Based on arguments in Marcus’s Number Fields and K. Conrad’s expository paper on ideal factorization. I assume familiarity with the concepts related to ideals and fractional ideals in a commutative ring.
Recall that an integral domain with the field of fractions is called a Dedekind domain if the following conditions hold:
- is Noetherian, so every nonempty set of ideals has a maximal element, or equivalently, every ideal is finitely generated,
- every prime ideal in is a maximal ideal, and
- is integrally closed in , so that every root of a monic polynomial from lying in lies in .
Our goal is to prove the following proposition:
Proposition: Assume is a Dedekind domain. For any nonzero (i.e. containing a nonzero element) fractional ideal in there exists a fractional ideal such that .
Proof: We can construct the inverse explicitly: define . It is straightforward to check that is a fractional ideal. From its definition we see immediately , so it is an ideal (since it’s a fractional ideal contained in ). If the equality holds, then we are home, since we can just take . Otherwise, it is a proper ideal. Hence we can apply the following lemma:
Lemma 1: If is a proper ideal in , then .
Before we prove it, let’s see why it is enough. We apply the lemma to . Taking we get , so that (recall the definition). We will reach a contradiction if we show is integral over , i.e. satisfies a monic polynomial equation over , since we assumed that is integrally closed. This proceeds by a standard linear algebra argument, which we include here for completeness.
Recall that every ideal of is finitely generated. It follows that every fractional ideal is also finitely generated (since it’s essentially just a scaled ideal). Say . Since , we can express as a linear combination of with coefficients in . We can arrange these into a square matrix , and setting to be the vector consisting of , we obtain , so matrix ( is the identity matrix here, not an ideal!) is singular. Expanding its determinant we find a monic polynomial equation satisfied by .
We only need to prove lemma 1 now. We will require two further lemmas.
Lemma 2: Let and be prime ideals. If contains the product , then for some .
Proof of lemma 2: Suppose for all . Since prime ideals are maximal, we can’t have , so there is an element . Then clearly , yet, since is prime and , , contradicting our assumption.
Lemma 3: Every nonzero fractional ideal contains a product of prime ideals.
Proof of lemma 3: Note that we can restrict our attention to ideals of , since every fractional ideal contains an ideal as a subset.
Suppose there is an ideal which doesn’t contain such a product. Because is Noetherian, there must be a maximal such ideal . clearly can’t be prime, so there are such that . Ideals must then contain properly, so by choice of they contain products of prime ideals, thus so does their product. But their product is , contradicting the fact that contains no product of prime ideals.
Proof of lemma 1: Choose any . Then, by lemma 3, contains a product of prime ideals, say . We may assume is minimal for which it is possible, i.e. doesn’t contain a product of any prime ideals. Since is Noetherian, is contained in a maximal ideal, which is also prime. Call it . Then , so, by lemma 1, for some , say . By our minimality assumption, , say . Then . However, , so that thus .
This completes the proof of .