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 \(R\) with the field of fractions \(K\) is called a Dedekind domain if the following conditions hold:

  • \(R\) is Noetherian, so every nonempty set of ideals has a maximal element, or equivalently, every ideal is finitely generated,
  • every prime ideal in \(R\) is a maximal ideal, and
  • \(R\) is integrally closed in \(K\), so that every root of a monic polynomial from \(R[x]\) lying in \(K\) lies in \(R\).

Our goal is to prove the following proposition:

Proposition: Assume \(R\) is a Dedekind domain. For any nonzero (i.e. containing a nonzero element) fractional ideal \(I\) in \(K\) there exists a fractional ideal \(J\) such that \(IJ=R\).

Proof: We can construct the inverse explicitly: define \(\widetilde I=\{\gamma\in K:\gamma I\subseteq R\}\). It is straightforward to check that \(\widetilde I\) is a fractional ideal. From its definition we see immediately \(I\widetilde I\subseteq R\), so it is an ideal (since it’s a fractional ideal contained in \(R\)). If the equality holds, then we are home, since we can just take \(J=\widetilde I\). Otherwise, it is a proper ideal. Hence we can apply the following lemma:

Lemma 1: If \(A\) is a proper ideal in \(R\), then \(\widetilde A\not\subseteq R\).

Before we prove it, let’s see why it is enough. We apply the lemma to \(A=\widetilde I I\). Taking \(\gamma\in\widetilde A\setminus R\) we get \(\gamma\widetilde I I\subseteq R\), so that \(\gamma\widetilde I\subseteq\widetilde I\) (recall the definition). We will reach a contradiction if we show \(\gamma\) is integral over \(R\), i.e. satisfies a monic polynomial equation over \(R\), since we assumed that \(R\) is integrally closed. This proceeds by a standard linear algebra argument, which we include here for completeness.

Recall that every ideal of \(R\) is finitely generated. It follows that every fractional ideal is also finitely generated (since it’s essentially just a scaled ideal). Say \(\widetilde I=(a_1,\dots,a_n)\). Since \(\gamma\widetilde I\subseteq\widetilde I\), we can express \(\gamma a_i\) as a linear combination of \(a_i\) with coefficients in \(R\). We can arrange these into a square matrix \(M\), and setting \(v\) to be the vector consisting of \(a_1,\dots,a_n\), we obtain \(Mv=\gamma v\), so matrix \(\gamma I-M\) (\(I\) is the identity matrix here, not an ideal!) is singular. Expanding its determinant we find a monic polynomial equation satisfied by \(\gamma\).

We only need to prove lemma 1 now. We will require two further lemmas.

Lemma 2: Let \(\frak p_1,\dots,\frak p_k\) and \(\frak p\) be prime ideals. If \(\frak p\) contains the product \(\frak{p}_1\dots\frak p_k\), then \(\frak{p}=\frak p_i\) for some \(i\).

Proof of lemma 2: Suppose \(\frak{p}\neq\frak p_i\) for all \(i\). Since prime ideals are maximal, we can’t have \(\frak p_i\subsetneq\frak p\), so there is an element \(a_i\in\frak p_i\setminus\frak p\). Then clearly \(a_1\dots a_k\in\frak{p}_1\dots\frak p_k\), yet, since \(\frak{p}\) is prime and \(a_i\not\in\frak p\), \(a_1\dots a_k\not\in\frak{p}\), contradicting our assumption. \(\square\)

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 \(R\), since every fractional ideal contains an ideal as a subset.
Suppose there is an ideal which doesn’t contain such a product. Because \(R\) is Noetherian, there must be a maximal such ideal \(M\). \(M\) clearly can’t be prime, so there are \(a,b\in R\setminus M\) such that \(ab\in M\). Ideals \(M+(a),M+(b)\) must then contain \(M\) properly, so by choice of \(M\) they contain products of prime ideals, thus so does their product. But their product is \((M+(a))(M+(b))=M+aM+bM+(ab)\subseteq M\), contradicting the fact that \(M\) contains no product of prime ideals. \(\square\)

Proof of lemma 1: Choose any \(\alpha\in A\). Then, by lemma 3, \((\alpha)\) contains a product of prime ideals, say \(\frak{p}_1\dots\frak p_k\). We may assume \(k\) is minimal for which it is possible, i.e. \((\alpha)\) doesn’t contain a product of any \(k-1\) prime ideals. Since \(R\) is Noetherian, \(A\) is contained in a maximal ideal, which is also prime. Call it \(\frak p\). Then \(\frak p\supseteq A\supseteq (\alpha)\supseteq\frak{p}_1\dots\frak p_k\), so, by lemma 1, \(\frak p=\frak p_i\) for some \(i\), say \(i=1\). By our minimality assumption, \(\frak{p}_2\dots\frak p_k\not\subseteq(\alpha)\), say \(\beta\in\frak{p}_2\dots\frak p_k\not\setminus(\alpha)\). Then \(\frac{\beta}{\alpha}\not\in R\). However, \(\beta A\subseteq \beta\frak p\subseteq\frak{p}_2\dots\frak p_k\frak p=\frak{p}_1\dots\frak p_k\subseteq(\alpha)\), so that \(\frac{\beta}{\alpha}A\subseteq R\) thus \(\frac{\beta}{\alpha}\in\widetilde A\setminus R\). \(\square\)

This completes the proof of \(I\widetilde I=R\). \(\square\)