MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
This has come up before – David Speyer Dec 20 '13 at 17:08
up vote 8 down vote accepted

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.

share|cite|improve this answer
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 $A$. – Edgardo Dec 16 '13 at 23:54
Sorry, I was confused. – Noah Snyder Dec 17 '13 at 0:13
In the infinite field case, you're using the separable hypothesis, right? – Ben Wieland Dec 17 '13 at 0:42
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. – Edgardo Dec 17 '13 at 0:44
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. – 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 $M=A$, $N=(x,y-1)\subset A$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.