Can the decomposition of a principal ideal in a Dedekind domain contain a non-principal ideal as a factor?

Question

If $I= \prod_{P_i\in Spec(R)}P_i$ is the decomposition into prime ideals of a principal ideal $I$ in a Dedekind domain $R$, can one of the $P_i$ be a non-principal ideal? I guess it can’t, but I don’t think I know a result that gives this.

Answer 1

Consider the case $R = \mathbb{Z}[\sqrt{-5}]$. Here, the decomposition of the principal ideal $\langle 2 \rangle$ is $$\langle 2 \rangle = \langle 2, 1 + \sqrt{-5} \rangle \cdot \langle 2, 1 - \sqrt{-5} \rangle.$$ However, neither $\langle 2, 1 + \sqrt{-5} \rangle$ nor $\langle 2, 1 - \sqrt{-5} \rangle$ is principal. (This can be shown by an argument involving norms; for example, if we had a single generator of $\langle 2, 1 + \sqrt{-5} \rangle$, then its norm would have to divide both $N(2) = 4$ and $N(1 + \sqrt{-5}) = 6$, so it would have to divide 2. But since $N(a + b\sqrt{-5}) = a^2 + 5 b^2$, that only leaves the possibility that the norm is 1 -- which would imply that the generator is in fact a unit. This gives a contradiction since the ideal $\langle 2, 1 + \sqrt{-5} \rangle$ is not the unit ideal.)

More generally, in any Dedekind domain which is not a UFD, I would expect that similarly, there should exist elements which are irreducible but not prime, and the decomposition of the corresponding principal ideal would necessarily have to include non-principal ideals (otherwise the factorization would have to include more than one ideal since the element is not prime, and the decomposition into principal ideals would show the element was in fact not irreducible).

Originally, the entire point of the prime decomposition was that it allows for a generalization of unique factorization, even in a non UFD, via replacing unique factorization into prime elements with unique factorization into prime "idealized elements".