Let $n = |\mathbb{Z}[i]/(1 + i)|$. Determine the value of $n$ and show that $\mathbb{Z}[i]/(1 + i)$ is isomorphic to a field of order $n$. Justify your answer.
I have an exam tomorrow so every minute counts.
Am I right in saying that $\mathbb{Z}[i]/(1+i)$ is just $(a+bi)(1+i)$ but then surely the cardinality is infinite. Also am I right in saying that the ideal is not principal due to its norm being root 2.
On
Think about the relationships that you know have to hold in $\mathbb{Z}[i]/(1+i)$.
On the one hand, the relationship $i^2=-1$ holds, because it is inherited from the defining relationship on $\mathbb{Z}[i]$.
On the other hand, the relationship $i=-1$ holds, because when we mod out by $1+i$ we set $1+i=0$.
Putting those two relationships together, you can conclude that $(-1)^2=-1$. Now can you tell what ring this is?
The norm of $a + bi$ is defined to be $(a + bi)(a - bi)$. It is also equal to $|\mathbf{Z}[i]/(a + bi)|$. This is part of a larger theory in Algebraic Number Theory. Let us see what happens in this specific example.
The elements of $\mathbf{Z}[i]/(1 + i)$ are the cosets of $(1 + i)$ and they are respectively $(1 + i)$ and $1 + (1 + i)$. Why is this? 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 $r = 0, 1$ or $i$. Hence the cosets are $0 + (1 + i)$, $1 + (1 + i)$ and $i + (1 + i)$.
Notice also that $1 + (1 + i) = i + (1 + i)$ since $1 - i = -i(1 + i) \in (1 + i)$.
Therefore $n = 2$ and you can show that $\mathbf{Z}[i]/(1 + i) \cong \mathbf{Z}/2\mathbf{Z}$.