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Let $A$ and $B$ be $\mathbb{C}$-algebras. Suppose that $M$ and $N$ are respectively simple $A$ and $B$ modules.

We can regard $M\otimes_{\mathbb C}N$ as $A\otimes_{\mathbb C} B$-modules in natural way.

${\bf My\ Question:}$ Is $M\otimes_{\mathbb{C}} N$ still a simple $A\otimes_{\mathbb C} B$-module? What conditions on $A$, $B$, $M$ and $N$ will make this result true? Thanks very much in advance!

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closed as unclear what you're asking by Johannes Hahn, Qiaochu Yuan, András Bátkai, Wolfgang, Chris Godsil Jan 17 '16 at 20:57

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  • $\begingroup$ Since you tagged commutative algebra as well as Lie algebras: what is your definition of an algebra? I deduce that you do not require $A$ and $B$ to have a unit, because otherwise $AN = 0$ would be impossible (as $1$ has to act as the identity). $\endgroup$ – R. van Dobben de Bruyn Jan 17 '16 at 17:27
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    $\begingroup$ @R.vanDobbendeBruyn $1_A$ is not the identity in $A\oplus B$. user85403: How do you define the tensor product? If the algebras are neither commutative nor Hopf algebras there is no canonical module structure on a tensor product. $\endgroup$ – Johannes Hahn Jan 17 '16 at 17:31
  • $\begingroup$ Apologies. But my first question still stands. $\endgroup$ – R. van Dobben de Bruyn Jan 17 '16 at 17:36
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    $\begingroup$ Crosspost: math.stackexchange.com/questions/1615711/… . As mentioned there, it is not true that $M \otimes N$ is a module over $A \oplus B$; it's a module over $A \otimes B$. $\endgroup$ – Qiaochu Yuan Jan 17 '16 at 17:56
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    $\begingroup$ @Johannes: the identity in $A \oplus B$ is $1_A \oplus 1_B$, and the proposed requirement on the action forces both $1_A$ and $1_B$ to act as zero. $\endgroup$ – Qiaochu Yuan Jan 17 '16 at 17:57
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Here's a partial answer: If $M$ and $N$ are finite-dimensional, then $M\otimes N$ is a simple $A\otimes B$ module. For a proof let $R= End_A(M)$, then by the Lemma of Schur we have $R={\mathbb C}\mathrm{Id}$. Replace $A$ with its image in $End_{\mathbb C}(M)$ and likewise for $B$, then by the Theorem of Wedderburn we have $A=End_R(M)=End_{\mathbb C}(M)$. As the same holds for $B$ we get that $A\otimes B=End(M)\otimes End(N)$, which by dimension reasons equals $End_{\mathbb C}(M\otimes N)$.

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