Let $A$ be an algebra over $\mathbb{C}$. Let $M$ be a left $A$-module, let $N$ be a right $A$-module and consider the tensor product $N \otimes\_A M$, which is a complex vector space.



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> **Q1:** Can this tensor product be the trivial vector spacefor non-zero $M$, $N$? If yes, what are examples where this happens?
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> **Q2:** Are there criteria on $M$ and $N$ which ensure that the tensor product is non-zero?
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*Edit: Thanks for the answers below. It seems that Q2 is too broad to answer in general. However, I am mostly interested in the case of an infinite-dimensional Clifford algebra; in other words, $A$ is central simple.*



Zero tensor product over a complex algebra?
I give an answer in case the modules are finite dimensional (not 100% sure whether the algebra also has to be finite dimensional but I think it is not needed). The base field can be arbitrary and does not need to be $\mathbb{C}$. Let the algebra $A$ be over a field $K$ and $D=Hom\_K(-,K)$. Then we have $N \otimes\_A M \cong D Hom\_A(N,D(M))$.


This makes it very easy to at least calculate the vector space dimension of $N \otimes\_A M$ and give alot of examples for Q1: Just take $M$ and $N$ simple such that $N$ is not isomorphic to $D(M)$ and by Schur's lemma $N \otimes\_A M \cong D Hom\_A(N,D(M))=0$.


Example: Let $A$ be a finite dimensional algebra with $n$ fixed idempotent $e\_1,...,e\_n$ and simple right modules $S\_1,...,S\_n$ and simple left modules $G\_1,...,G\_n$ (corresponding to the idempotents ). Then $D(G\_i) \cong S\_i$ and thus $S\_j \otimes\_A G\_i \neq 0$ if and only if $i=j$. 


At least when the algeba is a finite dimensional quiver algebra deciding when $D Hom\_A(N,D(M))=0$ can be reduced to linear algebra and answers Q2.


