Prime elements of the ring $\mathbb Z[i]$.

Question

I want to find the prime elements of the ring $\mathbb Z[i]$ of Gaussian integers.Define $d(a+ib)=a^2+b^2$ for all $a+ib\in \mathbb Z[i]-\{0\}$.I think if $d(a+ib)$ is prime in $\mathbb Z$ then and then only $a+ib$ is prime in $\mathbb Z[i]$,I am not sure whether the claim is true or not.I am not able to prove nor disprove this statement.Can someone help me to find a way out?

Answer 1

This is too long for a comment so I will post this as an answer. There is an exercise in Marcus' Number Fields:

Let $\alpha\in\mathbb{Z}[i]$. Show that if $N\left(\alpha\right)$ is a prime in $\mathbb{Z}$ then $\alpha$ is irreducible in $\mathbb{Z}[i]$. Show that the same conclusion holds if $N\left(\alpha\right)=p^{2}$, where $p$ is a prime in $\mathbb{Z}$, $p\equiv 3\pmod{4}$.

Proof. Note that $\alpha\in\mathbb{Z}[i]$ is an irreducible element if and only if $\alpha\neq 0$ and $\alpha\not\in\mathbb{Z}[i]^{\times}$, and for all $\beta,\gamma\in\mathbb{Z}[i]$ such that $\alpha=\beta\gamma$, one has $\beta\in\mathbb{Z}[i]^{\times}$ or $\gamma\in\mathbb{Z}[i]^{\times}$. Now suppose that $\alpha=\beta\gamma$ in $\mathbb{Z}[i]$. Then $N\left(\alpha\right)=N\left(\beta\right)N\left(\gamma\right)$ in $\mathbb{Z}$. If $N\left(\alpha\right)$ is prime in $\mathbb{Z}$, then either $N\left(\beta\right)=1$ or $N\left(\gamma\right)=1$ which implies that either $\beta$ or $\gamma$ is a unit in $\mathbb{Z}[i]$. Thus, $\alpha$ is irreducible in $\mathbb{Z}[i]$.

Now suppose that $N(\alpha)=p^{2}$ where $p\equiv 3\pmod{4}$. Then there are three cases. The first case when $N(\beta)=p^{2}$ implies that $\gamma$ is a unit and so $\alpha$ is irreducible in $\mathbb{Z}[i]$. The second case when $N(\gamma)=p^{2}$ implies that $\beta$ is a unit and so $\alpha$ is irreducible in $\mathbb{Z}[i]$. It remains to consider the case when $N(\beta)=N(\gamma)=p$. But this is impossible since if we write $\beta=c+di\in\mathbb{Z}[i]$, then \begin{align*} p=N(\beta)=c^{2}+d^{2} \end{align*} and after taking modulo $4$, we see that $3\equiv c^{2}+d^{2}\pmod{4}$. But the squared of any integer is either congruent to $0$ or $1$ modulo $4$, and hence, $N(\beta)\not\equiv 3\pmod{4}$, a contradiction. Thus, $\alpha$ is irreducible in $\mathbb{Z}[i]$.

Note that $\mathbb{Z}[i]$ is a Euclidean domain and so is a UFD (in particular an integral domain), so every irreducible is prime and vice versa.