# Localization commutes with arbitrary direct sums

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.

• I don't understand your sentence into parentheses. The projection $M\to M_i$ will do. Oct 9, 2019 at 13:04
• @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? Oct 9, 2019 at 13:12
• 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 $x\mapsto x_i$. Oct 9, 2019 at 14:05
• @GreginGre Can you please check the edit? Oct 9, 2019 at 16:36
• 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. Oct 9, 2019 at 19:12

## 3 Answers

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").

• 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... Oct 9, 2019 at 16:46
• Ah... sorry for that :P Yes, that proof is indeed very similar to the one here (playing with universal properties). Oct 9, 2019 at 16:49
• 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. Oct 9, 2019 at 16:58

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.

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

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.