Timeline for Are curves over imperfect fields defined over a smaller field?
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Apr 26, 2015 at 2:52 | comment | added | grghxy | Good point. Here you are using that failure of geometric reducedness over $L$ (as must be the case for any potential $C'$ since it cannot be generically $L$-smooth, as $C$ is not generically $K$-smooth) forces non-reducedness after scalar extension to $L^{1/p}=K$, by one of several equivalent characterizations of separability (or not) of field extensions. For general $K$ the extension $K/K^p$ might not be finite degree, but whatever. | |
Apr 26, 2015 at 2:40 | comment | added | Will Sawin | @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. | |
Apr 26, 2015 at 2:31 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 86 characters in body
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Apr 26, 2015 at 2:30 | comment | added | grghxy | 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. | |
Apr 26, 2015 at 2:30 | comment | added | Will Sawin | @grghxy Maybe the way I wrote my question was a little bit misleading. | |
Apr 26, 2015 at 2:26 | comment | added | grghxy | 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). | |
Apr 26, 2015 at 2:06 | comment | added | Will Sawin | @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 $y+ (x^3+tx) + (x^3+tx)^p + ... + (x^3+tx)^{p^{n-1}}$. | |
Apr 26, 2015 at 1:07 | history | asked | Will Sawin | CC BY-SA 3.0 |