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 $$I\widetilde I=R$$. $$\square$$