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.


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.

  • $\begingroup$ I don't think it uses commutativity anywhere (?) We are just identifying $\mathrm{Hom}_A(M,N)$ with the linear subspace of $\mathrm{Hom}_K(M,N)$ which commutes with $A$. $\endgroup$ – Edgardo Dec 16 '13 at 23:54
  • $\begingroup$ Sorry, I was confused. $\endgroup$ – Noah Snyder Dec 17 '13 at 0:13
  • $\begingroup$ In the infinite field case, you're using the separable hypothesis, right? $\endgroup$ – Ben Wieland Dec 17 '13 at 0:42
  • $\begingroup$ I don't think so. It comes down to this: Take a polynomial $f \in K[x_1, \dots, x_n]$. If there exists $(a_1, \dots, a_n) \in L^n$ such that $f(a_1, \dots, a_n) \neq 0$, then also there exists $(b_1, \dots, b_n) \in K^n$ with $f(b_1, \dots, b_n) \neq 0$. The existence of $a_i$ means that $f$ is not identically vanishing. $\endgroup$ – Edgardo Dec 17 '13 at 0:44
  • 2
    $\begingroup$ Sorry, here it is: Modules that become isomorphic to $M$ over the algebraic closure of our finite field $K$ are classified by $H^1(G, \mathrm{Aut}(M'))$, where $M'$ is $M$ base-changed to the algebraic closure, and $G$ is the absolute Galois group of $K$. Now $\mathrm{Aut}(M')$ is the set of $\bar{K}$-points of a connected algebraic $K$-group, namely, the automorphism group of $M$ (considered as a $K$-variety). There is a theorem of Lang and Steinberg that says $H^1$ always vanishes in this setting. $\endgroup$ – Edgardo Dec 17 '13 at 2:57

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 $M=A$, $N=(x,y-1)\subset A$.

  • $\begingroup$ This algebra isn't finite dimensional, though. $\endgroup$ – Dag Oskar Madsen Dec 16 '13 at 21:34
  • $\begingroup$ I interpreted the question as asking for an algebra of finite Krull dimension. Perhaps I misunderstood. $\endgroup$ – Julian Rosen Dec 16 '13 at 21:41
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    $\begingroup$ I meant finite dimensional over the field. But it's helpful to see an infinite dimensional example too, so leave this up. $\endgroup$ – Noah Snyder Dec 16 '13 at 21:44

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