Prove that $I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}$ is an maximal ideal of $ℤ[i]$. I'm making some exercises to prepare for my ring theory exam:

Prove that $I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}$ is an maximal ideal
  of $ℤ[i]$.

I know that $a+2l=b$ with $l\in ℤ$ (or should I say $l \in ℤ[i]$)? Therefore I can write $a+(a+2l)i=a+ai+2li$. So I don't have $0+i$ in my ideal. For even real parts I have even complex parts, and for odd real parts I have odd complex parts. But I don't see what I can conclude now. 
 A: Hint: (1) Check that $I$ is an ideal of $\mathbb Z[i]$, that is an additive subgroup with $\mathbb Z[i]\cdot I \subseteq I$.
(2) To show it is maximal, let $a+bi \in \mathbb Z[i] \setminus I$. Then $a-1 + bi \in I$ (why?). What can you conclude about the ideal $J = (I, a+bi)$, noting $a+bi - (a-1+bi) = 1$?
A: To prove that an ideal is maximal, it's enough to show that if you add anything to it, you get the full ring $\mathbb Z[i]$. So, let $z = c + id$ not be in your ideal, so that $c \not \equiv d \pmod{2}$, and look at $J = I + (z)$. We want to show that $J = \mathbb Z[i]$.
Subtracting the appropriate element of $I$ from $z$, you find that either $i \in J$ (if $c$ is even and $d$ is odd, so that $c + (d-1) i \in I$) or $1 \in J$ (if $c$ is odd and $d$ is even, so that $(c-1) + d i \in I$). If $1 \in J$, you are done, because $1$ belongs to no proper ideal. If $i \in J$, then $i \cdot (-i) = 1 \in J$, so you are done as well!
A: $I$ is an ideal because clearly $\mathbb Z[i]\cdot I=I$ (that's very easy to verify by definition). Then it's sufficient to say that $I$ has index $2$ as a subgroup and $2$ is a prime number therefore the only supergroup of $I$ inside $\mathbb Z[i]$ is the whole $\mathbb Z[i]$. The index is $2$ because there're $2$ congruence classes modulo $2$.
