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Let $A$ be a finite dimensional algebra. Call an $A$-bimodule $M$ tensor-indecomposable in case $M$ is not isomorphic to $X \otimes_K Y$ for a left $A$-module $X$ and a right $A$-module $Y$.

Question 1: In case $A$ is a connected (acyclic) quiver algebra, is it true that the bimodules $A$ and $D(A):=Hom_K(A,K)$ are tensor-indecomposable?

Edit: I guess in case they were not tensor-indecomposable $X$ or $Y$ must be 1-dimensional, else we would get a contradiction that $A$ (or $D(A)$) is not indecomposable. But then the right or left action would not be faithful in case $X$ or $Y$ is simple. Is this a correct proof that $A$ and $D(A)$ are tensor-indecomposable? (The proof should probably work for any bimodule $M$ that is indecomposable faithful as a left and right $A$-module.

Question 2: Is there a homological criterion when a bimodule $M$ is tensor-indecomposable?

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  • $\begingroup$ What is $D(A)$? $\endgroup$
    – Bugs Bunny
    Commented Mar 6, 2020 at 14:53
  • $\begingroup$ @BugsBunny I added the definition. $\endgroup$
    – Mare
    Commented Mar 6, 2020 at 14:56

1 Answer 1

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I think Q1 is clear as soon as $A\neq K$. BTW, $A=K$ is a silly counterexample.

Your quiver has a sink $a$ and a source $b$. There are no arrows from $a$ to $b$. Then $A$ has no element $x\neq 0$ such that $ax=x=xb$. But any tensor product, supported on all vertices, must have such an element.

A proof for $D(A)$ is similar and left to an interested reader :-))

No clue about Q2.

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