Prove an element in a ring is irreducible

Question

Prove that if element $a \in \mathbb{Z}[i]$ and $N(a) = \vert a \vert^2$ is a prime then $a$ is irreducible in $\mathbb{Z}[i]$.

My direction:

Suppose that $a$ is not irreducible in $\mathbb{Z}[i]$.

Give $b$ is a real divisor of $a$ in $\mathbb{Z}[i]$. Then, $\vert b \vert^2 \mid \vert a \vert^2 \Rightarrow \vert b \vert^2 \in \{1,\vert a \vert^2\}$.

Suppose that $b= x+iy (x,y\in \mathbb{Z})$.

1. Case 1: If $\vert b \vert^2 = 1$ then $x^2 +y^2 =1$.

Infer $(x,y) = (\pm 1,0)$ or $(x,y) = (0,\pm i)$

Infer $b=\pm1$ or $b=\pm i$

Both of them aren't real divisors of $a$ in $\mathbb{Z}[i]$. (conflict with the way choose $b$)

2. Case 2: $\vert b \vert^2 =\vert a \vert^2$. I stuck here now

Thanks for helping me.

Answer 1

It is not at all clear by what you mean by a 'real' divisor. Also, the notations $|a|^2$ and $|b|^2$ are rather deceptive, because $|a|,|b|\notin\Bbb{Z}[i]$ in general. Instead of your current approach, consider the following:

Given $a\in\Bbb{Z}[i]$ such that $N(a)$ is prime in $\Bbb{Z}$, let $b,c\in\Bbb{Z}[i]$ be such that $a=bc$. Then $$N(b)N(c)=N(bc)=N(a),$$ and hence either $N(b)=1$ or $N(c)=1$. Without loss of generality $N(b)=1$. You have already shown that then $b=\pm1$ or $b=\pm i$, so $b$ is a unit in $\Bbb{Z}[i]$. This proves that $a\in\Bbb{Z}[i]$ is prime.

To be fully rigorous, you might also want to show that $a$ is not a unit in $\Bbb{Z}[i]$.

Answer 2

Write $a=\zeta + \xi i; \zeta, \xi \in \mathbb Z $. Suppose $$(\alpha + \beta i)(\gamma + \delta i)=\zeta + \xi i;\alpha,\beta,\gamma,\delta \in \mathbb Z$$ Then $$(\alpha^2+\beta^2)(\gamma^2+\delta^2)=\zeta^2+\xi^2.$$ If $\zeta^2+\xi^2$ is prime, then $$\alpha^2+\beta^2=1, \text { and thus }\alpha=0, \beta= \pm 1 \text { or } \beta=0, \alpha= \pm 1$$ $$\text { or } \gamma^2+\delta^2=1, \text { and thus }\gamma=0, \delta= \pm 1 \text { or } \delta=0, \gamma= \pm 1$$. In any case, $\alpha + \beta i \text { or }\gamma + \delta i$ is invertible in $\mathbb Z[i]$, so $a=\zeta + \xi i$ is irreducible in $\mathbb Z[i]$.

Answer 3

Hint:

Write $a=bc$; then $N(a)=N(b)N(c)$, and use that $N(z)=1$ if and only if $z$ is a unit in $\mathbf Z[i]$ (which is true in the ring of integers of any finite algebraic extension of $\mathbf Q$).