MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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 $M$ a right simple module and $N$ be a left simple module over a ring $R$. I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}_R (M,N)$ replaced by $M \otimes_R N$. So my questions are:

Can we describe $M \otimes_R N$ explicitly?

In particular, for a fixed $M$, is $N$ such that $M \otimes_R N \neq 0$ unique up to isomorphism? If not, can we classify such $N$'s in a reasonable way?

share|cite|improve this question
If $R$ is an algebra over a field $k$ then $$ Hom_k(M\otimes_R N,k) = Hom_R(M,Hom_k(N,k)) = Hom_R(M,N^*), $$ so, for example if all simple $R$-modules are finite dimensional over $k$ then $M\otimes_R N \ne 0$ iff $M = N^*$. – Sasha Aug 23 '12 at 17:28
If $I_1$ and $I_2$ are two ideals of the ring $R$ then $\frac{R}{I_1} \otimes \frac{R}{I_2}$ is isomorphic to $\frac{R}{I_1 + I_2}$ .So if $R$ is commutative then $M \otimes N$ is $0$ or $M$ or $N$. – Ali Reza Aug 23 '12 at 18:04
@Sasha: Thank you. Actually, my intention was to understand this duality relation in a more general context... – Alexander Shamov Aug 23 '12 at 18:06
@AliReza: Right, but for commutative rings simple modules are just one-dimensional vector spaces over fields, so the answer is straightforward. :) In the noncommutative setting the $R/(I_1+I_2)$ description remains valid, but $I_1$ and $I_2$ are now ideals from different sides, and they are not even uniquely defined. Can we still get something useful from this sum of ideals? – Alexander Shamov Aug 23 '12 at 18:13
Yes, as abelian groups (or modules over the center of $R$). There is an additive homomorphism $R \to I_1\backslash R/I_2$, which is surjective and its kernel is the sum of $I_1$ and $I_2$ as abelian groups. Beyond that, it's at least as hard as double cosets, which are pretty annoying to work with. – Will Sawin Aug 23 '12 at 19:03
up vote 6 down vote accepted

Sasha's statement is true for any pair of modules.

The center of $R$ is a commutative ring $S$. Since the endomorphisms of a simple module are a division algebra, whose center is a field, the action of $S$ on every simple module factors through some field, so the action of $R$ of course factors through an algebra over that field.

The kenel of a map to a field is a prime ideal $p$, and the map to the field factors through the residue field $k_p$.

So if we have two finitely-generated modules $M$ and $N$, their annihilators in $S$ are two prime ideals of $S$, $p_1,p_2$. If the ideals are distinct, then $S$ annihilates $M \otimes_R N$ since the action of $S$ factors through $k_{p_1} \otimes_S k_{p_2}=0$. The tensor product is zero because one ideal necessarily contains an element $e$ not in the other. In the residue field that element, since it's not in the ideal, has an inverse. Then $1= 1\otimes 1= e^{-1}e\otimes 1=e^{-1}\otimes e=e^{-1} \otimes 0 =0$.

If they are the same ideal, set $R'= R\otimes_S k_p$. It is now an algebra over a field. Apply Sasha's statement.

share|cite|improve this answer
"If the module is finitely-generated then the field is finitely-generated over $S$..." In what sense do you want $M$ to be finitely generated? As an $S$-module? – Manny Reyes Aug 23 '12 at 19:28
Yes. If $R$ is reasonable you can get away with making it finitely-generated as an $R$-module but not all rings are reasonable. – Will Sawin Aug 23 '12 at 20:31
Perhaps you need $S$ to be Noetherian - otherwise I don't see how you show that the center of $\mathrm{End}_R M$ is a finitely generated $S$-module. Anyway, thank you! – Alexander Shamov Aug 23 '12 at 22:30
As I suspected, there's a way to fix it so there are no finiteness conditions needed. – Will Sawin Aug 24 '12 at 16:31
You forgot to invert $S - p$. – user91132 Aug 24 '12 at 17:33

I would guess the answer in general is hopeless, even for nice (noncommutative) algebras. For example, take $A_q := \mathbb C \langle X^{\pm 1}, Y^{\pm 1}\rangle / (XY=q^2YX)$, the quantum torus. If $q \in \mathbb C^*$ is not a root of unity, this is a simple algebra with trivial center. Let $M = P_k = \mathbb C[x^{\pm 1 }]$ as vector spaces with $1$ mapping to $m_0$ and $p_k$, respectively. Give $M$ and $P_k$ right and left $A_q$-module structures using

$f(x)m\cdot X = f(q^{-2}x) m$ and $f(x)m\cdot Y = xf(x)m$

$X\cdot f(x) p_k = xf(x)p_k$ and $Y\cdot f(x)m = q^{-k}x^k f(q^{-2}x)p_k$.

Claim: All the $P_k$ are non-isomorphic, each vector space $M \otimes_{A_q} P_k$ is 1-dimensional (spanned by $m\otimes p_k$), and if $q \in \mathbb C^*$ is not a root of unity, then $M$ and $P_k$ are simple.

share|cite|improve this answer
Wow. Thank you. Did you mean $Y \cdot f(x)p_k = q^{-k} x^k f(q^{-2} x) p_k$? – Alexander Shamov Sep 23 '12 at 18:43
Oops, sorry about that. – Peter Samuelson Sep 23 '12 at 21:46
BTW, for anyone that happens to read this, if there's a minor typo to correct in an answer, is it better to just correct it in the comments so that the question doesn't get bumped to the front page? – Peter Samuelson Sep 23 '12 at 21:54
Peter, I think the community standard is to correct the answer. See the meta threads and – B R Sep 24 '12 at 12:26

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.