Exercise 15.4.3

Proof. We prove that, assuming $\alpha$ is a prime in $\mathbb{Z}[i]$, that $\mathbb{Z}∕\mathit{\alpha \mathbb{Z}}[i]$ is a field.

Since $\mathbb{Z}∕\mathit{\alpha \mathbb{Z}}[i]$ is a ring, it is sufficient to prove that any nonzero $\bar{\beta }=\beta +\mathit{\alpha \mathbb{Z}}[i]$ has an inverse.

$\bar{\beta }\ne \bar{0}$ is equivalent to $\beta \notin \mathit{\alpha \mathbb{Z}}[i]$, so that $\alpha$ doesn’t divide $\beta$ in $\mathbb{Z}[i]$.

Since $\alpha$ is a prime in the principal ideal domain $\mathbb{Z}[i]$, $\alpha$ is relatively prime to $\beta$: if $\gamma$ divides $\alpha$ and $\beta$, then $\gamma$ is associate to $1$ or $\alpha$, but if $\gamma$ is associate to $\alpha$, then $\alpha$ divides $\beta$, and this contradicts the hypothesis, therefore $\gamma$ is associate to $1$. This proves that $\alpha$ is relatively prime to $\beta$.

Since $\mathbb{Z}[i]$ is a PID, this shows that there are some $\lambda ,\mu \in \mathbb{Z}[i]$ such that

thus, using $\bar{\alpha }=\alpha +\mathit{\alpha \mathbb{Z}}[I]=\mathit{\alpha \mathbb{Z}}[i]=\bar{0}$,

This proves that $\bar{\beta }=\beta +\mathit{\alpha \mathbb{Z}}[i]$ has inverse $\bar{\lambda }=\lambda +\mathit{\alpha \mathbb{Z}}[i]$ in $\mathbb{Z}[i]∕\mathit{\alpha \mathbb{Z}}[i]$.

If $\alpha$ is a prime in $\mathbb{Z}[i]$, then $\mathbb{Z}∕\mathit{\alpha \mathbb{Z}}[i]$ is a field.

Moreover, by Exercise 2, $|\mathbb{Z}∕\mathit{\alpha \mathbb{Z}}[i]|=N(\alpha )$, so that

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