The Integral Closure of $\mathbb{Z}_{(p)}$ in $\mathbb{Z}_p$ is (not) a Henselian Local Ring

Question

The original problem is to show that the integral closure of $\mathbb{Z}_{(p)}$ in $\mathbb{Z}_p$ is a Henselian local ring.

Update: this is not correct. See my answer below for a counter-example.

Original post:

Here is my attempt and the problem: let $A$ be the integral closure of $\mathbb{Z}_{(p)}$ in $\mathbb{Z}_p$ (also in $\mathbb{Q}_p$, since $\mathbb{Z}_p$ is integrally closed), then $K:=Frac(A)$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{Q}_p$. Let $v_p$ be the $p$-adic discrete valuation on $\mathbb{Q}_p$, I think $A$ should be $\{x\in{K}|v_p(x)\geq0\}$. If we can prove this then the problem can be solved easily. $A\subseteq\{x\in{K}|v_p(x)\geq0\}$ can be proved by taking valuation on the minimal polynomial, but the reverse side is hard. In other words, if $x\in\mathbb{Z}_p$ is algebraic over $\mathbb{Q}$, I don't know how to prove it is integral over $\mathbb{Z}_{(p)}$.

When we are considering a finite extension of complete discrete valuation fields, we can use the explicit formula of the unique extension of the valuation (and the fact that it is Galois invariant) to prove integral $\iff$ $v\ge0$. What makes it quite hard here is that $\mathbb{Q}$ is not complete, so there is no Hensel's lemma, no uniquely extended valuation or an explicit formula, which means that we cannot use similar method.

I will greatly appreciate either a proof or a hint, thanks!

Answer 1

Counter-example: take $p=5$. We prove $A$ is not local.

Assume that $A$ is local. Then $5\mathbb{Z}_5\cap{A}=:\frak{p}$ is a prime ideal of $A$ lying above the maximal ideal $5\mathbb{Z}_{(5)}$ in $\mathbb{Z}_{(5)}$. Since $A$ is integral over $\mathbb{Z}_{(5)}$, $\frak{p}$ must be a maximal ideal, so it is the unique maximal ideal in $A$. By Hensel's lemma, $X^2+1=0$ has a root $\alpha = 2+pa_1+p^2a_2+\dots$ in $\mathbb{Z}_p$. $2+\alpha\in A$ but $2+\alpha\notin5\mathbb{Z}_5$, which means that $2+\alpha$ is invertible in $A$, i.e, $\frac{1}{2+\alpha}=\frac{2-\alpha}{5}$ is integral over $\mathbb{Z}_{(5)}$. However, the minimal polynomial of $\frac{2-\alpha}{5}$ over $\mathbb{Q}$ is $X^2-\frac45X+\frac15$, a contradiction.

Answer 2

Here is a good to think about why the integral closure can't be a local ring: there are number fields that can be embedded in $\mathbf Q_p$ whose ring of integers has more than one prime ideal lying over $p$, so that is also true for the integral closure of $\mathbf Z_{(p)}$ in that number field, which implies your ring $A$ also must have more than one prime ideal lying over $p$.

Example. There are irreducible monic polynomials $f(x)$ in $\mathbf Z[x]$ with degree $n$ greater than $1$ such that $f(x)$ splits completely mod $p$, and thus $f(x)$ has all of its roots in $\mathbf Z_p$. An example is $f(x) = x^2 - a$ where $a$ is a nonsquare integer such that $a \bmod p$ is a square, like $x^2 + 1$ with $p = 5$ (taking $a = -1$). Letting $\alpha$ be a root of $f(x)$ in $\mathbf Q_p$, so $\alpha \in A \subset \mathbf Z_p$, the number field $K := \mathbf Q(\alpha)$ is contained in $\mathbf Q_p$ and its ring of integers $\mathcal O_K$ has $n$ prime ideals lying over $p$. Localizing, the integral closure of $\mathbf Z_{(p)}$ in $K$ also has $n$ prime ideals over $p$, and that ring is inside $A$, so $A$ has at least $n$ prime ideals over $p$.
Thus $A$ is not a local ring.

The example in your own answer uses $f(x) = x^2 + 1$ with $p = 5$: there are two primes over $5$ in $\mathbf Z[i]$, which can be viewed as a subring of $\mathbf Z_5$. Localizing, there are two primes over $5$ in $\mathbf Z_{(5)}[i]$, so $A$ in $\mathbf Z_5$ has more than one prime over $5$. Thus $A$ is not a local ring.