Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$. Let $\alpha\in \Bbb{Z}$ such that $\sqrt{\alpha}\notin{\Bbb{Q}}$, and let $N:\Bbb{Z}[\sqrt{\alpha}]\to \Bbb{Z}$ be the norm function. Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$.
What I did so far is: Let $z=a+b\sqrt{\alpha}.$ Then $N(z)=a^2-\alpha b^2$. I am having trouble making conclusions so as to find out things about $z$ itself.
The definition I was given: An element in a ring $A$ is called "irreducible" if its divisors are associated to it or are reversible.
I would appreciate your help.

Note that $N$ is multiplicative, i.e. $N(z_1z_2)=N(z_1)N(z_2)$. If $z$ is reducible in $\Bbb{Z}[\sqrt{\alpha}]$, then $z=z_1z_2$ for some nonunits $z_1,z_2\in\Bbb{Z}[\sqrt{\alpha}]$. Thus, $N(z)=N(z_1)N(z_2)$. But $z_1$ and $z_2$ are nonunits, so their norm is not $\pm1$, so $N(z)$ is reducible in $\Bbb{Z}$.
