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On page 33 of Vakil's book on Algebraic Geometry, he shows how one can define the localization of modules purely in terms of universal property and later shows that a specific definition satisfies the property. Basically he says that if $M$ is an $A$-module and $S$ a multiplicative subset of $A$, define a map $\phi:M\rightarrow S^{-1}M$ as being initial among $A$-module maps $M\rightarrow N$ such that all elements of $S$ are invertible in $N$ i.e. $s\times\cdot: N\rightarrow N$ is an isomorphism for all $s$.

So far, so good. However, he then proceeds to make two assertions which confuse me:
(i) The definition determines $\phi:M\rightarrow S^{-1}M$ up to unique isomorphism.
(ii) By definition, $S^{-1}M$ can be extended to a $S^{-1}A$-module.

I tried to prove the first one by first assuming that there's another map $\psi:M\rightarrow B$ that also satisfies the universal property. I was tempted to then say, by the universal property defined, that there is a unique map from $f:B\rightarrow S^{-1}M$ such that $\phi = f\circ \psi$. However, I have no way of knowing that elements of S are invertible in $S^{-1}M$. Do note that we haven't explicitly defined $S^{-1}M$ yet and so we can't conclude that elements in $S$ are invertible in $S^{-1}M$

The second assertion is related to the first one. How does one conclude from the definition that $S^{-1}M$ can be extended to form a $S^{-1}A$-module? I tried using the invertibility of $S$ in $N$ or choosing a specific $N$ but invertibility in $N$ says nothing about invertibility in $S^{-1}M$ and choosing a specific $N$ (localized module $M$ according to the actual definition for instance) seems like cheating since we're supposed to get the assertion purely via universal property.

EDIT: So it seems like I didn't fully understand what "initial" meant. I didn't count $\phi:M\rightarrow S^{-1}M$ to be in the class of maps that have elements of $S$ to be invertible in the range. The problem is pretty straightforward after realizing that.

Akash Kulkarni
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    An $S^{-1} A$-module is precisely an $A$-module such that every element of $S$ acts invertibly. – Qiaochu Yuan Aug 10 '20 at 21:33
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    Yes you have, being the “initial object” means that it is itself an object in that category ( even though the A-module maps from M s.t bla bla bla is not really a category unless you define morphisms as making commutative triangles ). – LBE Aug 10 '20 at 21:38

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The assumption that $M\to S^{-1}M$ is initial among maps where elements of $S$ act invertibly on the target includes the assumption that the elements of $S$ act invertibly on $S^{-1}M$ (if it didn't, $M$ would always be the appropriate universal object, which is clearly wrong). This takes care of all issues relating to (1) and you can apply the standard technique about universal properties uniquely determining objects up to unique isomorphism.

As for (2), an $S^{-1}A$-module is an $A$-module where every element of $S$ acts invertibly. Once you know that $S$ acts invertibly on $S^{-1}M$, this is clear.

KReiser
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  • Thanks! I am kind of new to category theory and apparently wasn't fully acquainted with the term "initial". The rest follows easily once that was cleared up. – Akash Kulkarni Aug 10 '20 at 21:56