Is there a non-advanced theorem (or some other simple way) to show that $\langle 1+i \rangle$ is prime in $\mathbb{Z}[i]$? I'm somewhat stuck. I've so far argued that any even integer and any number of the form $a+ib$, where $a,b$ are either both even or both odd, is in this ideal. But any number of this form that is divisible into some number $\ne 1+i$ is also of the same form, so it must still be in this ideal. Hence, this ideal is prime.
But I don't feel this is quite rigorous.
He asked for the simplest way, so probably he does not want to use quotients. Let $a+bi, c+di \notin \langle 1+i \rangle$, i.e. by the definition of a prime ideal we have to show that the product is also not contained in the ideal.
You have already figured out that an element is contained in the ideal if and only if the parity of the real and imaginary part are the same.
This translates into $a-b = c-d = 1 \mod 2$. Note that there are no signs mod 2, i.e. also $c+d=a+b=1 \mod 2$.
We have $$(a+bi)(c+di)=ac-bd + (ad+bc)i$$ and compute $$ac-bd-(ad+bc) = a(c-d)-b(c+d)=a-b=1 \pmod 2,$$ hence $$(a+bi)(c+di)=ac-bd + (ad+bc)i \notin \langle 1+i \rangle.$$
HINT: Consider the map $\phi : \Bbb Z[i] \to \Bbb Z_2$ by $a+ib \to a-b$. See that the Kernel of the map $\phi$ is $(1+i)$.
Now $\Bbb Z[i]/(1+i) \simeq Z_2$.
Since $\Bbb Z_2$ is a field we conclude that $(1+i)$ is a maximal ideal and hence a prime ideal.
A proper ideal $I$ of a ring $R$ is prime if and only if $R/I$ is an integral domain (i.e. it has no zero divisors).
One of the most useful calculation devices for computing rings is that the "evaluation at $a$" map is an isomorphism
$$ R[x] / (x-a) \to R $$
We can use this if we convert the problem to one of polynomial rings:
$$ \mathbb{Z}[i] \cong \mathbb{Z}[x] / (x^2 + 1) $$
and consequently,
$$ \begin{align*} \mathbb{Z}[i] / \langle i+1 \rangle &\cong \left( \mathbb{Z}[x] / (x^2 + 1) \right) / \langle x+1 \rangle \\&\cong \mathbb{Z}[x] / (x^2 + 1, x+1) \\&\cong \left( \mathbb{Z}[x] / (x + 1) \right) / \langle x^2 + 1 \rangle \\&\cong\mathbb{Z} / \langle (-1)^2 + 1 \rangle \\&\cong \mathbb{Z} / \langle 2 \rangle \\&\cong \mathbb{F}_2 \end{align*}$$
Since $\mathbb{F}_2$ is an integral domain, conclude $\langle i+1 \rangle$ is prime. In fact, we can say more: because the result is a field, we know $\langle i+1 \rangle$ is a maximal ideal.
For any positive integer $n$ we consider the homomorphism $\chi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/ I, z\mapsto z+ I$, where $I$ is the principal ideal of $\mathbb{Z}[i]$ generated by the element $n+i$ of $\mathbb{Z}[i]$. A simple calculation shows that $\ker\chi= J$, where $J$ is the principal ideal of $\mathbb{Z}$ generated by the element $n^2+1$. Moreover, $\forall a, b\in \mathbb{Z}$ it holds $(a+bi)+ I=(a-nb)+I$. Hence, $\chi$ is an epimorphism. Now the first isomorphism theorem tell us that $\mathbb{Z}/J\cong \mathbb{Z}[i]/ I$.
For $n=1$, we have $J=(1+i)$, $I=(2)$ and $\mathbb{Z}/(2)\cong \mathbb{Z}[i]/ (1+i)$ Since $\mathbb{Z}/(2)$ is a field $(1+i)$ is maximal ideal and hence prime.
Generally, $(n+i)$ is a maximal ideal of $\mathbb{Z}[i]$ iff $n^2+1$ is a prime integer.
