Localization of modules as a left adjoint

Let be a commutative ring with unit and let be a multiplicative submonoid of . Localization at is a functor

Writing for the canonical ring localization homomorphism, we can use it to put an -module structure on any -module in a functorial way. In other words, we have a functor (restriction of scalars) Since is isomorphic to the full subcategory of consisting of modules in which the action of every is invertible, we can think of as a forgetful functor!

Claim. Localization is left adjoint to restriction of scalars: . Spelled out, this means there is a natural isomorphism of sets

Thinking of as a forgetful functor, this shines new light on localization of modules as a sort of “best inverse” to forgetting that one can multiply by -fractions, or more precisely some sort of free construction (recall that free constructions are left adjoints to forgetful functors). In other words, and as expected, the localization is the “smallest” or “free-est” module in which one can divide by elements of .

Proof of the claim. Recall that we characterized localization via universal arrows: for any -linear map , there exists a unique -linear map such that , where is the universal arrow.

We define by sending some -linear map to . The previous paragraph tells us that is a bijection.

The last thing to check is naturality. It suffices to check that for any -linear map , we have . However, this is immediate because we defined as the unique -linear map such that this is verified!

Now recall that extension of scalar has a more “usual” left adjoint, called extension of scalars, denoted in our case by Its action on objects is given by . Because of the unicity of adjoints up to a unique canonical isomorphism, there is a natural isomorphism of functors So localization enjoys the same universal property as this tensor product, and similarily this tensor product enjoys the same universal property as the localization.

No comment found.

Add a comment

You must log in to post a comment.