Let $k$ a field and $A$, $B$ finite dimensional associative $k$-algebras. If $M$ is an irreducible $A$-module, and $N$ is an irreducible $B$-module, then $M\otimes_kN$ is an irreducible $A\otimes B$-module. Is it also true that $M\otimes N$ is indecomposable as an $A\otimes B$-module, if $M$ and $N$ are indecomposable? If not, what is an example?
1 Answers
The statement is false even for irreducible modules. Take $k=\mathbb{R}$, and $A=B=\mathbb{C}$ viewed as $\mathbb{R}$-algebras. Then the tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$ is isomorphic to the algebra $\mathbb{C}\times \mathbb{C}$ (see, for instance, this question).
Now, take $M=N=\mathbb{C}$. Then $M$ and $N$ are irreducible $\mathbb{C}$-modules, but their tensor product $M\otimes_\mathbb{C} N = \mathbb{C}\otimes_\mathbb{R} \mathbb{C}$ is not an irreducible $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$-module. It is, in fact, the direct sum of the submodule generated by $1\otimes 1 + i\otimes i$ and the one generated by $1\otimes 1 - i\otimes i$. (If you view $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$ as $\mathbb{C}\times \mathbb{C}$ instead, then these two submodules will correspond to $\mathbb{C}\times \{0\}$ and $\{0\}\times \mathbb{C}$).

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