Recall that given an integral domain, there is a canonical way to construct the “smallest field containing it”, its field of fractions. Here, we will generalize this construction to arbitrary rings.
We let A be a fixed ring throughout.
A subset is said to be multiplicative if
- if , then .
Our objective is to construct a ring by inverting all elements of S. We will, in fact, define this via its universal property.
For a multiplicative , its localization comprises of a pair
where is a ring and is a ring homomorphism such that is a unit for any , with the following property.
- For any pair , where is a ring and is a homomorphism such that is a unit for any , there is a unique ring homomorphism such that .
In summary, any ring homomorphsim which maps S into the units of B must factor through the localization.
By definition, we obtain a bijection for any ring B
Intuitively, we may imagine as the “smallest” ring extension of A in which every element of S becomes a unit. [Technically, this is wrong since is not injective in general, but this is just for intuition.]
Prove that if and are both localizations, there is a unique ring isomorphism such that .
Prove that if A is a domain and , then is the field of fractions of A.
We will prove that exists by construction.
Take with the equivalence relation:
- if there exists such that .
and be the set of its equivalence classes; write for the equivalence class containing . Then is a ring under the following operations.
The proof is tedious but straight-forward. For example, for transitivity, if and , then there exist such that
To prove that addition and product are well-defined, suppose so that for some ; we wish to show and . Now
and we are done. It remains to show that these operations turn into a ring, which we will leave to the reader. ♦
The ring together with the homomorphism
gives the localization.
Note that is a unit for each since . Now for any ring B and homomorphism such that is a unit for each , we set
This map is well-defined since if then for some so and since are all units we have .
It is also straight-forward to show that f is a ring homomorphism and .
To show that f is unique, we have for all . Hence for all , we have
From this concrete description of , we see that
- if S has no zero-divisors of A, then is injective;
- in particular if A is an integral domain and , then , the field of fractions of A;
- as a consequence, if A is an integral domain so is .
Some Properties of Localizations
If and are multiplicative subsets and is a ring homomorphism such that , we obtain a ring homomorphism
Easy exercise: can be done directly or by universal property. ♦
Some common instances of localizations are as follows.
Example 1: Inverting One Element
The easiest way to get a multiplicative set is to pick and set . We write for the resulting .
For example if , then
the -subalgebra of generated by .
Prove that , where is the polynomial ring.
Example 2: Localization at a Prime
Let be a prime ideal. Then is multiplicative by the definition of prime ideals. We write , the localization of A at the prime ideal .
For example, suppose . Then
Usually denotes the ring of 2-adic integers (which we will cover much later). Hence we will write for the first example and for the second (note the brackets).
Ideals of Localized Ring
Given a general ring homomorphism , we can “push” an ideal through f to obtain an ideal of . Recall that the product notation means we take the set of all finite sums
One can also think of it as the ideal of B generated by .
Conversely, an ideal of B gives us an ideal of A. For convenience we write for and for if f is implicit.
Note that this definition of is consistent with our earlier definition of where M is an A-module.
Main Case of Interest
Now we apply this to our map , .
is exactly the set of elements of which can be written as for some .
It is possible for to hold in with but .
Let be the set of elements as described. It suffices to show:
- is an ideal of .
For the first claim, let . Then so . Also . Hence .
For the second, let , and . Then
since and . So is an ideal of . ♦
The following gives a relationship between ideals of and those of .
For an ideal , let , an ideal of A. Then
(⊆) Suppose , i.e. . Since is an ideal of we have .
(⊇) Pick where . Then so . We see that . ♦
Hence we obtain the following correspondence.
One can imagine this as: the ideals of form a subset of those of A. Clearly this subset is proper in general (e.g. take and ).