Local property of localization and tensor product

Question

Let $M$ be an $R$-module, and let $U\subset R$ be a multiplicatively closed subset. Let $M_U$ be the localization of $M$ at $U$, and let $R_U$ be the localization of $R$ at $U$. Then we have the "well known" isomorphism $$ M_U\cong M\otimes_R R_U. $$

I could prove this by constructing concrete isomorphism with both direction.

But I heard that this isomorphism can be shown by universal property of localization and tensor product.

How can I do that? Thank you for your help.

Local property which I want to use is,

The ring homomorphism j : R → R maps every element of S to a unit in R* = S^ −1R. The universal property is that if f : R → T is some other ring homomorphism into another ring T which maps every element of S to a unit in T, then there exists a unique ring homomorphism g : R* → T such that f = g∘j.*

How can I show

$$ M\otimes_R R_U→M_U $$

and

$$ M×R_U→M\otimes_R R_U $$ preserves units in $U$?

Answer 1

You can't show this with this universal property, as it only refers to rings.

However there is a universal property of $M_U$ which is:

The map $M\to M_U$ induces an isomorphism $\hom_R(M_U,N)\to \hom_R(M,N)$ for any $N$ such that every element of $U$ acts invertibly on $N$ (and of course every element of $U$ acts invertibly on $M_U$).

In other words, if $M\to N$ is a morphism, and every element of $U$ acts invertibly on $N$, then it extends uniquely to a morphism $M_U\to N$.

Now you can use this universal property to show that $M\to M\otimes_R R_U$ satisfies the same one. You'll have to use :

• the universal property of the tensor product;
• the universal property of $R_U$ as an $R$-module, not as a ring;
• the universal property of $R$ as an $R$-module;
• basic properties of the tensor product.