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Let C be a curve (so a $1$-dimensional, proper $k$-scheme).

$C$ is called geometrically regular if $C \otimes_k k'$ is regular for all finite extensions $k \subset k'$.

Assume $char(k)=p>0$ and let $k^{\text{ perf}}$ the perfect closure of $k$.

My question is why if $C \otimes _k k^{\text{ perf}}$ is regular then it follows that $C$ is geometrically regular?

Remark: This are equivalent statements but the converse implication is trivial.

user267839
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  • I don't think I have an answer but note that your definition of "geometrically regular" is non-standard (you say it for all field extensions, others usually for finitely generated field extensions). Your definition would not make sense for general Noetherian schemes over a field (since regularity is traditionally only defined for Noetherian schemes, and when you base change to an infinitely generated field, you lose Noetherianity) but in your case it it OK because schemes are of finite type. Whether or not your definition is equivalent to the usual one, I do not know. –  Apr 30 '19 at 07:47
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    @AknazarKazhymurat: what do you mean by by a "general" Noetherian scheme in contrast to "common" Noeterian scheme (so locally Noetherian & quasi compact)? Naively I would guess that if we consider a nice enough class of schemes than the conventional definition should coincide with the one given in the thread if we restrict the field extensions to algebraic ones (since every (possibly infinite) alg extension is always direct limit of finite extensions). But do you know an argument if we would instead work with the conventional definition of geometrical regularity which you have mentioned above? – user267839 Apr 30 '19 at 10:43
  • maybe my English is bad. What I mean by "general Noetherian scheme"? I mean a scheme bad enough to show the truth of a particular statement I have made (that the definition "geometrically regular iff regular for all field extensions" does not make sense for Noetherian schemes over a field, since regularity does not make sense for non-Noetherian rings; see here https://math.stackexchange.com/a/338437/631975). I feel that in some sense which may or may not be easily formalized, most Noetherian schemes over a field are like that. –  Apr 30 '19 at 11:30
  • yes yes now I see the problem. Have now modified the question. Thank you for this remark – user267839 Apr 30 '19 at 11:58
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    the argument here (https://mathoverflow.net/a/330397/749000) shows, I believe, that the previous definition that was used and the usual definition are equivalent for finite type algebras over a field. –  Apr 30 '19 at 19:15

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See EGA IV, seconde partie, Proposition 6.7.7. Assume you have a scheme locally of finite type over a field $k$ and a point $x\in X$. Take $Q(k')$ in the proposition to mean "$k'$ is an extension of $k$, and $X\times_{k}k'$ is regular at every point in the preimage of $x$ under the canonical projection $X\times_{k}k'\rightarrow X$". The proposition says that if $Q(k')$ is true for a single perfect extension of $k$, then it is true for all extensions of $k$. Now note that a scheme is regular iff it is regular at every point. Game over.