# Fractional Ideals

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*.

Definition.A

fractional idealof A is an A-submodule such that there is an such that .

**Note**

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 .

**Easy Exercise**

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.

Definition.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 .

Proposition 1.If M and N are fractional ideals of A, so are , , and; also is a fractional ideal if .

**Proof**

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

♦

**Exercise A**

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:

Definition.Let M be a fractional ideal of A.

We say M is

principalif it is of the form for some ; we say it isinvertibleif MN = A for some fractional ideal N, in which case we write .An ideal of A is said to be

invertibleif it is invertible as a fractional ideal.

**Note**

Clearly principal fractional ideals are invertible. The following examples show that the converse is not true, which is why the theory is interesting.

**Examples**

1. Let , a non-UFD. The prime ideal is not principal but it is invertible because

2. Let . The maximal ideal is invertible because

Proposition 2.An invertible fractional ideal of A is finitely generated as an A-module.

**Proof**

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.

Proposition 3.Let N be an invertible fractional ideal of A. For any fractional ideal M, we have

.

**Proof**

We have and thus by definition. Conversely and multiplying both sides by gives . ♦

**Note**

Two special cases are important here.

- Let
*M*=*A*: we have . - If
*M*and*N*are invertible, so is .

**Exercise**

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 .

# Picard Group

Definition.The set of principal ideals forms a subgroup of the group of invertible fractional ideals. The resulting quotient is called the

Picard groupof A, i.e..

**Note**

Every element of the Picard group can be represented by an invertible *ideal* .

We immediately have:

Proposition 4.The Picard group of a UFD is trivial.

**Proof**

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 . ♦

# Localization

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:

Proposition 5.For any fractional ideals M, N of A, we have

- ;
- ;
- ;
- if is finitely generated, then .

**Note**

In particular, if *M* is an invertible fractional ideal of *A*, then is an invertible fractional ideal of .

**Proof**

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

. ♦

Corollary 1.We obtain a group homomorphism

where M is an invertible ideal of A representing an element of .

**Example**

Let with , a non-principal invertible ideal. If then takes since is generated by .

I guess the “M” in ” By proposition 2, we can find a finite set of generators x_1, \ldots, x_n of M” should be .

Is it true that $M

In the definition of Picard group, should it be “invertible fractional ideals” instead of just “fractional ideals”? Also, can you give an example of a fractional ideal that is not invertible?

Thanks corrected. For an example, let . The maximal ideal of A is then not invertible.

For an even easier example, take the maximal ideal of the coordinate ring . 😀