invertible iff x\notin p\mathbb{Z}_p iff a_0\ne 0.

Let G=U(\mathbb{Z}_p) be the group of units in \mathbb{Z}_p. We are now going to find elements of finite order in G.

1. Assume that a\in\mathbb{Z} with (a,p)=1. Let

x_a=\{a^{p^n}+p^{n+1}\mathbb{Z}\}_{n\geq 0}.

(a) x_a is a p-adic integer, since

a+p^n\mathbb{Z}\in U_n=U(\mathbb{Z}/p^n\mathbb{Z}), with |U_n|=p^n-p^{n-1}.

We have

a^{p^n-p^{n-1}}\equiv 1~ mod ~p^n\Rightarrow a^{p^n}\equiv a^{p^{n-1}}~ mod ~ p^n

(b) We have

(x_a)^p=(a^{p^{n+1}}+p^{n+1}\mathbb{Z})=\{a^{p^n}+p^{n+1}\mathbb{Z}\}=x_a.

So (x_a)^p=x_a\Rightarrow x_a(x_a^{p-1}-1)=0\Rightarrow x_a^{p-1}=0 since x_a\ne 0 and \mathbb{Z}_p is a domain.

If a_1\ne a_2~mod ~p\Rightarrow x_{a_1}\ne x_{a_2}.

Thus we find (p-1) roots of 1. They are all the (p-1)-roots of 1.

2. (a) If p is odd, then \mathbb{Z}_p does not contain nontrivial p-roots of 1.

For x=\sum_ia_ip^i with x^p=1. Then

1=x^p\Rightarrow a_0^p\equiv 1~ mod ~p\Rightarrow a_o=1.

Assume x\ne 1, then

x=1+a_ip^i+\cdots with a_i\ne 0.

We can write as x=1+\alpha p^i+zp^{i+2} where \alpha=a_i+a_{i+1}p.

So x^p=(1+\alpha p^i)^p ~mod~ p^{i+2}

= 1+\alpha p^{i+1}~ mod ~p^{i+2} since p is odd.

= 1+a_ip^{i+1}~ mod~p^{i+2} where 1\leq a_i\leq p-1, which is a contradiction.

(b) p=2, then if z\in G has order a power of 2, then z\in\{1,-1\}.

By contradiction, assume z^4=1, so z^2=-1. We have

z=1+2a+4b\in\mathbb{Z}_2, where a,b\in\{0,1\}, So

z^2=(1+2a+4b)^2\equiv 1 ~mod ~ 4\mathbb{Z}_2 which is a contradiction.

3. x\in G where |x|=m with (m,p)=1, then m|(p-1).

Assume x=(x_n+p^n\mathbb{Z})_{n\in\mathbb{N}}.

For any n, we have

x_n^m\equiv 1 ~mod ~ \mathbb{Z}/p^n\mathbb{Z}

\Rightarrow x_n^{p^n-p^{n-1}}\equiv 1 ~mod ~\mathbb{Z}/p^n\mathbb{Z}.

\Rightarrow x_n^{(m,p^n-p^{n-1})}\equiv 1~mod~\mathbb{Z}/p^n\mathbb{Z}

\Rightarrow x_n^{(m,p-1)}\equiv 1~mod~ \mathbb{Z}/p^n\mathbb{Z} Since (m,p)=1.

Let k=(m,p-1), then we have

x^k\equiv 1 ~mod~ \mathbb{Z}/p^n\mathbb{Z}, with k|(p-1).

So we have the following theorem

Theorem

Let H be the group of units of finite order in \mathbb{Z}_p. Then

  1. H\cong C_{p-1} if p is odd.
  2. H\cong C_2 if p=2.

Proof.

h\in H, let m be such that |h|=m=p^ab where (b,p)=1.

We have

(h^{p^a})^b=1, then b|(p-1).

If p=2 then b=1, so |h|=2^a, so h=\{1,-1\}, which proves that H\cong C_2.

If p is odd, then we have

(h^b)^{p^a}=1, so from 2(a) above, we have h^b=1 and b|(p-1). So they are all (p-1)-roots of unity, which proves that H\cong C_{p-1}.

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