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I have found this online notes from Columbia University. However, I have a question (maybe very silly) about the proof.

It says that by the universal property, the map $f'$ induces a unique $S^{-1}R$-module homomorphism. But should it be an $R$-module homomorphism?

I think maybe I messed up with the universal property. I always assume that if we have an $R$-bilinear map from $M\times N$ to $L$, then it induces an $R$-module homomorphism from $M\otimes_RN$ to $L$. How could we induce a $S^{-1}R$ homomorphism if we have $\otimes_R$ instead of $\otimes_{S^{-1}R}$?

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PS It says that the map $f'$ is bilinear. In the sense of $S^{-1}R$-bilinear or $R$-bilinear? I suppose it should be the latter, is it?

Thank you very much in advance!

user26857
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Tortuga
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1 Answers1

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You are 100% right that the map induced by the universal property is, a priori, just an $R$-linear map.

However, the following lemma shows it doesn't matter:

Lemma. Let $f: M \to N$ be an $R$-module map, where $M$ and $N$ are $S^{-1}R$-modules. Then in fact $f$ is $S^{-1}R$-linear.

This is because $s f(\frac{r}{s} m) = rf(m) $ in $N$, so, multiplying both sides by $1/s$ gives that $f$ is $S^{-1}R$-linear.

user26857
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hunter
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  • ah thank you ! I was being stupid! – Tortuga Oct 23 '18 at 02:35
  • not stupid -- I think the book is saying something misleading. – hunter Oct 23 '18 at 02:55
  • @hunter Can you please also explain why $S^{-1}A\otimes_A M$ is an $S^{-1}A$-module? – Atom Apr 26 '23 at 18:32
  • @Atom good question, and actually that doesn't have anything to do with localization.

    If $B$ is an $A$-algebra -- which means it is an $A$-module that is also a ring, with the compatibility condition that $a(bb') = (ab)b'$ -- then $B \otimes_A M$ is a $B$-module. The $B$-module structure is just coming from by multiplication on the left-hand-side.

     – hunter Apr 26 '23 at 18:44