Is $\langle1-i\rangle$ a maximal ideal in $\mathbb{Z}[i]$?

Question

I'm trying to show that $\langle1-i\rangle$ is a maximal ideal of $\mathbb{Z}[i]$. I started by assuming there is some ideal $A$ that properly contains $\langle1-i\rangle$, and then I want to show that $1 \in A$, and then I will have that $A= \mathbb{Z}[i] $, so $\langle1-i\rangle$ is maximal. Is this the right approach?

Thanks

Second Attempt: Because $\forall a+bi \in \mathbb{Z}[i]$, we have $a+bi = (a+b) -b(1-i)$, then we have $a+bi - (a+b) = -b(1-i) \in \langle1-i\rangle$, which implies $a+bi +\langle1-i\rangle = a+b +\langle1-i\rangle$. So every coset is of the form $c +\langle1-i\rangle$, where $c\in \mathbb{R}$. Furthermore, $-i(1-i)^2=2 \in \langle1-i\rangle$. Write $c = 2q + r$, where $q,r\in \mathbb{R}$ and $r=0,1$. Then because $2 \in \langle1-i\rangle$, we have $2q\in \langle1-i\rangle$ and therefore $2q= 2q+r-r = c-r\in \langle1-i\rangle$, so $c+\langle1-i\rangle = r+\langle1-i\rangle$, and every coset is of the form $0 + \langle1-i\rangle$ or $1 +\langle1-i\rangle$. Thus, $1 +\langle1-i\rangle$ is the only nonzero element, and is invertible with itself as its inverse. Thus, $\mathbb{Z}[i]/\langle1-i\rangle$ is a field with two elements, and by a theorem, $\langle1-i\rangle$ must be maximal.

Is this correct?

Answer 1

One way to do this is the following: An ideal $J$ containing $I:= \langle 1-i\rangle$ is also a principal ideal (since $\mathbb{Z}[i]$ is a Euclidean domain, and hence a PID), and so must be generated by an element $x = a+bi \in \mathbb{Z}[i]$. Since $I\subset J$ it follows that $x\mid 1-i$, so to prove your claim, it suffices to show that $1-i$ is an irreducible element of $\mathbb{Z}[i]$.

Suppose $1-i = zw$, then taking the norm-squared (in $\mathbb{C}$) on both sides, you see that $$ 4 = |z|^2|w|^2 $$ So consider the cases $|z|^2 \in \{1,2,4\}$. Prove that $2$ is impossible, and so conclude that $1-i$ is irreducible.

Answer 2

Here's another way to do it: We know that an ideal $I \subset R$ is maximal $\iff R/I$ is a field (Prove this).

Now let's define a homomorphism $\phi:\mathbb{Z}[i] \rightarrow \mathbb{Z}_2$ the following way: Given any $(a+bi) \in \mathbb{Z}[i]$:

$$\phi(a + bi) = [a+b]_2$$

You should show that this is indeed a (surjective) homomorphism. Noting that $\ker(\phi) = \langle1-i\rangle$, we simply apply the isomorphism theorem:

$$Z[i]/\langle1-i\rangle \cong \mathbb{Z}_2$$

We conclude $\langle1-i\rangle$ is maximal since $\mathbb{Z}_2$ is a field.

Answer 3

The norm of the ideal $(1-i)$ is the ideal $(2)$ of $\mathbf{Z}$.

If $(1-i)$ factored into (proper) ideals as $IJ$, then the norms of $I$ and $J$ would be a factorization of $(2)$, which is impossible because that would mean one of the two ideals has norm $(1)$, and thus would be the unit ideal.