# Local property of localization and tensor product

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 $$U$$?

## 1 Answer

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.
• Thank you very much!! Sorry to bother again, but does the isomorphism still hods as R-mod? Commented Jan 4, 2021 at 4:22
• Sorry for some reason I wasn't pinged from this comment ! The answer is yes because all of the universal properties I mention are universal properties of $R$-modules - at least in the commutative case Commented Jan 17, 2021 at 15:34