$0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$

Question

I am looking at the proof of the sentence:

$\mathbb{Z}_p$ contains only the ideals $0$ and $p^n \mathbb{Z}_p$ for $n \in \mathbb{N}_0$. It holds that $\bigcap_{n \in \mathbb{N}_0 p^n \mathbb{Z}_p}=0$ and $\mathbb{Z}_p/p^n \mathbb{Z}_p \cong \mathbb{Z}/p^n \mathbb{Z}$. Especially, $p\mathbb{Z}_p$ is the only maximal ideal.

The ideal $0$ is contained in each ring, right?

To prove that $p^n \mathbb{Z}_p$ is the only other ideal that is contained in $\mathbb{Z}_p$ we do the following:

Let $I \neq 0$ any ideal.

From $\bigcap_{ \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ we have that there is a $n \in \mathbb{N}_0$ with $I \subseteq p^n \mathbb{Z}_p$, but $I \not \subseteq p^{n+1}\mathbb{Z}_p$. We assume $I=p^n \mathbb{Z}_p$. $I \subseteq p^{n+1}\mathbb{Z}_p$ implies the existence of a $x=p^nu \in I$ with $u \in \mathbb{Z}_p^{\star}$. Then it also holds that $p^n \in I$ and so $p^n \mathbb{Z}_p \subseteq I$. Since the other inclusion is based on the choice of $n$, $I=p^n \mathbb{Z}_p$.

Could you explain me why from $\bigcap_{n \in \mathbb{N}_0}p^n \mathbb{Z}_p=0$ we conclude that there is a $n \in \mathbb{N}_0$ such that $I \subseteq p^n \mathbb{Z}_p$ but $I \subsetneq p^{n+1}\mathbb{Z}_p$ ?

PS: $$\mathbb{Z}_p=\{ (\overline{x_k}) \in \prod_{k=0}^{\infty} \mathbb{Z}/p^{k+1}\mathbb{Z} | x_{k+1} \equiv x_k \mod{p^{k+1}}\}$$

Answer 1

Since

$$\bigcap_{k \in \mathbb{N}_0} p^k \mathbb{Z}_p = 0,$$

for an arbitrary ideal $I \neq 0$

$$\bigcap_{k \in \mathbb{N}_0} I \cap p^k \mathbb{Z}_p = 0.$$

Since $I \cap p^0 \mathbb{Z}_p = I$, but the intersection over all $k \in \mathbb{N}_0$ is $0$, for some finite $n+1 \in \mathbb{N}$ the intersection must become $\subsetneq I$.

Let this $n$ be minimal, i.e. $I \cap p^{n} \mathbb{Z}_p = I$ but $I \cap p^{n+1} \mathbb{Z}_p \subsetneq I.$ From the latter, you can write $x = p^{n} u \in I$ for some unit $u$, thus $p^{n} \in I$ and $p^{n} \mathbb{Z}_p \subseteq I$.

Answer 2

Any nonzero element of $\mathbb{Z}_p$ can be written in a unique way as $$ x=up^n $$ for an invertible $u$ and $n\in\mathbb{N}$. This follows easily from the definition of $\mathbb{Z}_p$ as inverse limit.

If $I$ is a nonzero ideal of $\mathbb{Z}_p$, consider the least $n$ such that $p^n\in I$. Then $p^n\mathbb{Z}_p\subseteq I$. Assume $x\in I$; it is not restrictive to assume $x=p^m$ for some $m$. Now, by definition, $m\ge n$, so $x=p^m=p^n\cdot p^{m-n}\in p^n\mathbb{Z}_p$.