## Principal ideals generated by irreducible elements need not be prime

Suppose $\pi$ is an irreducible element in an algebraic integer ring $\mathcal{O}$. Must $(\pi)$ be prime as an ideal in $\mathcal{O}$?

Recall that $1+\sqrt{-5}$ is irreducible in $\mathcal{O}_K$, where $K = \mathbb{Q}(\sqrt{-5})$, since no element in this ring has norm 3. Moreover, $1 - \sqrt{-5} \notin (1+\sqrt{-5})$, as we show. If to the contrary we have $(a+b\sqrt{-5})(1+\sqrt{-5}) = 1 - \sqrt{-5}$, then comparing coefficients we have $a-5b = 1$ and $a+b = -1$, so that $3b = -2$ for some $b \in \mathbb{Z}$, a contradiction. Using this previous exercise, we have $(1+\sqrt{-5}) \subsetneq (1+\sqrt{-5}, 1-\sqrt{-5}) \subsetneq \mathcal{O}$. In particular, $1+\sqrt{-5}$ is irreducible, but $(1+\sqrt{-5})$ is not maximal and thus not prime.

So it need not be the case that $(\pi)$ is prime if $\pi$ is irreducible.