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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?

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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
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@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

2 Answers 2

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.

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"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 $S - p$. –  Konstantin Ardakov 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.

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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
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Peter, I think the community standard is to correct the answer. See the meta threads tea.mathoverflow.net/discussion/906/minor-andor-repeated-edits and tea.mathoverflow.net/discussion/394/minor-edits. –  B R Sep 24 '12 at 12:26

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