# Do all varieties have only finitely many etale covers of fixed degree

I've been wondering about the following "finiteness statement" concerning etale covers for a while.

Let $K$ be a field of characteristic zero, not necessarily algebraically closed. A variety over $K$ is a smooth quasi-projective geometrically connected scheme over $K$.

Let $X$ be a variety, and let $d$ be an integer.

Question. Are there only finitely many varieties $Y$ over $K$ such that there exists a finite etale morphism $Y\to X$ of degree $d$ (up to $K$-isomorphism)?

This is true if $K$ is algebraically closed. In fact, by a standard Lefschetz principle argument, we may and do assume $K$ is the field of complex numbers. Then, the statement follows from GAGA and the fact that the topological fundamental group of $X$ is finitely generated.

In general, there might be problems with twists, but I'm not sure.

Note that if $\dim X = 0$ then the only variety $Y$ satisfying the above hypotheses is $Y=$ Spec $K$, because we require our varieties to be geometrically connected (over $K$).

If it helps, in my set-up, we may assume $K$ is finitely generated over $\mathbf Q$, or that $K$ is even a number field.

• There is indeed a problem with twists. For example, consider the degree $2$ covers of $\mathbb{A}^1 - \{0\}$ given by adjoining a square root of $cX$ to $K(X)$, where $X$ is the standard coordinate on $\mathbb{A}^1$ and $c \in K$.
– naf
Sep 12, 2013 at 13:07
• Is there also an example with $X$ a projective curve of genus at least two? The fact that $\mathbf G_m$ might fail to have this property could be related to the fact that its Euler characteristic is zero. Sep 12, 2013 at 17:19
• Yes, there is. Take any étale double cover $\pi\colon Y \rightarrow X$ of smooth projective curves over a field $k$. For example, we may take $X$ to have genus $2$ and $Y$ to have genus $3$. Then $\pi$ is an unramified covering map, and $Y$ has a unique structure of $\mathbf{Z}/2\mathbf{Z}$-torsor over $X$ (with $\mathbf{Z}/2\mathbf{Z}$ just interchanging the sheets of the covering). We can twist $\pi \colon Y \rightarrow X$ by any Galois cohomology class $c \in \operatorname{H}^1(k,\mathbf{Z}/2\mathbf{Z})$, of which there are infinitely many if e.g. $k$ is a number field.
– R.P.
Sep 12, 2013 at 23:11

Just to add one word to ulrich's answer: for a geometrically irreducible $K$-variety $X$, there is a short exact sequence of etale fundamental groups. $$0 \to \pi_1^{\text{et}}(X\otimes_K \overline{K}) \to \pi_1^{\text{et}}(X) \to \text{Gal}(\overline{K}/K) \to 0.$$ An étale degree $d$ cover is equivalent to a homomorphism $\phi:\pi_1^{\text{et}}(X)\to \mathfrak{S}_d$. This homomorphism gives an irreducible cover of $X\otimes_K \overline{K}$ if the image of $\pi_1^{\text{et}}(X\otimes_K \overline{K})$ is a transtive subgroup of $\mathfrak{S}_d$. But for a given homomorphism $\phi_{\overline{K}}:\pi_1^{\text{et}}(X\otimes_K \overline{K}) \to \mathfrak{S}_d$ with transitive image, there may be many extensions of this to a homomorphism $\phi$. That does not in itself prove that there are infinitely many étale covers of, say $\mathbb{G}_{m,K}$. But it does help organize the infinitely many covers that ulrich listed.