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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!

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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.

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  • $\begingroup$ 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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