Maximal Ideals in $R=\{a+bi:a,b\in \mathbb Z\}$ I've read similar question but please this is not duplicate of Maximal ideals in the ring of Gaussian integers because the answer to it contain PID which I've not yet done etc.

$R=\{a+bi:a,b\in \mathbb Z\}$ be a subring of $\mathbb C$ and $I=(7)$, $J=(13)$ in $R$. Is $I$ or $J$  maximal?

How to go with this?
 A: $(3 + 2i)(3 - 2i) \in J$, but $3+2i, 3-2i \not\in J$. This shows that the ideal $J$ is not prime. Therefore it cannot be maximal.
We prove that $R/I$ is a field. Let $a + bi \in R - I$, thus $a$ and $b$ cannot both be divisible by $7$. It is enough to show that $a + bi$ is invertible in $R/I$. But $(a+bi)(a-bi) = a^2 + b^2$. The number $a^2 + b^2$ can be shown to be invertible in $\mathbf{Z}/(7)$, hence it is also invertible in $R/I$.
To prove this, write $x = \bar{a}, y = \bar{b}$ in $\mathbf{Z}/(7)$. We have $x, y \ne 0$. Now $x^2 + y^2 = x^2[1 + (y/x)^2]$. Thus it is enough to check that $1 + z^2 \ne 0$ for all $z \in \mathbf{Z}/(7)$, which is easy to do.
NOTE
A less elementary proof can be given by showing that the rings $R/I$ and $R/J$ are isomorphic respectively to $F_7[X]/(X^2 + 1)$ and $F_{13}[X]/(X^2 + 1)$. The first is a field because $X^2 + 1$ is irreducible in $F_7[X]$. The second is not because $X^2 + 1 = (X+5)(X-5)$ in $F_{13}[X]$. These methods rely however on the fact that $K[X]$ is a PID for any field $K$.
