# Do regular noetherian schemes of dimension one only have finitely many etale covers of bounded degree

Let $X$ be a regular noetherian scheme of dimension one. Let $d$ be an integer.

Question. Are there only finitely many finite etale morphisms $Y\to X$ of degree $d$?

I want to exclude finite etale morphisms obtained from the base field (if there is any), i.e., if $X$ is of finite type over a field $k$, I don't want to consider the finite etale covers $X_K\to X$ obtained from finite separable field extensions $k\subset K$.

Even if we exclude these finite etale covers, the answer is negative when $X=\mathbf{A}^1_{\mathbf F_p}$.

Then again, the answer is positive if $X$ is

1. an open subscheme of Spec $O_K$, with $O_K$ the ring of integers of a number field;
2. a (not necesarily compact) algebraic curve over an algebraically closed field of characteristic zero.
3. An algebraic curve over a field $k$ of characteristic zero (if we exclude the finite etale covers coming from finite extensions $k\subset K$).

In general the answer is negative as shown above. Which condition on $X$ can we impose to obtain a positive answer.

I was thinking about $X$ has to be of characteristic zero. Does this suffice?

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As far as I know, the only known proof of this result for curves goes through the comparison with the topological fundamental group over $\mathbb{C}$. –  Piotr Achinger Jan 7 '13 at 10:16
Are you looking for "interesting" examples of connected Dedekind $\mathbf{Q}$-schemes admitting infinitely many non-isomorphic connected finite etale covers of some bounded degree? If $R$ is a 2-dimensional normal local noetherian domain (not necessarily henselian) whose residue field $k$ has characteristic 0 and admits infinitely many finite Galois extensions of degree $\le d$ then the complement of the closed point in ${\rm{Spec}}(R)$ is such a non-affine example. Such an $R$ need not admit a $k$-algebra structure (since no henselian hypothesis), so how would you rule it out? –  user30379 Jan 7 '13 at 14:58
Your counterexample (an open affine subscheme of Spec $R$) is two-dimensional, no? Also, I'm not sure how to exclude such examples. It might be possible to do a case-by-case analysis. –  Masse Jan 7 '13 at 18:21
@Masse: pranavk's counterexample is dimension one and not affine. –  Qing Liu Jan 8 '13 at 15:07
@Masse: What is the goal? It seems misguided to do a "case-by-case analysis" when no goal has been articulated; one is just wandering around in the wilderness of general connected Dedekind schemes, which is a zoo. Why not guide yourself via something that actually uses modern algebraic geometry (such as the later parts of Qing Liu's excellent textbook about arithmetic curves, or the classification of smooth projective surfaces over an algebraically closed field, or Mumford's book on abelian varieties, or Neron models, ...)? You'll develop good taste for useful generalizations that way. –  user30379 Jan 9 '13 at 3:18