Showing that a number is prime if norm is prime

Question

Suppose that $\alpha\in\mathbb{Z}[i]$, and $N(\alpha)=\alpha .\bar{\alpha} =p$ (norm), a prime in $\mathbb{Z}$. Then show that $\alpha$ is a prime in $\mathbb{Z}[i]$.

My attempt: Let $\alpha|ab$ for some $a,b\in\mathbb{Z}[i]$ and if we assume that, $\alpha\not| b$ and $\alpha\not| a$ then $N(\alpha)\not|N(a)$ and $N(\alpha)\not|N(b)$ (I showed this using the fact that $\mathbb{Z}[i]$ is a euclidean domain). But this is a contradiction as $\alpha|ab\Rightarrow N(\alpha)|N(a)N(b)\Rightarrow N(\alpha)|N(a)$ or $N(\alpha)|N(a)$.

Is the approach correct? Can anyone suggest some alternatives?

Thank You.

Answer 1

Suppose $a=xy$ then $p=N(a)=N(x)N(y)$ this means that $N(x)=1$ or $N(y)=1$ thus $a$ is irreducible in $\mathbb{Z}[i]$ and since it is a PID it is also prime.

Answer 2

Hint:

$$p=ab\implies p^2=\mathcal N(p)=\mathcal N(a)\mathcal N(b)\iff\mathcal N(b)\,\mid\,p\;\;\text{or}\;\;\mathcal N(a)\,\mid\,p$$

Take it from here.