Proposition-Set of integer p-adics

[evinda]

Hello! (Wave) Proposition:
"$\mathbb{Z}_p$ contains only the ideals $0$ and $p^n \mathbb{Z}_p$ for $n \in \mathbb{N}_0$. It holds $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ 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."Proof:

We will show that $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$.
If $x=(\overline{x_k}) \in p^n \mathbb{Z}_p$, then it is $0=x_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ for $k<n$. Therefore it holds for a $x$ $x_k=0$ for all $k \in \mathbb{N}_0$, thus $x=0$.
Now let any ideal $I \neq 0$. From $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ it follows that there is a $n \in \mathbb{N}_0$ with $I \subseteq p^n \mathbb{Z}_p$, but $I \nsubseteq p^{n+1}\mathbb{Z}_p$.
We claim that $I=p^n \mathbb{Z}_p$. From $I \nsubseteq p^{n+1}\mathbb{Z}_p$ follows the existence of a $x=p^nu \in I$ with $u \in \mathbb{Z}_p^{\star}$. Thus it holds $p^n \in I$ and so $p^n \mathbb{Z}_p \subseteq I$.
Since the other inclusion depends on the choice of $n$, we have $I=p^n \mathbb{Z}_p$.

Now we consider the canonical projection

$$\pi: \mathbb{Z}_p \subseteq \Pi_{k \in \mathbb{N}_0} \mathbb{Z}/p^{k+1}\mathbb{Z} \rightarrow \mathbb{Z}/ p^n \mathbb{Z}$$

It is obviously surjective since $\overline{x_{n-1}} \in \mathbb{Z}/p^n \mathbb{Z}$ is the image of $x_{n-1} \in \mathbb{Z}_p$. The kernel of $\pi$ is an ideal and thus of the form $ker \pi=p^l \mathbb{Z}_p$. Since $\pi(p^k)=0$ exactly then when $k \geq n$, it holds $ker \pi=p^n \mathbb{Z}_p$. From the fundamental theorem on homomorphisms we get that $\mathbb{Z}_p/ p^n \mathbb{Z}_p \cong \mathbb{Z}/p^n \mathbb{Z}$. Could you maybe explain to me the above proof? (Thinking)

 

[jvicens]

Sure! Let's break it down step by step.

First, we want to show that the only ideals in $\mathbb{Z}_p$ are $0$ and $p^n \mathbb{Z}_p$ for $n \in \mathbb{N}_0$. To do this, we will show that the intersection of all of these ideals is just $0$. In other words, any element that is in all of these ideals must be $0$.

Next, we assume that $x$ is an element in $p^n \mathbb{Z}_p$, which is the ideal generated by $p^n$ in $\mathbb{Z}_p$. This means that $x$ can be written as $x = (\overline{x_k})$ where $\overline{x_k}$ is the image of $x_k$ in $\mathbb{Z}/p^{k+1}\mathbb{Z}$ (the quotient ring of $\mathbb{Z}$ modulo $p^{k+1}$). We know that $x_k = 0$ for all $k < n$, otherwise $x$ would not be in $p^n \mathbb{Z}_p$. Therefore, $x_k$ must be $0$ for all values of $k$, meaning that $x$ itself must be $0$.

Now, let's consider any ideal $I$ that is not $0$. From our previous result, we know that there exists some $n \in \mathbb{N}_0$ such that $I \subseteq p^n \mathbb{Z}_p$. However, we also know that $I \nsubseteq p^{n+1}\mathbb{Z}_p$. This means that there must be an element in $I$ that is not in $p^{n+1}\mathbb{Z}_p$. We can call this element $x = p^nu$, where $u$ is a unit (meaning that $u$ has an inverse in $\mathbb{Z}_p$). This means that $p^n$ must be in $I$.

Since $p^n$ is in $I$, we can say that $p^n \mathbb{Z}_p \subseteq I$. The other direction, $I \subseteq p^n \mathbb{Z}_p$, is a bit more tricky. It depends on the choice of $