A counterexample to a statement Question Give an example that $z\in\mathbb{Z}[\sqrt{-d}]$, $d\geq1$, $|z|^2$ is a prime number in $\mathbb{Z}$ but $z$ is not prime in $\mathbb{Z}[\sqrt{-d}]$. 
Problem I understand that if $\mathbb{Z}[\sqrt{-d}]$ is a UFD then the above is NOT true since $z=xy\implies|z|^2=|x|^2|y|^2$ so $|x|=1$ or $|y|=1$ therefore $x$ or $y$ is invertible in $\mathbb{Z}[\sqrt{-d}]$ hence $z$ is irreducible in $\mathbb{Z}[\sqrt{-d}]$ then prime in $\mathbb{Z}[\sqrt{-d}]$. So $\mathbb{Z}[\sqrt{-d}]$ cannot be a UFD, for example, $\mathbb{Z}[\sqrt{-d}]$ can be $\mathbb{Z}[\sqrt{-3}]$ or $\mathbb{Z}[\sqrt{-5}]$. But what about a specific example?
 A: There is no such example. In particular, if $|\alpha|^2$ is a prime number, then $\alpha\mathbb Z[\sqrt{-d}]$ is a prime ideal, and hence $\alpha$ is prime.
If you've seen any algebraic number theory, this follows from the fact that the ideal norm $N(\alpha\mathbb Z[\sqrt{-d}])$ is equal to $|\alpha|^2$. 
Otherwise, let $\alpha = x+y\sqrt{-d}$ where $x^2+dy^2=p$ for some prime $p$. First observe that as a group under addition, $\mathbb Z[\sqrt{-d}]\cong\mathbb Z^2$ via the isomorphism $a+b\sqrt{-d}\mapsto (a,b)$.
Under this isomorphism, the map
$$\mathbb Z[\sqrt{-d}]\to \mathbb Z[\sqrt{-d}]\\a+b\sqrt{-d}\mapsto \alpha(a+b\sqrt{-d})$$
corresponds to 
$$\mathbb Z^2\to\mathbb Z^2\\\begin{pmatrix}a\\b\end{pmatrix}\mapsto\underbrace{\begin{pmatrix} x&-dy\\y &x\end{pmatrix}}_{\Gamma}\begin{pmatrix}a\\b\end{pmatrix}$$
from which it follows that 
$$[\mathbb Z[\sqrt{-d}]:\alpha\mathbb Z[\sqrt{-d}]] = [\mathbb Z^2:\Gamma\mathbb Z^2] =\det\Gamma = p$$
so that $$\mathbb Z[\sqrt{-d}]/\alpha\mathbb Z[\sqrt{-d}]\cong \mathbb F_p.$$
Since the quotient is a field, $\alpha$ must be prime.
