Let $A^n$ be an affine space over $\mathbb{F}_q$. Let $F_q$ be the absolutely Frobenius of $A^n$. Let $\bar{A^n}$ be the base change to $\bar{\mathbb{F}_q}$ and $F_q×1$ be $F_q\times_{\mathbb{F}_q}id_{\bar{\mathbb{F}_q}}$. If an affine subspace $B$ of $\bar{A^n}$ is $F_q×1$-stable, must it come from an affine subspace of$A^n$ by base change -$×\bar{\mathbb{F}_q}$?
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3$\begingroup$ Hint: do the case of a one-dimensional linear subspace first (you will need Hilbert's theorem 90). For a linear subspace of dimension $d$, reduce to the previous case using $\bigwedge^d$. For affine subspaces, add one extra coordinate $x$ and put everything in the $x=1$ plane. $\endgroup$– Piotr AchingerCommented Jun 8, 2022 at 10:36
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$\begingroup$ @PiotrAchinger : do we really need Noether-Hilbert's 90 in the 1-dimensional case? This question mathoverflow.net/questions/418349/… shows that a fairly trivial proof exists in this case. $\endgroup$– LibliCommented Jun 8, 2022 at 21:16
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1$\begingroup$ @Libli right, it does not. And one can probably also imitate proof (T) here directly, by picking a basis of $A$ and $B$ in such a way that the matrix of $B\to A$ is of the form $[ {\rm Id}_d | \ast ]$. In more sophisticated language, the previous hint uses the Plucker embedding of the Grassmannian, while this one uses standard affine charts on the Grassmannian. $\endgroup$– Piotr AchingerCommented Jun 9, 2022 at 5:39
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