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I am wondering if this is valid reasoning:

Let $N$ and $N'$ be $R$-modules. I know that $0\rightarrow N\rightarrow N'$ is exact and want to show for a given multiplicative subset $S$ and that $0\rightarrow (S^{-1}R)\otimes_R N\rightarrow (S^{-1}R)\otimes_R N'$ is exact. Is valid to say that I if I show that that $ (S^{-1}R)\otimes_R N$ is isomorphic to $S^{-1}(R\otimes_R N)$, that because I know that $0\rightarrow R\otimes N\rightarrow R\otimes N'$ is exact, and therefore $0\rightarrow S^{-1}(R\otimes_R N)\rightarrow S^{-1}(R\otimes_R N')$ is exact, so by the isomorphism $0\rightarrow (S^{-1}R)\otimes_R N\rightarrow (S^{-1}R)\otimes_R N'$ is exact.

(I guess my question is when I say by isomorphism is this OK. Can I mess up the functions associated with the arrows?)

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1 Answer 1

up vote 1 down vote accepted

First, you can simplify $R \otimes N \cong N$ etc.

What you want is that there is a natural isomorphism $S^{-1}R \otimes_R N \cong S^{-1} N$. This means that for every morphism $N \to N'$ of $R$-modules the diagram

$$\begin{array}{cc} S^{-1}R \otimes_R N & \rightarrow & S^{-1} N\\ \downarrow & & \downarrow \\ S^{-1}R \otimes_R N' & \rightarrow & S^{-1} N' \end{array}$$

commutes. Finally, if a sequence is isomorphic to an exact sequence (i.e. we have a commutative ladder whose steps are isomorphisms), then this sequence is also exact.

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