# Invertibility of ideals in Dedekind domains

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 $\gamma\widetile I I\substeq R$, 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 $I\widetilde I=R$. $\square$