Prove that $\mathbb{Z}[i]/\langle 1+i \rangle \cong \mathbb{Z}/2\mathbb{Z}$ [duplicate]
Prove that: $\mathbb{Z}[i]/\langle 1+i \rangle \cong \mathbb{Z}/2\mathbb{Z}$.
This is my first time using the First Isomorphism Theorem for Rings and I am looking for feedback for whether some steps in this proof could have been done better. From my understanding, I need to:
Define a surjective homomorphism $\phi$ between $\mathbb{Z}[i]$ and $\mathbb{Z}/2\mathbb{Z}$.
Show that $\ker(\phi) = \langle 1+i \rangle$.
Define $\phi: \mathbb{Z}[i] \to \mathbb{Z}/2\mathbb{Z}$ by $\phi(a+bi)=[a+b]_2$.
$\boldsymbol {\phi} \textbf{ is surjective:}$
Given $a+bi$, we have $a,b \in \mathbb{Z}$. So take $b$ to be $0$, then we have $\phi (a+0i)=[a]_2$, and clearly we can get $0$ and $1$.
$\boldsymbol{\phi} \textbf{ is a homomorphism:} $
$$\begin{align} \phi((a+bi)+(c+di)) &=\phi((a+c)+(b+d)i) \\ &=[(a+c)+(b+d)]_2+[c+d]_2 \\ &=[(a+b)+(c+d)]_2 \\ & =[a+b]_2+[c+d]_2 \\ &= \phi(a+bi)+\phi(c+di).\end{align}$$
$$\begin{align} \phi((a+bi)(c+di)) &= \phi((ac-bd)+(ad+bc)i) \\ &=[(ac-bd)+(ad+bc)]_2 \\ &=[(ac+bd)+(ad+bc)]_2 \\ &=[(a+b)(c+d)]_2 \\ &=[a+b]_2[c+d]_2 \\ &=\phi(a+bi)\phi(c+di). \end{align}$$
$\boldsymbol{\ker(\phi)=\langle 1+i \rangle}$:
We have that $\ker(\phi) = \{ (a+bi) \in \mathbb{Z}[i]: \phi(a+bi) =[a+b]_2 =[0]_2\}$.
From my understanding, elements in $\langle 1+i \rangle$ are just of the type $(1+i)(c+di)$ where $c,d \in \mathbb{Z}$.
I've tried to show that one is a subset of the other. Another idea I have is in noting that $[a+b]_2=[0]_2 \iff 2|(a+b)-0$ and somehow link it to showing that these are the same elements in $\langle 1+i \rangle$, but I could not do it. I have not managed to come up with a reasonable proof in any of these approaches. How does one actually show this?
The construction of the homomorphism is good. Since $\phi(1+i)=[0]_2$, you know that $\langle 1+i\rangle\subseteq\ker\phi$. Conversely, if $a+bi\in\ker\phi$, you have $a+b=2c$ for some $c$ and $$ a+bi=(1+i)(x+yi) $$ where $x+y=b$, $x-y=a$ (and the solution exists) because $a+b$ and $a-b$ are even.
However, you're not using the homomorphism theorem. For this you can proceed in several steps. Consider the unique homomorphism $\alpha\colon\mathbb{Z}[X]\to\mathbb{Z}/2\mathbb{Z}$ such that $\alpha(X)=[1]_2$ ($X$ is an indeterminate). The ideal $\langle X^2+1\rangle$ is contained in the kernel of $\alpha$, since $\alpha(X^2+1)=[1]_2^2+[1]_2=[0]_2$.
The homomorphism theorem provides a surjective homomorphism $\beta\colon\mathbb{Z}[X]/\langle X^2+1\rangle\to\mathbb{Z}/2\mathbb{Z}$, that is, $\beta\colon\mathbb{Z}[i]\to\mathbb{Z}/2\mathbb{Z}$.
You can notice that this homomorphism is exactly your $\phi$. Then you can proceed as before.
If you already know that $\langle 1+i\rangle$ is maximal, the proof ends with the observation that $\langle 1+i\rangle\subseteq\ker\beta$.
You are almost there. Elements in $\langle 1+i \rangle$ are exactly those that you described: those of the form $(1+i)(c+di)$. These are clearly in the kernel (because $1+i$ is in the kernel and you already showed $\phi$ takes products to products).
The other containment is more interesting. Given $a + bi$ in the kernel, you need to show that you can divide it by $1+i$. More precisely, you need to show that there exists $c + di \in \mathbb{Z}[i]$ such that $a + bi = (1+i)(c+di)$.
My hint would be that you can guess the formula for $c+di$ by simplifying the fraction $\frac{a+bi}{1+i}$. Be sure to verify that everything that shows up in your simplified fraction is an integer. To do this you will use that $a+bi$ is in the kernel, i.e. that $a+b$ is even.
There is a common sense way of answering this question. Let's find representatives for the quotient.
Clearly $(1+i)(1-i)= 2$ is in the ideal. Thus any element of $\mathbb{Z}$ is equivalent to one of $0,1,i,1+i$ modulo $1+i$.
But clearly $1+i$ is equivalent to $0$ and $1$ is equivalent to $i$. Also $0,1$ are inequivalent mod $1+i$.
Thus the ring $\mathbb{Z}[i]/\langle 1+i\rangle$ has two elements, hence must be isomorphic to $\mathbb{Z}/2\mathbb{Z}$.