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Let $M_i$ be a arbitrary colection of $A$-modules and $S$ a multiplicative subset of $A$. I want to show that $$S^{-1}\left(\bigoplus_i M_i\right)\cong \bigoplus_i S^{-1}M_i$$ as $A$-modules and as $S^{-1}A$-modules. I know how to explicitly write an isomorphism between them but I want to do it using the universal properties to understand better how they work.

Here's what I've done:

Let $M=\bigoplus_i M_i$ with the canonical injections $\iota_i:M_i\to M$. Also let $\Phi:M\to S^{-1}M$ be the canonical morphism associated with the localization. Composing these morphisms we get $$\Phi\circ\iota_i:M_i\to S^{-1}M.$$ By the universal property of the localization, we obtain a unique morphism $$\overline{\Phi\circ\iota_i}:S^{-1}M_i\to S^{-1}M$$ such that $\overline{\Phi\circ\iota_i}\circ\Phi_i=\Phi\circ\iota_i$, where $\Phi_i:M_i\to S^{-1}M_i$ is the localization morphism.

Finally, by the universal property of the coproduct we obtain a morphism $$\bigoplus_i S^{-1}M_i\to S^{-1}M.$$ I think this morphism might be the desired isomorphism but I don't know how to prove it.

(Maybe a good idea would be to find its inverse but for it I need some kind of morphism $M\to M_i$, which is not available when the direct sum is infinite.)

Also, I know there are a couple questions here about similar things but they do either the explicit isomorphism or the finite case.

Edit: after @GreginGre commentaries, I have a morphism $S^{-1}M\to S^{-1}M_i$ but I don't know neither how to obtain a morphism $S^{-1}M\to\bigoplus S^{-1}M_i$ (since this is the wrong side for the universal property of coproducts) nor how to show that these two morphisms are inverses of each other.

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  • $\begingroup$ I don't understand your sentence into parentheses. The projection $M\to M_i$ will do. $\endgroup$
    – GreginGre
    Oct 9, 2019 at 13:04
  • $\begingroup$ @GreginGre If I understood corectly, we only have projections in the finite case as (in this case) the coproduct coincides with the product. If the direct sum is not a product, how do I have projections? $\endgroup$
    – Gabriel
    Oct 9, 2019 at 13:12
  • $\begingroup$ Your assumption $M=\bigoplus_i M_i$ means that any element $x\in M$ may be written in a unique way $x=\sum_i x_i, x_i\in M_i$ , where almost all $x_i's$ are $0$ (except a finite number). In other words, you have $M\simeq \coprod_i M_i$. The projection is $x\mapsto x_i$. $\endgroup$
    – GreginGre
    Oct 9, 2019 at 14:05
  • $\begingroup$ @GreginGre Can you please check the edit? $\endgroup$
    – Gabriel
    Oct 9, 2019 at 16:36
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    $\begingroup$ Localization is a tensor product and hence a left adjoint, so it preserves arbitrary colimits. Somewhat harder to see abstractly is that localization is a tensor product with a flat module, so also preserves finite limits. $\endgroup$ Oct 9, 2019 at 19:12

3 Answers 3

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Might not be exactly what you want, but you only need to understand the following fact:

The localization $S^{-1}M$ is canonically isomorphic to the tensor product $S^{-1}A \otimes_A M$, as $S^{-1}A$-modules.

The question is then nothing but the fact that tensor product commutes with arbitrary direct sums (see e.g. this wiki page "distributive property").

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  • $\begingroup$ Yeah, this is in the next page of the book (Vakil's Fundaments of Algebraic Geometry) but I imagine that proving that tensor products commute with arbitrary direct sums will be very similar to the one here... $\endgroup$
    – Gabriel
    Oct 9, 2019 at 16:46
  • $\begingroup$ Ah... sorry for that :P Yes, that proof is indeed very similar to the one here (playing with universal properties). $\endgroup$
    – WhatsUp
    Oct 9, 2019 at 16:49
  • $\begingroup$ Also to your original question: you might try to prove directly that your map from $\bigoplus S^{-1}M_i$ to $S^{-1}M$ is both injective and surjective. $\endgroup$
    – WhatsUp
    Oct 9, 2019 at 16:58
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Similarly as you have constructed the map, you can show that $\bigoplus_i S^{-1}M_i$ satisfies the universal property of $S^{-1}\bigoplus_i M$, which is just the combination of the universal property of localisation and the direct sum. This can be done 'by hand', which usually involves a large diagram, or as follows, if you are confident with functors:

We have the following sequence of natural equivalences of functors from the category of $S^{-1}A$-modules to the category of sets, given by various universal properties: $$\begin{align*} \hom_{S^{-1}A}(S^{-1}M,-) &= \hom_A(M,-|_A)\\ &=\prod\nolimits_i \hom_A(M_i,-|_A)\\ &=\prod\nolimits_i \hom_{S^{-1}A}(S^{-1}M_i,-|_A)\\ &= \hom_{S^{-1}A}\left(\bigoplus\nolimits_i S^{-1}M_i,-\right),\\ \end{align*}$$ where $-|_A$ means restriction of the $S^{-1}$-module structure to $A$.

This is the same as saying that their universal properties are equivalent; formally, one should invoke the Yoneda Lemma to conclude the claim. Tracing through the isomorphisms also tells you how to construct the isomorphisms if you plug in $S^{-1}M$ or $\bigoplus\nolimits_i S^{-1}M_i$ and look where the identity goes. And this shows that the map you've constructed is indeed the natural isomorphism.

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I did (almost) what @Ben said and I think it worked well! It's not as pretty as working with functors, but it works for a beginner. I'll write here what I've done so it may help someone in the future.

Let $M=\bigoplus_i M_i$ be an arbitrary direct sum of $A$-modules with canonical injections $\iota_i:M_i\to M$. Composing with $\Phi:M\to S^{-1}M$ we get morphisms $M_i\to S^{-1}M$. By the universal property of localization, there are maps $h_i:S^{-1}M_i\to S^{-1}M$. These are the only maps such that $h_i\circ\Phi_i=\Phi\circ\iota_i$, where $\Phi_i:M_i\to S^{-1}M_i$ is the canonical map of the localization.

I affirm that $S^{-1}M$, along with these morphisms, satisfies the universal property of coproducts for $\bigoplus_i S^{-1}M_i$. Let $N$ be an $S^{-1}A$-module and $f_i:S^{-1}M_i\to N$ be morphisms. We want to show that there exists a unique morphism $f:S^{-1}M\to N$ such that the diagram diagram

commutes.

Since $f_i\circ\Phi_i$ are morphisms from $M_i\to N$, we use the universal property of coproducts to obtain a morphism $\tilde{f}:M\to N$. Finally, by the universal property of localization, we obtain our morphism $f:S^{-1}M\to N$ which satisfies $f_i\circ\Phi_i=f\circ\Phi\circ\iota_i$. This means that $f_i\circ\Phi_i=f\circ h_i\circ\Phi_i$. By the uniqueness in the universal property of $S^{-1}M_i$, $f_i=f\circ h_i$ and so our diagram commutes. The uniqueness follows from the fact that we had unique morphisms all along the way.

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