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Suppose $R$ is a commutative ring with unity, and $R'$ an $R$-algebra with structure map $\phi: R\to R'$. Let $M,N$ be two $R'$-modules. Then there exists a natural $R'$-linear (hence $R$-linear) map $$ \tau: M\otimes_{R}N\to M\otimes_{R'}N $$ the map is simply $m\otimes n\mapsto m\otimes'n$. The map $\tau$ is always surjective. We also know that if $\phi$ is surjective then $\tau$ is also injective. In fact, the kernel of $\tau$ is the submodule $K\subset M\otimes_{R}N$ generated by differences $$ (x'm)\otimes n-m\otimes (x'n) $$ with $x'\in R'$, $m\in M$ and $n\in N$ (proving this part is quite long but I know it). The kernel vanishes when $\phi$ is surjective. My question is:

Is $\tau$ an isomorphism only if $\phi$ is surjective? I don't think so, for example I suspect $$S^{-1}M\otimes_{R}S^{-1}N\simeq S^{-1}M\otimes_{S^{-1}R}S^{-1}N$$ to be valid, although $R\to S^{-1}R$ is not necessarily surjective. So is there a set of nice conditions under which $\tau$ is an isomorphism?

user26857
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Hamed
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    (1) Note that $R \to S^{-1}R$ is an epimorphism in the category of rings. That's probably the better notion. (2) $M$ and $N$ could be stupid, so I think you need to decide whether you want this to work for all modules, whether there can be some dependence, etc. – Hoot Jun 14 '16 at 08:30
  • Maybe it is better if I ask when are the two bifunctors $\otimes_R$ and $\otimes_{R'}$ naturally isomorphic if $R'$ is an $R$-algebra? Would you think this is the better question? – Hamed Jun 14 '16 at 10:12
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    For your last question, clearly you need the natural map $R'\otimes_R R'\to R'$ to be isomorphic. Assuming that, check that the rest follows. – Mohan Jun 14 '16 at 16:05

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