Arbitrary Collection of Modules
Finally, we consider the case where we have potentially infinitely many modules.
For a collection of A-modules , we have
First claim: we will show that the LHS satisfies the universal property for direct sums. Now for any -module N, we have:
- The first correspondence follows from the universal property of direct sums.
- The second follows from the universal property of localization .
- The third follows from the universal property of direct sums.
- The last follows from the universal property of localization .
This is a natural isomorphism, with both sides functorial in N. Now apply the following to get an isomorphism of A-modules
Prove the following corollary of Yoneda lemma: if are objects such that there is a natural isomorphism
then in .
In general, localization does not commute with direct products:
To be specific, the projection maps for indices j induce , and by the universal property of products
which is not an isomorphism in general. For example, we can set and . If , then in the RHS is not in the image of the map.
1. Find A-submodules such that as submodules of .
2. Prove that for a collection of A-submodules of M, we have
as submodules of . [Hint: use a straightforward proof.]
Another important property we wish to emphasize is locality. As a motivation suppose V is an irreducible k-variety (k algebraically closed) with , which is a domain.
- Let be a function such that for each , is regular at P.
Thus V is a union of open subsets such that each restriction can be locally written as where and for all . We claim that this implies , or to be specific, there exists such that for all .
Indeed, the preceding condition says for all , and since all maximal ideals of are of the form , we only need to show:
For any integral domain A, , where intersection occurs in the field of fractions of A.
(⊆) is obvious. For (⊇), suppose lies in all . Let . Note that is an ideal of A, since if then , and if then for any we have as well.
If , then for some maximal ideal . But so for some we have . Thus , a contradiction. ♦
When we look at , this becomes quite obvious: when p is prime, is the ring of where a, b are integers and b is not a multiple of p. Hence if (reduced) lies in all it just means b is not divisible by any prime, so .
More Local Properties
In all the following results, denotes a maximal ideal of A.
Let M be an A-module. Then if and only if for all .
(⇒) is obvious. For (⇐) fix and let . It is easy to show that this is an ideal of A. Now for any maximal ideal , since is zero there exists such that . Thus . Since is an ideal not contained in any maximal ideal, it is (1), i.e. so . ♦
The ideal above is called the annihilator of m; we will have more to say about it later.
Let be a homomorphism of A-modules. Then is zero if and only if is zero for all .
(⇐) : for each we have since . Thus by proposition 3, and we have . ♦
Let be homomorphisms of A-modules. Then if and only if for all .
Apply proposition 4 to . ♦
Let be A-linear maps. The sequence is exact if and only if
is exact for each .
(⇒) We saw in theorem 1 here that localization is an exact functor.
(⇐) First we show that , or equivalently . But for each so .
For the reverse inclusion, consider the module . We have
Since this is zero for all , we have . ♦
An A-linear map is injective (resp. surjective) if and only if is injective (resp. surjective) for each .
Apply proposition 5 to and respectively. ♦
As a general statement, we say the above properties are local. Philosophically, this means in order to check a certain property, it suffices to check at each maximal ideal so we can restrict ourselves to the case where the base ring is local. With Nakayama’s lemma, we will see that finitely generated modules over local rings are quite well-behaved.
Geometrically, this says we only need to look at each point of the variety to check the property. For example, if we interpret modules over the coordinate ring as vector bundles over the variety, then locality says: a bundle map is injective if and only if is injective at every point P.
Recall that the localization of a reduced ring is reduced, and that of an integral domain is an integral domain. Decide if each of the following is true and justify your answer.
- If is reduced for all , then so is A.
- If is an integral domain for all , then so is A.