Specifically, the problem I'm trying to solve is
Let $d>2$ be a square-free integer. Show that in the ring $\mathbb{Z}[\sqrt{-d}]$ the element $2$ is irreducible, but the ideal $(2)$ is not prime.
I believe for the first part that I want to assume $2$ is reducible, apply the norm map (which is multiplicative so it preserves irreducibility), and contradict the fact that $d$ is square-free. I.e. suppose $4=(a_1^2+db_1^2)(a_2^2+db_2^2)=\cdots$ and deduce that this implies $d$ is not square free. Unfortunately I'm not seeing how the details fit together here.
For the second part, $(2)$ is prime if whenever $xy\in(2)=\{2r:r\in\mathbb{Z}[\sqrt{-d}]$ then $x\in(2)$ or $y\in(2)$. It seems like the best approach to this would be to use the contrapositive: show that for some $x,y\notin(2)$, we can have $xy\in(2)$. Once again, I'm getting stuck on the details here.
Beyond the elementary argument for this problem, I'm trying to gain a deeper understanding of the concepts at play here. The fact that an irreducible element can generate an ideal that isn't prime is fascinating. If anyone has insight on when the is true and when it's not, I would greatly appreciate that.
Best Answer
You do not need to do it by contradiction. Assume that $2=(a+b\sqrt{-d})(x+y\sqrt{-d})$. Then applying the norm map, we get $4 = (a^2+db^2)(x^2+dy^2)$. Thus, up to reordering, either $a^2+db^2=4$ and $x^2+dy^2=1$ (in which case, $(x+y\sqrt{d})$ is a unit), or else $a^2+db^2=x^2+dy^2=2$. But $a^2+db^2=2$ has no solution if $d\gt 2$.
For the second part, note that $a+b\sqrt{-d}\in (2)$ if and only if $2|a$ and $2|b$. Now look at $$(a+b\sqrt{-d})(x+y\sqrt{-d}) = (ax+dby) + (ay+bx)\sqrt{-d}.$$ Can you arrange things so that both $ax+dby$ and $ay+bx$ are even, but you do not have $a$ and $b$ both even, nor $x$ and $y$ both even? Your computations may depend on the parity of $d$.