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Let $X$ be a connected, integral curve over a field $k$, and let $Y \rightarrow X$ be a finite etale cover. Corresponding to this cover there is a finite extension of function fields $k(Y): k(X)$.

Question: is it possible to characterize the finite extensions of $k(X)$ which arise in this way in a purely field theoretic manner? In other words, given any finite $F: k(X)$, is there some field theoretic property that $F$ satisfies iff $F=k(Y)$ for some finite etale $Y \rightarrow X$?

Feel free to assume whatever else you'd like about $X$ if it helps.

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Yes, this is possible. For simplicity, compare first to number theory: given a finite field extension $E / \mathbf Q_p$ you want to know whether $\mathcal O_E / \mathbf Z_p$ ($\mathcal O_E$ is the integral closure of $\mathbf Z_p$ in $E$) is etale. This is precisely the case if the original extension is unramified (and etale, of course, so in this case just normal)

The same will work for function fields, except that you have more than one place to consider. I did not check the details, but I suppose it will be easier if $X$ is smooth proper to begin with. In this case let $Y$ be the smooth curve that corresponds to $F$ by the bijection of function fields vs. smooth proper curves. Then, replace $\mathcal O_E$ above by the valuation rings corresponding to all the closed points of $Y$. Because of the smoothness assumption, these rings will be DVR's, so the question whether the extension of the local DVR's $\mathcal O_y$ of $Y$ and $X$ is etale is determined by studying the splitting of the (ideal generated by) a uniformizing element $\pi \mathcal O_y$ (here $\pi \in \mathcal O_x$ is the uniformizer = generator of the maximal ideal).

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  • $\begingroup$ Is it then also possible to detect these valuation rings purely on the field level too? Perhaps if you make the correspondence between field extensions and covers explicit that's possible? $\endgroup$
    – KristianJS
    Commented May 23, 2013 at 23:09
  • $\begingroup$ Yes, think of a Riemann surface: a point $x \in X$ gives rise to a valuation $v_x : k(X) \rightarrow \mathbf Z$ on rational functions. Namely, $v_x(f)$ is the pole/zero order of $f$ at $x$. The valuation ring $\mathcal O_x \subset k(X)$ consists of those $f$ satisfying $v_x(f) \geq 0$, i.e., the ones not having a pole at this point. The same works in general for smooth curves (i.e., not over $\mathbf C$, but a general field.) $\endgroup$
    – Jakob
    Commented May 24, 2013 at 1:51
  • $\begingroup$ You shouldn't forget to check the residue extensions to be separable. (You don't need to worry about this if your residue fields are all perfect.) $\endgroup$
    – Ariyan Javanpeykar
    Commented May 24, 2013 at 7:16
  • $\begingroup$ Thanks, I'll think about this a bit first and get back to you if it's what I want. $\endgroup$
    – KristianJS
    Commented May 26, 2013 at 20:58

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