Prove $(7)$ is maximal in Gaussian Integers

Question

I would like to prove that $(7)$ is a maximal principal ideal in Gaussian Integers. I approached this problem by trying to prove that $7$ is prime in $\Bbb Z[i]$. However, I am not sure how to proceed from here.

Suppose $7 = (a+bi) (c+di)$. Then, \begin{align*} \begin{bmatrix} c & -d \\ d & c \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} &= \begin{bmatrix} 7 \\ 0 \end{bmatrix} \\ \begin{bmatrix} a \\ b \end{bmatrix} &= \frac{1}{c^{2} + d^{2}} \begin{bmatrix} c & d \\ -d & c \end{bmatrix} \begin{bmatrix} 7 \\ 0 \end{bmatrix} \\ \begin{bmatrix} a \\ b \end{bmatrix} &= \frac{7}{c^{2} + d^{2}} \begin{bmatrix} c \\ -d \end{bmatrix} \end{align*}

This does not seem to lead me anywhere.

Answer 1

The maximal ideals of $\mathbb Z[x]$ are of the form $(p, f(x))$ where $p$ is a prime and $f(x)$ is an irreducible polynomial mod $p$.

Given that description, since $\mathbb Z[i]/(7)\cong \mathbb Z[x]/(7, x^2+1)$, you can conclude that the denominators are maximal ideals after confirming $x^2+1$ is irreducible mod $7$.

Notice by the same token $(2)$ is not a maximal ideal in the Gaussian integers.

Answer 2

Prime element corresponds to prime ideal. Irreducible element corresponds to maximal ideal. So let's prove that $7$ is irreducible in $\mathbb Z[i]$.

If $7= \alpha\beta$, then $49=N(7)=N(\alpha\beta)=N(\alpha)N(\beta)$ and so $N(\alpha),N(\beta)\in \{1,7,49\}$. We can't have $N(\alpha)=7$ and so $N(\alpha)=1$ or $N(\beta)=1$, that is, $ \alpha$ or $\beta$ is a unit, as required.