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.
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.