Ideal as kernel of a homomorphism in Gaußian integers Consider the ring $\mathbb{Z}[i]$ of Gaußian integers. The principal ideal $(1+i)$ is maximal ideal in this ring. Since ideals are kernels of some homomorphisms, I would like to see a homomorphism from $\mathbb{Z}[i]$ to some ring whose kernel is $(1+i)$.
Of course, it is kernel of the natural homomorphism

$$f:\;\mathbb{Z}[i]\rightarrow \mathbb{Z}[i]\big/(1+i).$$

However, I tried to define some nice map $f$ whose kernel will be $(1+i)$. I tried this one: $f\colon \mathbb{Z}[i]\rightarrow \mathbb{Z}$, $m+ni\mapsto m-n$, but this $f$ is not a homomorphism. Can you help me to modify (or to define another homomorphism) whose kernel is the ideal $(1+i)$. 
 A: For completeness, I'll do everything from scratch in case future viewers are wondering how you know $(1+i)$ is maximal.
You can note that $(1+i)$ is maximal because it has a prime norm (hence it is irreducible) and $\Bbb Z[i]$ is a Euclidean ring under the norm (hence also a PID), so prime ideals are maximal and so the quotient is a field.
We note that $(1+i)|2$--indeed $(1+i)(1-i)=2$--hence $2\equiv 0\pmod{1+i}$, i.e. the field has characteristic $2$. Indeed, we can see even more, since $2\equiv 0\pmod{1+i}$ we may reduce $m,n\mod 2$ to one of four total choices (in the quotient)

$$\begin{cases} m\equiv n\equiv 0\mod 2 \\ m\equiv n+1\equiv 0\mod 2 \\ m\equiv n+1\equiv 1\mod 2 \\ m\equiv n\equiv 1\mod 2\end{cases}$$

So the field has at most $4$ elements. Noting that also (and even more trivially)
$$1+i\equiv 0\pmod{1+i}$$
we may identify the first and last case as well as the second and third, so the image is the field with two elements, namely $\Bbb Z/2\Bbb Z$. Any homomorphism which sends things of the same parity to $0\mod 2$ must agree with the canonical projection
$$\Bbb Z[i]\to\Bbb Z[i]/(1+i)\cong\Bbb Z/2\Bbb Z$$
so clearly

$$m+ni\mapsto m+n\mod 2$$

is a formula for the homomorphism onto the quotient, and since $1\not\equiv 0\pmod{1+i}$ we have that we may choose the image of $1, i\in \Bbb Z[i]$ to be $1$ in the quotient and the image of $0, 1+i$ to be $0$, and all other cases reduce to these by the earlier $\mod 2$ discussion.
Note: Your choice of $m+ni\mapsto m-n$ is actually not a bad one, the only thing you forgot is that you must take $m,n\mod 2$. We see that

$$1\equiv-1\mod 2\implies m+n\equiv m-n\mod 2$$

so this is just another way of expressing the same homomorphism.
A: You can also think of this in the following way:
Look at the composition
$\mathbb Z[i] \to \mathbb Z[i]/(1+i) \cong \mathbb Z[X]/(X^2+1,X+1) = \mathbb Z[X]/(2,X+1) \cong (\mathbb Z/2\mathbb Z[X])/(X+1) \cong \mathbb Z/2\mathbb Z$
The first isomorphism is given by $a+bi \mapsto a+bX$ and the last isomorphism is given by $X \mapsto -1$. Hence the composition is given by $a+bi \mapsto a-b \mod 2$.
The same calculation works for every maximal ideal $(c+di) \subset \mathbb Z[i]$ with $c \neq 0 \neq d$ (this imples $c^2+d^2=p$ for some prime $p$ and we have the equality of ideals $(X^2+1,dX+c)=(p,dX+c)$ in $\mathbb Z[X]$). We get an explicit isomorphism
$\mathbb Z[i]/(c+di) \to \mathbb Z/p\mathbb Z, a+bi \mapsto a-bcd^{-1} \mod p$
