In general, we can do this for a domain

Let be a ring, and a multiplicative subset, i.e., and implies . This leads to

under the equivalence relation if and only if there exists such that

Now, is a ring with the operations

and are the additive and multiplicative units respectively of the ring .

Let and . If we take the “naïve” relation on , i.e., if and only if , then we don’t actually get an equivalence relation—transitivity fails.

For a multiplicative subset of a ring ,

is a ring homomorphism such that . Further, the ring is universal with this property. That is, for all maps such that , there exists a unique map such that the following diagram commutes

Addition is similar.

If then

So .

Given , we set

We first check that this is well-defined. If , then there exists such that . Since is a ring homomorphism, we apply to this equation to get

But since , , so we can invert to find that

Therefore, since , and

So is well-defined as a map of sets.

So the diagram commutes.

As for uniqueness, if is another map with

then we must have

Hence, if and , then

So, recalling that so that ,

So .

Let be an integral domain. Then

since multiplication by a nonzero element in an integral domain is injective. So if is multiplicative, we actually have

Let be a ring and . Define the multiplicative set

Define (alternatively, we write ). Then

We check the universal property. Let be a ring homomorphism such that (equivalent to . Now consider the map

making the following diagram commute:

Then

So descends to a unique map such that the following diagram commutes:

This shows that satisfies the universal property of localisations. Hence, we get two unique maps

so we get an isomorphism.

For , we have

Consider the map

is surjective with

So descends to a unique map

Now consider the following commutative diagram, where we have inverted , the image of , in , inducing a unique surjective homomorphism

by the universal property of localisations:

But the kernel of is

in since is a unit. So is an isomorphism.

Let be prime. Then is a multiplicative set (by definition of being prime). Let

Then

is the unique maximal ideal of . We define

to be the residue field. In particular, .

Let be a ring and a multiplicative subset. For an ideal , its extension is defined to be

For an ideal , its contraction is defined to be