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Let $p$ be a prime number, and $k$ a finite field with $q=p^n$ elements. Let $X$ be a scheme over $k$ and denote by $X'$ its base change to an algebraic closure $\bar k$ of $k$. Denote by $F_X:X\rightarrow X$ the Frobenius endomorphism of $X$ relative to $k$, and by $F'_X$ its base change to $X'$.

Let $f:Z\rightarrow X'$ be a closed subscheme, over $\bar k$, that is invariant under $F'_X$, meaning that there exists a $k'$-morphism $G:Z\rightarrow Z$ so that $f\circ G=F'_X\circ f$. Is it then true that $Z$ admits a descent to $k$?

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If $Z$ is reduced, then the answer is yes, and the proof is rather elementary.

First, we can suppose $X$ is affine. Indeed, cover $X$ by affine open subsets $X_i$'s. Then $Z\cap X'_i$ is also stable by $F'_X$ because both $X'_i$ and $Z$ are stable by $F'_X$. If we can show that $Z\cap X'_i$ is defined over $k$, then $Z$ is clearly defined over $k$ too.

So suppose $X=\mathrm{Spec} A$ is affine. The closed subscheme $Z$ is defined over some finite extension $K/k$ of degree $r$. Let $I$ be the defining ideal of $Z$ in $X_K$. We have to show that $I$ is stable by $\mathrm{Gal}(K/k)$. Consider the generator $\sigma: \lambda \mapsto \lambda^{q^{r-1}}$ of $\mathrm{Gal}(K/k)$. Let $b=\sum_i a_i\otimes \alpha_i\in I$ with $a_i\in A$ and $\alpha_i\in K$. Denote by $(F_{X})_K$ the relative Frobenius $F_X$ of $X$ extended to $K$. Then $$\sigma(b)^q=(\sum_{i} a_i\otimes \alpha_{i}^{q^{r-1}})^q=\sum_{i} a_i^q\otimes \alpha_{i}=(F_{X})_K(b)\in I$$ Therefore $\sigma(b)\in \sqrt{I}=I$ and we are done.

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The reducedness hypothesis is necessary. Just consider $X$ to be the affine plane over $k$ and $Z$ a copy of $\alpha_p$ inside the kernel of the relative $q$-Frobenius over $\overline{k}$ ($q = |k|$) such that $Z$ corresponds to a tangent direction not rational over $k$. Then $Z$ is trivially preserved by the $q$-Frobenius endomorphism of $X$, yet not defined over $k$. – BCnrd Nov 11 '10 at 5:32

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