Tensor product of simple modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:03:09Z http://mathoverflow.net/feeds/question/105339 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105339/tensor-product-of-simple-modules Tensor product of simple modules Alexander Shamov 2012-08-23T16:33:13Z 2012-09-23T21:47:25Z <p>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:</p> <p>Can we describe $M \otimes_R N$ explicitly?</p> <p>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?</p> http://mathoverflow.net/questions/105339/tensor-product-of-simple-modules/105344#105344 Answer by Will Sawin for Tensor product of simple modules Will Sawin 2012-08-23T19:19:48Z 2012-08-24T16:30:34Z <p>Sasha's statement is true for any pair of modules.</p> <p>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.</p> <p>The kenel of a map to a field is a prime ideal $p$, and the map to the field factors through the residue field $k_p$.</p> <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 $1= 1\otimes 1= e^{-1}e\otimes 1=e^{-1}\otimes e=e^{-1} \otimes 0 =0$.</p> <p>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.</p> http://mathoverflow.net/questions/105339/tensor-product-of-simple-modules/107518#107518 Answer by Peter Samuelson for Tensor product of simple modules Peter Samuelson 2012-09-19T00:06:35Z 2012-09-23T21:47:25Z <p>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</p> <p>$f(x)m\cdot X = f(q^{-2}x) m$ and $f(x)m\cdot Y = xf(x)m$</p> <p>$X\cdot f(x) p_k = xf(x)p_k$ and $Y\cdot f(x)m = q^{-k}x^k f(q^{-2}x)p_k$.</p> <p>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.</p>