# Prove that $(1+i)⊂\mathbb{Z}[i]$ is a maximal ideal.

For $(1+i)$ to be a maximal ideal, I know that $\mathbb{Z}[i]/(1+i)$ needs to be a field. I am just not sure how to show every element has a multiplicative inverse. Or is there another way to show it is a field?

• How many elements does $\mathbb Z[i]/(1+i)$ have? How many rings with that number of elements are there? – Henning Makholm Oct 24 '17 at 16:28
• If you know that $\mathbb{Z}[i]$ is a PID, then it suffices to show that $1+i$ is irreducible - and that follows from the fact that $|1+i|^2 = 2$ is prime. – Daniel Schepler Oct 24 '17 at 19:48

Note that in your quotient ring, $2=(1+i)(1-i)$ goes to $0$, as does $2i$. So if it is to be a field, then it is either the field with two elements or the field with four elements. Now you can just check whether your quotient ring is isomorphic to either of them, and you're done.
Alternatively, $\Bbb Z[i] \cong \Bbb Z[x]/(x^2 + 1)$ by definition. By the third isomorphism theorem, $\Bbb Z[i]/(1+i)\cong \Bbb Z[x]/(x^2 + 1, x+1)$. Now we have that $$(x^2 + 1, x + 1) = ((x+1)(x-1) + 2, x+1) = (2, x+1)$$ Divide out by this ideal instead, and the result might be clearer.
Given $a+bi\in\mathbb{Z}[i]$, consider $$a+bi=(a-b)+b(1+i) \tag{*}$$ Observe that $(1+i)(1-i)=2$, so if $a-b=2q+r$ with $0\le r<2$, we have $$a+bi=r+((1-i)q+b)(1+i)$$ so $a+bi\in(1+i)$ if and only if $r=0$ (that is, $a-b$ is even, but it's not really relevant).
Therefore the number of cosets in $\mathbb{Z}[i]/(1+i)$ is two, because $r$ can only take the values $0$ and $1$ and a ring with two elements is a field.
This may seem a posteriori knowledge, but the information that $(1+i)(1-i)=2$ is very basic and the step (*) is the first thing to do anyway.
Here is an uninformed approach: check that the map $$\mathbb{Z}[i]\ni a+bi\mapsto (a+b) \bmod{2}\in \mathbb{Z}/2$$ is a surjective morphism of rings with kernel exactly the ideal $(1+i)$.