# The ring of $p$-adic integer is a local ring

The ring of $$p$$-adic integer is a complete local ring.

I already know the ring of $$p$$-adic integer is complete. Problem is to show it's a local ring. I know this answer is already answered here.

In that answer, it uses one useful fact: If $$P\subsetneq R$$ is a proper ideal such that $$x\in R\setminus P\Rightarrow x\in R^\times$$, then $$P$$ is a unique maximal ideal i.e. $$R$$ is a local ring. Then it says in the case of power series $$f\notin (X)\Rightarrow f\in R^\times$$ so $$(X)$$ is the unique maximal ideal. Applying this to may case, $$x\notin (p\Bbb Z_p)$$ then $$x\in (\Bbb Z_p)^\times$$. This is not clear to me. Could you explain?

Edit: I already know $$p\Bbb Z_p$$ is a maximal ideal. But don't know why it's a unique maximal ideal.

Edit: I know there is some lecture notes that shows $$\Bbb Z_p$$ is a local ring using DVR argument but I don't want to do that. Or is that the simplest way?

• The is a typo in the last sentence: the claim is that if $x \not\in (p\mathbb{Z}_p)$ then $x \in (\mathbb{Z}_p)^\times$. The much smaller ring $(\mathbb{Z}/p\mathbb{Z})$ does not play a role here Commented Jun 1, 2021 at 9:26
• That said: what needs to be done is that for some abstract element $f$ of $\mathbb{Z}_p$ that is not in $p\mathbb{Z}_p$ you want to construct some element $h$ of $\mathbb{Z}_p$ such that $fh = 1$. This construction can follow all the steps in the other answer. Commented Jun 1, 2021 at 9:30
• Ah wait... there is one thing. In general we have that $f$ is of the form $a + pg$ for some element $a \in \mathbb{Z}$ and $g \in \mathbb{Z}_p$. Now in the power series case we can easily assume without loss of generality that $a = 1$. Here that is less obvious, it seems we need a 'separate' proof for the special case that $f \in \mathbb{Z}$ first. For that step it seems it helps that $a$ does have in inverse in the smaller ring $(\mathbb{Z}/p\mathbb{Z})$, so it does play some role after all... Commented Jun 1, 2021 at 9:31
• Viewing $\mathbb{Z}_p$ as the inverse limit of $\mathbb{Z}/p^n\mathbb{Z}$ it is immediately clear that some sequence of compatible residues is invertible in each component if and only if its first component is non-zero, hence the claim Commented Jun 1, 2021 at 9:41
• I think @MikeDaas 's argument is the simplest and best argument Commented Jun 2, 2021 at 11:44

Let $$f$$ be an arbitrary element of $$\mathbb{Z}_p$$. We can expand it as $$a_0 + a_1p + a_2p^2 + \ldots$$ with the $$a_i$$ integers (which we can even take in $$\{0, \ldots, p-1\}$$ but that is irrelevant for the answer).
When $$a_0 = 0$$ we have $$f \in p\mathbb{Z}_p$$ and there is nothing to prove.
When $$a_0 = 1$$ we can copy the steps in the other answer to construct an element $$f^{-1}$$ (as an infinite series of the above form) satisfying $$ff^{-1} = 1$$.
What remains is the case that $$a_0 \neq 0$$ but we do not necessarily know that it equals $$1$$ either. Now in this case there is a $$b \in \mathbb{Z}$$ such that $$a_0b \equiv 1 \mod p$$. ($$b$$ is of course not unique, just pick one). This uses that $$\mathbb{Z}/p\mathbb{Z}$$ is a field.
Now it is easy to check that the element $$fb \in \mathbb{Z}_p$$ is of the form $$1 + b_1p + b_2 p^2 + \ldots$$. So by the earlier answer we can construct an element $$(fb)^{-1}$$ such that $$(fb)(fb)^{-1} = 1$$.
The last step is then observing that $$f(b(fb)^{-1}) = 1$$ as well, so we can take $$f^{-1} = b(fb)^{-1}$$.