# Quotient vs Localization

Taking the quotient and localization are two sides of the same coin when we look at .

- Quotient removes the “small” prime ideals in – it only keeps the prime ideals containing .
- Localization removes the “large” prime ideals in – it only keeps prime ideals disjoint from
*S*.

It turns out these two operations commute.

Proposition 1.Let be an ideal and a multiplicative subset.

- Let , a multiplicative subset of .
- Let , an ideal of .
Then we have an isomorphism

.

**Proof**

We will use the universal properties here.

The canonical homomorphism gives us . Thus we get a ring homomorphism which takes . This map takes every element of to a unit so it factors through . The latter map takes and thus .

The reverse map is left as an exercise. [Hint: start with .] ♦

Definition.Let be a prime ideal. The

residue fieldis defined equivalently as:

- the field of fractions of , or
- the quotient of local ring by its unique maximal ideal.

Thanks to proposition 1, the definition is sensible (let and ).

**Question to Ponder**

If is the coordinate ring of a variety over algebraically closed *k*, what do elements of correspond to?

# Localization of Modules

Let *M* be an *A*-module and *S* a multiplicative subset of *A*.

Proposition 2.The

localized moduleis defined as with and under the equivalenceNow becomes an -module via:

**Proof**

Easy exercise (but tedious!). ♦

Proposition 3.The -module satisfies the following universal property. We have the pair

where is an -module, is A-linear, such that for any pair

where is an -module and is A-linear, there is a unique -linear map such that .

**Note**

Universal property thus says we get a bijection for any -module *N*:

**Proof**

Existence: given , we define by . It is well-defined: if then for some , and so , which gives since *t* acts bijectively on *N*. It is easy to check *f* is -linear and .

Uniqueness: suppose is -linear and for every . Then

so . ♦

# Functoriality

Definition.Suppose is an A-linear map. We have an -linear map

**Proof**

Let us check the map is well-defined. If then for some . Applying *f* gives and hence . Linearity is an easy exercise. ♦

Clearly and for any and . Thus localization gives a functor

Furthermore, the functor is *A*-linear, i.e. for *A*-modules *M* and *N*,

In particular, it is additive.

## Generalization

One can generalize the localization functor.

Definition.Let be an -algebra and an -module. The B-module

inducedfrom M comprises of a pairwhere is a B-module, is A-linear, such that for any pair

where is a B-module and is A-linear, there is a unique B-linear such that .

**Note**

Again universal property says the following is a bijection for any *B*-module *N*:

**Exercise A**

1. Prove uniqueneess of *f* in the above example.

2. Prove that for an ideal and , the *B*-module induced from *M* is . This explains what we meant when we said the construction is canonical.

**Question to Ponder**

Does give an additive functor from the category of *A*-modules to the category of *B*-modules? Can all the necessary constructions and properties be deduced just from the universal property of induced modules?

In later articles, we will give an explicit construction of the induced module.

# Exactness

This result is of paramount importance.

Theorem 1.The localization functor is exact, i.e. if is exact, so is

.

**Proof**

For any we have and so .

Conversely suppose . Then there exists , so and . We thus have for some which gives us . ♦

Now all the properties we proved about exact additive functors can be ported over here.

1. If is injective, so is . Hence for *A*-submodule we may regard as an -submodule of .

2. If is surjective, so is .

3. For *A*-submodule , we have .

4. For an *A*-linear map we have

5. For *A*-submodules we have, as submodules of ,

**Exercise B**

Prove that for an ideal ,

- the localized module is the ideal we saw earlier,
- for any
*A*-module*M*we have as -submodules of .

Do we not require that the ideal does not intersect in Proposition 1?

If they intersect, we get a trivial ring on both sides so it works out.