Theorem   Let $I$ be an invertible ideal in an integral domain $R$ , and that \begin{equation*} I=\mathfrak{p}_1\mathfrak{p}_2\cdots\mathfrak{p}_m=\mathfrak{q}_1\mathfrak{q}_2\cdots\mathfrak{q}_n \end{equation*}are two factorizations of $I$ into a product of prime ideals. Then $m=n$ and, up to reordering of the factors, $\mathfrak{p}_k=\mathfrak{q}_k$ ($k=1,2,\ldots,n$ ).

Proof. We use induction on $m+n$ . First, the case with $m+n=0$ is trivial, so suppose that $m+n>0$ . As the set of prime ideals $\mathfrak{p}_k$ , $\mathfrak{q}_k$ is partially ordered by inclusion, there must be a minimal element. After reordering, without loss of generality we may suppose that it is $\mathfrak{p}_1$ . Then \begin{equation*} \mathfrak{q}_1\mathfrak{q}_2\cdots\mathfrak{q}_n\subseteq\mathfrak{p}_1, \end{equation*}so $n\ge 1$ . Furthermore, as $\mathfrak{p}_1$ is prime, this implies that $\mathfrak{q}_k\subseteq\mathfrak{p}_1$ for some $k$ . After reordering the factors, we can take $k=1$ , so that $\mathfrak{q}_1\subseteq\mathfrak{p}_1$ .

As $\mathfrak{p}_1$ is minimal among the prime factors, we have $\mathfrak{q}_1=\mathfrak{p}_1$ . Also, $\mathfrak{p}_1$ is a factor of the invertible ideal $I$ and so is itself invertible. Therefore, it can be cancelled from the products, \begin{equation*} \mathfrak{p}_2\cdots\mathfrak{p}_m=\mathfrak{q}_2\cdots\mathfrak{q}_n. \end{equation*}The induction hypothesis gives $m=n$ and, after reordering, $\mathfrak{p}_k=\mathfrak{q}_k$ for $k=2,\ldots,n$ .