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Suppose that $F$ is a subfield of a field $E$ and, for
$n\times n$ matrices $A_1,\dots,A_m, B_1,\dots,B_m$ over $F$, there exists a matrix $T\in{\rm GL}_n(E)$ such that $T^{-1}A_iT=B_i$ for all $i$.

Does this imply that such a matrix $T$ can be chosen from ${\rm GL}_n(F)$?

It is easy to see that the answer is

  • yes if $m=1$;
  • and yes if the field $F$ is infinite.
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    $\begingroup$ I have edited that. In these cases the answer is yes. $\endgroup$ – Anton Klyachko Aug 14 '20 at 11:12
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    $\begingroup$ @FedorPetrov :It's Schur's Lemma I think. The $A_{i}$ span $M_{n}(F)$ over $F$ and certainly span $M_{n}(G)$ over $G$. Hence the $B_{i}$ span $M_{n}(G)$ as well. Thus, over $G$ , if there is a non-zero matrix $S$ with $A_{i}S = SB_{i}$ for each $i$. Then $ImS$ is an invariant subspace of the vectors space $G^{n}$ of column vectors., so is the whole space $G^{n}$. Thus $S$ is invertible, and the we have $TS^{-1}A_{i}ST^{-1} = A_{i}$ for each $i$, so $ST^{-1}$ is scalar. $\endgroup$ – Geoff Robinson Aug 14 '20 at 12:45
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    $\begingroup$ Doesn't the Noether-Deuring theorem work anyway? The hypotheses give us two finite-dimensional modules over the free algebra $F\langle x_1,\dots,x_m\rangle$ that become isomorphic when you extend scalars to $G$, so by Noether-Deuring they are already isomorphic over $F$. $\endgroup$ – Jeremy Rickard Aug 14 '20 at 12:50
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    $\begingroup$ @GeoffRobinson ah, $Im(S)$ is invariant for all $A_i$, thus for all matrices, thus it is the whole space $F^n$, and $S$ is invertible. This seems to avoid the uniqueness argument. $\endgroup$ – Fedor Petrov Aug 14 '20 at 13:35
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    $\begingroup$ I like this question, but I have never before seen someone so notationally bold as to use $G$ for a field. (My personal field alphabet has $E$ or $K$ as the next letter after $F$.) Now what do we call algebraic groups over $G$? // Also, is it any help to look at a matrix $\widetilde T$ over $F$ conjugating $A_1 \oplus \dotsb \oplus A_m$ to $B_1 \oplus \dotsb \oplus B_m$? $\endgroup$ – LSpice Aug 14 '20 at 14:14
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This question is answered in comments:

"As everyone is saying, this follows from Noether-Deuring. See mathoverflow.net/questions/28469/hilbert-90-for-algebras for a quick proof. I also asked this question a while back math.stackexchange.com/questions/305696."

        – David E Speyer

Thanks to all!

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