Prove $R = \{ a + bi : a,b \in \Bbb Z \}$ is a subring in $\Bbb C$ and that $M = \{ a + bi : 3 \mid a \text{ and } 3\mid b\}$ is a maximal ideal in $R$. I was able to prove that $R$ is a subring of $\Bbb C$ easily, since $\forall $ $x,y \in $ R,$x - y \in R$, and $xy \in R$. So indeed $R$ is a subring of $\Bbb C$. Now, it was also easy to show that $M$ is an ideal in $R$, since $M \subseteq R$ and, $\forall $ $c,d \in M$, $c - d \in M$, and , $\forall r \in R$, $cr=rc \in M$.
By a hint, I was guided to show that if $r + si \notin M$, then $3 \not\mid $ r or $3 \not\mid s $. Now, it is easy to show this implies that $3 \not\mid ( r^2 + s^2 )$. Then, I need to show that any ideal $I$ containing $r + si$ and $M$ must also contain $1$. I would be very grateful for any ideas. Much obliged
Here is a hint, based on what you have already shown: if $I$ contains $M$ and $r+si$, then in particular $I$ contains $3$ and $r^{2}+s^{2}$. Since $3$ does not divide $r^{2}+s^{2}$ and $3$ is a prime number, it follows that $\gcd(r^{2}+s^{2}, 3) = 1$. Now use Bezout's lemma; feel free to comment if you need more assistance from here.