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On pg 34 of Ravi Vakil's Foundations of Algebraic Geometry he says that

an $S^{-1}A$-module is the same thing as an $A$-module for which $s \; \times\; \cdot :M\rightarrow M$ is an isomorphism for all $s \in S$.

I can show explicitly that this is true: an $S^{-1}A$-module is an $A$ module where $s \; \times\; \cdot :M\rightarrow M$ has inverse $\frac{1}{s} \; \times\; \cdot :M\rightarrow M$. In the other direction, if $s \; \times\; \cdot :M\times M\rightarrow M$ has inverse $g_s$ then we define $\frac{a}{s} \cdot m := a \cdot g_s(m)$, which gives us the $S^{-1}A$-module structure.

But I don't know how to show abstractly that this is the same as the universal property. I wanted to show that $M$ was a $B$-module for some $A$-algebra $B$ in which the image of $S$ was invertible. Then the universal property gives us a unique map $S^{-1}A \rightarrow B$ which gives us the module structure. But I'm not sure whether this is right or how to do this.

porkramen
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  • Another universal property for the module $S^{-1}M$ can be seen by writing $S^{-1}M = M \otimes_{R} S^{-1}R$, and then you can apply the universal properties of the tensor product and the localization $S^{-1}R$. – Elchanan Solomon Oct 28 '13 at 01:00
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    I don't understand what you mean by the map $s \times \cdot$. Presumably you just mean multiplication by $s$, in which case that is a map $M \to M$, not $M\times M \to M$? – Bruno Joyal Oct 28 '13 at 04:47

1 Answers1

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The following are equivalent (categories):

  • $S^{-1} A$-modules
  • $A$-modules $M$ equipped with with a homomorphism $S^{-1} A \to \mathrm{End}_A(M)$ of $A$-algebras
  • $A$-modules such that the unique homomorphism of $A$-algebras $A \to \mathrm{End}_A(M)$ sends $S$ to units
  • $A$-modules such that for every $s \in S$ the map $s \cdot - : M \to M$ is an isomorphism.
Martin Brandenburg
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