Prove that the quotient ring $\frac{\mathbb{Z[i]}}{(1+i)}$ is a field of order $2$
My attempt : I found the answer here see page no $1$
It is written that
By computing the norm we know $1+i$ is irreducible in $\mathbb{Z[i]}$, so $\mathbb{Z[i]}/(1+i)$ is indeed a field. Any element in $\mathbb{Z}[i]$ has the form $a+bi$ for some $a, b \in\mathbb{ Z}$. If $a ≡ b \mod 2$, then $(1+i)( (a+b)/2 + (b−a)i/2) =a + bi$, so $a + bi = 0$ in the quotient field.If $a\neq b \mod 2$, then $a − b ≡ b \mod 2$, in which case by a similar computation we have $(a + bi) − 1 = 0 $in the quotient field. Thus the only elements in the quotient field is $0$ and $1$, so we obtain a finite field of order $2.$
My confusion :Im not getting this statement
If $a\neq b \mod 2$, then $a − b ≡ b \mod 2$, in which case by a similar computation we have $(a + bi) − 1 = 0 $in the quotient field.
First confusion :If $a\neq b \mod 2$, then why $a − b ≡ b \mod 2$?
2nd confusion:why do we have $(a + bi) − 1 = 0 $ in the quotient field?.
My thinking :If $a ≡ b \mod 2$,then $a+b$ and $b-a $ are even $\implies a+bi \in (1+i)$
The elements of $\mathbf{Z}[i]/(1 + i)$ are the cosets of $(1 + i)$ and they are respectively $(1 + i)$ and $1 + (1 + i)$. Take an element $a + bi \in \mathbf{Z}[i]$. By the division theorem, we can write $a + bi = q(1 + i) + r$ with $N(r) < N(1 + i) = 2$. That is $r = 0, 1$ . Hence the cosets are $0 + (1 + i)$, $1 + (1 + i)$
Therefore every element of $\mathbb{Z}[i] $ is in either $(1+i)$ or $1+ (1+i)$ so $\frac{{Z}[i]}{(1+i)}$ is a finite ring of order $2$,which must be a field
