Let L/K be a (separable?) field extension, let A be a finite dimensional algebra over K, and let M and N be two A-modules. Let $A' = L \otimes\_K A$ be the algebra given by extension of scalars, and let $M' = L \otimes\_K M$ and $N' = L \otimes\_K N$ be the A'-modules given by extension of scalars.


Does $M' \cong N'$ (as A'-modules) imply that $M \cong N$ (as A-modules)?


([This question](https://mathoverflow.net/questions/17532/does-linearization-of-categories-reflect-isomorphism) is obviously related. Note that just as for that question it is easy to see that base extension reflects isomorphisms in the sense that if a map $f: M \rightarrow N$ has the property that $f' : M' \rightarrow N'$ is an isomorphism then f is an isomorphism. This is asking about the more subtle question of whether it reflects the property of being isomorphic.)


I apologize if this is standard (I have a sinking suspicion that I've seen a theorem along these lines before), but I haven't been able to find it. There's a straightforward proof in the semisimple setting, but I have made no progress in the non-semisimple setting.



Does base extension reflect the property of being isomorphic?
I hope I'm not misunderstanding the question. Here goes:


We'll show that if $M,N$ are finite-dimensional over $K$, 
then they are isomorphic over $K$. 


Think of the linear space $X=\mathrm{Hom}\_{A}(M,N)$
as a variety over $K$. Inside $X$ look at 
the $K$-subvariety $X'$ of maps that are not isomorphisms $M \rightarrow N$.
Now $X' \neq X$,
because there is an $L$-point of $X$ not in $X'$. Therefore, 
over an infinite field $K$, there will certainly exist a $K$-point of $X$
that doesnt lie in the proper subvariety $X'$. 


If $K$ is finite: 
$M,N$ are both $K$-forms of the same module $M'$ over $L$. 
The $L$-automorphisms of $M'$ are a connected group, because
they amount to the complement of the hypersurface $X'$ inside the linear space $X$. So its Galois cohomology vanishes, thus the
same conclusion.


