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Let $f: R \to S$ be a morphism of Noetherian rings (or more generally $S$ can just be an $R-R$ bimodule with a bimodule morphism $R \to S$). Suppose $f$ is faithfully flat on both sides, so $M \to M \otimes_R S$ is injective for any right $R$-module $M$, and $N \to S \otimes_R N$ is injective for any left $R$-module $N$.

Is it then true that $M \otimes_R N \to M\otimes_R S \otimes_R N$ is injective for any pair $(M, N)$ of a right and a left $R$-module? This seems way too optimistic, but I can't seem to find a counterexample.

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up vote 8 down vote accepted

Here's an example. Let $R = {\mathbb C}[x]$ and let $S = {\mathbb C}\langle x,y\rangle/(xy-yx-1)$, i.e. the first Weyl algebra $A_1$. Then $S$ is free as both a left and right $R$-module, and comes equipped with the natural ($R$-bimodule) inclusion of $R$. On the other hand, if you take $M = {\mathbb C}[x]/(x) = N$, you'll get for $M\otimes_R S\otimes_R M$ the zero module. Indeed, any element of $S$, i.e. any differential operator with polynomial coefficients (writing $\partial = \partial/\partial x$ in place of $y$) can be written in the form $\sum_i p_i(x) \partial^i$, so any element in $M\otimes_R S = {\mathbb C}[x]/(x) \otimes S$ is represented by an expression $\sum_i c_i \partial^i$ where the $c_i$ are constants, and now an induction on $k$ shows (I believe, my brain is a little fuzzy at this hour) that, for the right $R$-module structure on $S/xS$, one has $\partial^k \cdot x = k \partial^{k-1}$. One can conclude that $M\otimes_R S\otimes_R M = 0$, whereas of course $M\otimes_R M\cong M$ in this example...

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