Etale fundamental group of an order in a number field

Question

Let $\mathcal{O}$ be an order in a number field $K$, that is a subring of $K$ with rank as abelian group equal to $[K:\mathbb{Q}]$. What is known about the SGA3-étale fundamental group of $X=\mathrm{Spec}(\mathcal{O})$ ? Are there example where it is not profinite ? My motivation for asking this question is for computing the étale cohomology group $H^1(X,\mathbb{Z})=\mathrm{Hom}_{cont}(\pi_1^{SGA3}(X),\mathbb{Z})$.

If X is geometrically unibranch then the SGA3 étale fundamental group equals the étale fundamental group, hence is profinite and the cohomology group vanishes.

Answer 1

I only vaguely know about étale fundamental groups at the moment so I went the other way and computed $H^1(X,\mathbb{Z})$. Let $\pi : Y=\mathrm{Spec}(\mathcal{O}_K) \to X$ denote the normalization. Denote $Z$ the singular locus and put $s_v=\#\pi^{-1}(v)-1$ for every $v\in Z$.

Claim : $H^1(X,\mathbb{Z})$ is a free abelian group of rank $\sum_{v\in Z}s_v$.

Proof : Consider the cokernel $F$ of the injection $0\to \mathbb{Z} \to \pi_\ast \mathbb{Z}$. Then $F$ is a skyscraper sheaf supported on singular points and we can compute by Galois cohomology that $H^0(X,F)\simeq \oplus_{v\in Z}\mathbb{Z}^{s_v}$ (this comes from the vanishing of the $H^1$ in Galois cohomology with coefficients in $\mathbb{Z}$). Because $H^1(Y,\mathbb{Z})=0$ since $Y$ is normal (see etale-cohomology-with-coefficients-in-the-integers), the long exact sequence in cohomology reads $$ 0 \to \mathbb{Z} \xrightarrow{\mathrm{Id}} \mathbb{Z} \to H^0(X,F) \to H^1(X,\mathbb{Z}) \to 0 $$ which proves the claim. $\blacksquare$

In particular, as soon as there is a singular point with at least two primes above it in the normalization, the (SGA3) étale fundamental group of $X$ be not profinite.