Throughout this article, let A be an integral domain and K its field of fractions. We do not assume the ring to be noetherian. The objective here is to develop the theory of multiplying and dividing certain classes of non-zero ideals so that we get a group structure.
A fractional ideal of A is an A-submodule such that there is an such that .
The condition ensures that the elements of M do not have “too many denominators” occurring in them. Thus is not a fractional ideal of because all positive integers can appear as denominators among .
1. Let be an A-submodule; prove that M is a fractional ideal if and only if there exists such that .
2. Find an example of A where K is a fractional ideal of A.
We can copy and paste many of the ideal constructions here.
Let M, N be fractional ideals of A. We have , which are A-submodules of K. We also define the following.
- is the set of finite sums , where .
- is the set of all such that .
As before, represents the “lcm” of M and N, represents their “gcd”, while , represent “product” and “division” respectively.
Be aware that unlike the case of ideals, in the definition of we have to consider all .
If M and N are fractional ideals of A, so are , , and; also is a fractional ideal if .
It is easy to show they are all A-submodules of K. Now pick such that and . Then
For the last case pick any ; then because
Write down and prove as many relations as you can for the above four operations. E.g.
Invertible Fractional Ideals
To avoid complications, henceforth we will only consider non-zero fractional ideals.
Note that the set of fractional ideals of A forms a commutative monoid under product, with identity A. In other words, it satisfies all axioms of a group except the existence of inverse. Hence we define:
Let M be a fractional ideal of A.
We say M is principal if it is of the form for some ; we say it is invertible if MN = A for some fractional ideal N, in which case we write .
An ideal of A is said to be invertible if it is invertible as a fractional ideal.
Clearly principal fractional ideals are invertible. The following examples show that the converse is not true, which is why the theory is interesting.
1. Let , a non-UFD. The prime ideal is not principal but it is invertible because
2. Let . The maximal ideal is invertible because
An invertible fractional ideal of A is finitely generated as an A-module.
Suppose ; write for . We claim generate M as an A-module. Indeed let . Then
and each . ♦
Clearly the set of invertible fractional ideals of A forms a group under product, for if M and N are invertible so is MN. The identity of this group is A. The inverse is given by the following.
Let N be an invertible fractional ideal of A. For any fractional ideal M, we have
We have and thus by definition. Conversely and multiplying both sides by gives . ♦
Two special cases are important here.
- Let M = A: we have .
- If M and N are invertible, so is .
Prove that a fractional ideal is free if and only if it is principal.
Prove that in the ring , where k is a field, the maximal ideal is invertible. Prove also that is invertible if .
The set of principal ideals forms a subgroup of the group of fractional ideals. The resulting quotient is called the Picard group of A, i.e.
Every element of the Picard group can be represented by an invertible ideal .
We immediately have:
The Picard group of a UFD is trivial.
Let be an invertible ideal of a UFD A. By proposition 2, we can find a finite set of generators of M. We claim that generates .
Since for each i we have and thus .
Conversely let so . For each i, and thus . So . Thus and so . ♦
Now suppose is a multiplicative subset not including 0.
We get a map from the set of fractional ideals of A to that of . Indeed if is an A-submodule satisfying then is an -submodule satisfying . Clearly if M is invertible (resp. principal), so is .
Localization preserves the four operations we defined:
For any fractional ideals M, N of A, we have
- if is finitely generated, then .
In particular, if M is an invertible fractional ideal of A, then is an invertible fractional ideal of .
The first two properties follow from that of general submodules. The third is easy. For the last, we have
For the reverse inclusion suppose for . Write ; then for each i. Write for and ; we see that for each i where . [ Recall that A is an integral domain. ] Hence
We obtain a group homomorphism
where M is an invertible ideal of A representing an element of .
Let with , a non-principal invertible ideal. If then takes since is generated by .