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Let $C$ be regular projective curve defined over a field $K$. Let $K/L$ be a totally inseparable finite extension. Does there exist a regular projective curve $C'$ over $L$ such that that the pullback of $C'$ to $K$ is birational to $C$?

The answer is obviously no, in fact I think it's obvious that any curve can't be defined over a field smaller than its field of moduli in the moduli space of curves. But this isn't true. Take the curve over $\mathbb F_p(t)$ ($p\neq 3$) defined by:

$y^p-y = x^3+tx$

Then that curve is birational to the curve defined by:

$y^p-y =x^{3 p^n} + t^{p^n} x^{p^n}$

which is defined over $\mathbb F_p(t^{p^n})$, because, by Artin-Schreier theory, they both represent the same cover of $\mathbb A^1$. But it's easy to see that the curve over $\mathbb F_p(t)$ corresponds to a non-isotrivial family of smooth curves and hence is defined in the moduli space of curves by a finite degree subfield of $\mathbb F_p(t)$. (The way I see it is by observing that the monodromy of the cohomology sheaf of this family of curves isn't finite.)

Of course the problem is that the moduli space of smooth curves is a Deligne-Mumford stack but the moduli space of regular curves is not.

So does my intuition fail for every curve, or just some of them?

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  • $\begingroup$ @grghxy In my example, there is an isomorphism between the affine curves, hence a birational equivalence between their regular compactifications, that fixes $x$ and sends $y$ to $y+ (x^3+tx) + (x^3+tx)^p + ... + (x^3+tx)^{p^{n-1}}$. $\endgroup$
    – Will Sawin
    Commented Apr 26, 2015 at 2:06
  • $\begingroup$ OK, now I understand (and I apologize for the erroneous example, now deleted). I am well-aware that regularity is usually destroyed by non-separable ground field extension, but I had misunderstood the question to ask that $C'_K$ be isomorphic to $C$ (which would force $C'$ to be $L$-smooth when $C$ is $K$-smooth) rather than merely birational (which doesn't require $C'_K$ to even be regular). $\endgroup$
    – grghxy
    Commented Apr 26, 2015 at 2:26
  • $\begingroup$ @grghxy Maybe the way I wrote my question was a little bit misleading. $\endgroup$
    – Will Sawin
    Commented Apr 26, 2015 at 2:30
  • $\begingroup$ If $C$ is generically $K$-smooth (equivalently, its function field is $K$-separable) then the answer is affirmative. Indeed, we can then find a dense open $U \subset C$ admitting a finite etale map onto a dense open $V \subset \mathbf{P}^1_K$. By pure inseparability of $K/L$ we have $V=V'_K$ for a dense open $V' \subset \mathbf{P}^1_L$. By topological invariance of the etale site (applied to $V=V'_K \rightarrow V'$), $U$ descends to a finite etale cover $U'$ of $V'$, so the regular compactification $C'$ of $U'$ does the job. $\endgroup$
    – grghxy
    Commented Apr 26, 2015 at 2:30
  • $\begingroup$ @grghxy Good point! If the function field of $C$ is not $K$-separable, I guess the answer is no. Take $L$ to be the field of $p$th powers of $K$. Then if the function field of $C'$ is spearable, the function field of $C'_K$ will also be separable, and if the function field of $C'$ is not separable, then $C'_K$ will not be reduced. $\endgroup$
    – Will Sawin
    Commented Apr 26, 2015 at 2:40

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