Checking if $\langle 2 \rangle$ is a maximal ideal in $\mathbb{Z}[i]$ Is $\langle 2 \rangle$ a maximal ideal in $\mathbb Z[i]$?
Solution: We know that $\mathbb Z[i]$ is an Euclidean domain and hence a principal ideal domain.
Consider $2 \in \mathbb Z[i]$.                           
Then $N(2) = 2^2 = 4$. (NOTE: $N$ is norm).
Since $N(2)$ is not prime, this implies that $2$ is reducible in $\mathbb Z[i]$.
And this implies that $\langle 2 \rangle$ is not maximal.
I want to know is there any mistake in this?
Thank you.
 A: As Marc indicates, your reasoning is incorrect: $N(z)$ composite does not justify saying that the number $z$ is reducible. Consider Mark's example; $N(3)=3^2$ is composite but $3$ is irreducible.
There are two ways you could go about this problem: factor $2$ (hint: it's associate to a perfect square of a Gaussian integer; find out the absolute value of this Gaussian integer and then go from there), or show the quotient ${\Bbb Z}[i]/(2)\cong{\Bbb Z}[x]/(2,x^2+1)\cong{\Bbb F}_2[x]/(x^2+1)$ is not a field (which means showing that $x^2+1$ is reducible over ${\Bbb F}_2$) since $M\trianglelefteq R$ is maximal $\Leftrightarrow R/M$ is a field.
(It's no coincidence that $2$ is (associate to) a perfect square in ${\Bbb Z}[i]$ and $x^2+1$ is a perfect square in the ring ${\Bbb F}_2[x]$; as a bonus/challenge exercise, find the connection between these two facts.)
A: Although you have arrived at the correct conclusion, you got there prematurely. Though there is more than one way to properly arrive at the conclusion, some more "elementary" than others.
The fact that $N(2) = 4$ is composite suggests that 2 is composite. But to be sure, we need to find at least one prime factor $p$, and then we'll observe that $1 < N(p) < N(2)$ (remember that $N(u) = \pm 1$ means that $u$ is a unit, not a prime; I make sure to address the issue of signs because there are domains that have numbers with negative norms).
Another helpful observation is the fact that the norm is multiplicative. If $2 = pq$, then $N(p) N(q) = N(2) = 4$. If 2 indeed has a nontrivial prime factor, then $N(p) = 2$ is the only possibility. This should soon lead to the discovery that $(1 - i)(1 + i) = 2$.
Therefore $\langle 2 \rangle \subset \langle 1 + i \rangle \subset \mathbb Z[i]$; no subsets can be inserted here, only subset-or-equals.
