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Let $R$ be a commutative $k$-algebra. In fact, I'm interested in the case when $k=\mathbb{Q}$ and $R$ is of the form $\mathbb{Q}[x_1,...,x_n]/I$ where $I$ is finitely generated.

Recall that $R$-bimodules are the same thing as $R \otimes_k R$-modules. I'll write $S=R \otimes_k R$.

Let $M$ and $N$ be finitely presented bimodules. To be clear, this means that there is a surjective map (of $S$-modules) $S^{\oplus n} \mapsto M$, whose kernel is a finitely generated $S$-submodule of $S^{\oplus n}$.

The question is: must $M\otimes_R N$ be a finitely presented $S$-module?

Optional follow-up for the affirmative case: writing $M=S^{\oplus a}/I$ and $N=S^{\oplus b}/J$, how can I write $M \otimes_R N$ as $S^{\oplus c}/K$? By this I mean: what are the generators of $K$ in terms of those of $I$ and $J$?

Motivation: I'm trying to define certain $R$-bimodules on Macaulay2. As far as I know, M2 (understandably) can only handle finitely presented $S$-modules. The main thing I want to work with is tensors of these bimodules over $R$. This is not implemented (so I'm trying to implement it by hand). However I'm starting to think the resulting bimodules may not be finitely presented.

Eric Wofsey
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Alvaro Martinez
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1 Answers1

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No, you can pretty much never expect this. For instance, suppose $R=k[x]$ and let $M=N=S\cong k[x,x']$ (with one variable being the left action of $R$ and the other variable being the right action of $R$). Then $M\otimes_R N$ will be $k[x,y,z]$, with $R$ acting on the left by $x$ and on the right by $z$ (with $y$ being in the "middle" of the tensor product, the right action on $M$ and the left action on $N$ that have been identified). This is not finitely generated as an $S$-module (it is an entire polynomial ring $S[y]$).

Eric Wofsey
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