The quotient ring $\mathbb{Z}[i]/(1 + i)$

Question

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.

Please try searching. I searched site:math.stackexchange.com gaussian integers quotient and got many results: 1, 2, 3, 4 — Viktor Vaughn, Jun 04 '17 at 18:03
If you look more closely at $(a + bi)(1 + i)$, you should find a way to see that both $2a$ and $2bi$ are in $(1 + i)$, for any $a, b \in \Bbb Z$, hinting that some even-versus-odd stuff might happen... — pjs36, Jun 04 '17 at 18:09
See here for an elementary proof that it's isomorphic to $,\Bbb Z/2.\ $ — Bill Dubuque, Jun 04 '17 at 18:50

score 6 · Accepted Answer · answered Jun 04 '17 at 18:04

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 $\mathbf{Z}[i]/(1 + i) \cong \mathbf{Z}/2\mathbf{Z}$.

the ^absolute value of the norm of $a+bi$ equals $|\mathbb Z[i]/(a+bi)|$ — cat, Jun 04 '17 at 19:22
@cat In general, the product of the Galois-conjugates of $a$ and the size of $\mathcal O_K/(a)$ are both called the "norm of $a$" (although possibly with some adjective like "ideal norm" or "field norm") and are equal up to sign. For the ring of Gaussian integers they are the same because $a^2 + b^2$ is always non-negative. — Trevor Gunn, Jun 04 '17 at 19:42

score 1 · Answer 2 · answered Jun 04 '17 at 18:10

1

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?

answered Jun 04 '17 at 18:10

mweiss

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There are many rings satisfying $1=-1$, not only $\mathbb{Z}/2$. – Dietrich Burde Jun 04 '17 at 18:53
True, but this one is generated by a single element. – mweiss Jun 04 '17 at 19:01

The quotient ring $\mathbb{Z}[i]/(1 + i)$

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