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.

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.