Does base extension reflect the property of being isomorphic? 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 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.
 A: Here's a counterexample to the same statement for infinite dimensional algebras:
Take $K=\mathbb{R}$, $L=\mathbb{C}$, $A=\mathbb{R}[x,y]/(x^2+y^2-1)$. Then $A$ is a Dedekind domain with class group cyclic of order 2, and $A'=A\otimes\mathbb{C}$ is a PID. We can take $M$ and $N$ to be non-isomorphic projective rank 1 modules over $A$, which both necessarily become free after tensoring with $\mathbb{C}$.
Explicitly, we can take $M=A$, $N=(x,y-1)\subset A$.
A: 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.
