# Etale fundamental group of an order in a number field

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.

• Sure, you can create a nodal nodal order by gluing points together, just the same the way you produce a nodal curve. The ring of functions on the nodal affine line is $\{f\in \mathbb k[x]\;|\;f(0)=f(1)\}$. Similarly, take the order $\{x\in\mathbb Z[i]\;|\;x+(1-2i)\equiv x+(1+2i)\in \mathbb Z/5\}$ using the identifications $\mathbb Z[i]/(1+2i)=\mathbb Z/5=\mathbb Z[i]/(1-2i)$. – Ben Wieland Oct 8 '20 at 0:49

## 1 Answer

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.