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.
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 $S[y]$).