$\def\Hom{\mathrm{Hom}}$Let $k$ be a field, $A$ a $k$-algebra and let $M$ and $N$ be $A$-modules, finite dimensional over $k$. Let $K$ be an extension of $k$, so $A \otimes K$ is a $K$-algebra and $M \otimes K$ and $N \otimes K$ are $A \otimes K$ modules.
I would like to say that, if $M \otimes K \cong N \otimes K$, then $M \cong N$.
When $k$ is infinite, there is a very easy proof. The $K$-vector space $\Hom_{A\otimes K}(M\otimes K, N \otimes K)$ is simply $\Hom_{A}(M,N) \otimes K$. Saying $M \cong N$ means that there is a matrix in $\Hom_{A}(M, N)$ with nonzero determinant. If the polynomial $\det( \ )$ is nonzero somewhere after extending scalars, then it is already nonzero over $k$.
But, over a finite field, this argument is broken. If $M$ and $N$ were dimension $3$ over $k = \mathbb{F}_2$, and $\Hom(M,N)$ consisted of those matrices of the form $\left( \begin{smallmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & x-y \end{smallmatrix} \right)$, then this matrix is singular for any $(x,y) \in k^2$, but in extension fields it can be invertible.
Nonetheless, I think I have an argument that the statement is true for finite fields as well. Has anyone seen this statement before? Is there an easy proof I missed?
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