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.
Let be an ideal and a multiplicative subset.
- Let , a multiplicative subset of .
- Let , an ideal of .
Then we have an isomorphism
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 .] ♦
Let be a prime ideal. The residue field is 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.
The localized module is defined as with and under the equivalence
Now becomes an -module via:
Easy exercise (but tedious!). ♦
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 .
Universal property thus says we get a bijection for any -module N:
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 . ♦
Suppose is an A-linear map. We have an -linear map
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.
One can generalize the localization functor.
Let be an -algebra and an -module. The B-module induced from M comprises of a pair
where 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 .
Again universal property says the following is a bijection for any B-module N:
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.
This result is of paramount importance.
The localization functor is exact, i.e. if is exact, so is
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 ,
Prove that for an ideal ,
- the localized module is the ideal we saw earlier,
- for any A-module M we have as -submodules of .