Assume that $R$ and $S$ are associative $\mathbb{C}$-algebras with unit $1_R$ and $1_S$, respectively. In addition, assume that $_RM$ is a simple left $R$-module and $_SN$ is a simple left $S$-module. Furthermore, we assume that $M,N$ are finite dimensional. It is easy to see that $M \otimes_{\mathbb{C}}N$ is an $R\otimes_{\mathbb{C}} S$-module.

Is it true that $\text{End}_{R\otimes S}(M\otimes_{\mathbb{C}} N) \cong \text{End}_{R}(M) \otimes_{\mathbb{C}} \text{End}_{S}(N)$ as vector spaces?

This just means that they have the same dimension, however, certain canonical isomorphism is expected. Thanks!


By Schur's Lemma, the two endomorphism algebras on the right are isomorphic to $\mathbb{C}$ (so long as $\dim M$ and $\dim N$ are finite, which you assumed in the question). So in that case you only need that $M\otimes N$ is a simple $R\otimes S$-module. Let $\sum_i m_i \otimes n_i$ be a non-zero element of a submodule $U$ of $M\otimes N$ with the $n_i$ linearly independent and the $m_i$ nonzero. By the density theorem there is an $r\in R$ with $rn_i = \delta_{1i}n_i$. So $m_1\otimes n_1 \in U$. So $Sm_1\otimes Rn_1 \subset U$, i.e. $M\otimes N \subset U$ and $M\otimes N$ is simple.

But we don't need simplicity of $M$ and $N$ for this result on endomorphism algebras to be true, so I drop that in what follows. First note that in the case $R,S = \mathbb{C}$, the natural map $\Phi_\mathbb{C}: \operatorname{End}_\mathbb{C}(M)\otimes \operatorname{End}_\mathbb{C}(N) \to \operatorname{End}_\mathbb{C}(M\otimes N)$ is an isomorphism. So for any $R,S$, the natural map $\operatorname{End}_R(M)\otimes \operatorname{End}_S(N) \to \operatorname{End}_{R\otimes S}(M\otimes N)$ is injective, being the restriction of $\Phi_\mathbb{C}$ to a subspace. We need only show surjectivity.

Let $\alpha \in \operatorname{End}_{R\otimes S}(M\otimes N)$. Since $\Phi_\mathbb{C}$ is onto we can write $$ \alpha=\sum_i f_i\otimes g_i$$ where $f_i \in \operatorname{End}_\mathbb{C}(M)$ and $g_i \in \operatorname{End}_\mathbb{C}(N)$ and the $g_i$ are linearly independent.

Since $\alpha$ is a module map, for all $r\in R,s\in S, m\in M, n \in N$ we have $$ \sum_i f_i(rm)\otimes g_i(sn)= \sum_i rf_i(m)\otimes sg_i(n)$$ Take $s=1$. Linear independence of the $g_i$ shows the $f_i$ are $R$-module homomorphisms. This means we can re-write $\alpha$ as $\sum_i f_i' \otimes g_i'$ where the $f_i'$ are linearly independent $R$-homomorphisms. The same trick now shows that the $g_i'$ are $S$-homomorphisms. Thus $\alpha$ is in the image of the natural map.

  • $\begingroup$ Oh! Thanks! This beautiful proof is that I want. I understand it now. $\endgroup$ – Shevlev May 7 '16 at 5:42

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