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

Question

When I first looked at this problem, I thought that $\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/5\mathbb{Z}$, but apparently the correct answer is $\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/2\mathbb{Z}$.

Here's where I'm confused: saying that $\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/2\mathbb{Z}$ is saying that there is one element in $\mathbb{Z}[i]$ that is divisible by $1+i$, which is the $0$ element in $\mathbb{Z}[i]/(1+i)$, and there there's only ONE other element in ${\mathbb Z}[i]$ that is not divisible by $1+i$, thus it is isomorphic to $\{0,1\}$. I'm just not seeing how this isomorphism makes any sense, whatsoever...

A complete proof would be helpful, but I guess I'm more looking for intuition.

Bill Dubuque · Accepted Answer · 2014-03-18T21:17:43.567

There is a natural ring image of $\,\Bbb Z\,$ in $\,R = \Bbb Z[i]/(1\!+\!i)\,$ by mapping integer $\,n\,$ to $\ n \pmod{1\!+\!i}$ by composing the natural maps $\,\Bbb Z\to \Bbb Z[i]\to \Bbb Z[i]/(1+i).\,$ This map $\, h\color{#0a0}{ \ {\rm is\ surjective\ (onto)}}$ since $\,{\rm mod}\ 1\!+\!i\!:\ \,1\!+\!i\equiv 0\,\Rightarrow\,i\equiv -1\,\Rightarrow\, a+bi\equiv a-b\in\Bbb Z.\,$ Finally, let's compute the kernel $\,I\,$ of the ring homomorphism $\,h.\,$ By rationalizing a denominator, $\,I = \color{#c00}{\ker h = 2\,\Bbb Z}\,$ as follows

$$ n\in I\iff (1\!+\!i)\mid n\ \ {\rm in}\ \ \Bbb Z[i]\iff \dfrac{n}{1\!+\!i}\in \Bbb Z[i]\iff \dfrac{n(1\!-\!i)}{2}\in\Bbb Z[i]\iff \color{#c00}{2\mid n}$$

Therefore, applying the First Isomorphism Theorem, $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{2\,\Bbb Z}.$

More generally see here – Bill Dubuque Oct 15 '22 at 16:51 — Bill Dubuque, Oct 15 '22 at 16:51

score 3 · Answer 2 · edited May 06 '21 at 11:32

3

Let us construct a bijection $$ \Bbb Z[i]/(1+i)\ni [x+y i]\longmapsto [x-y] \in \Bbb Z/2\Bbb Z $$ (note: modulo $(1+i)$ means you let $i=-1$.)

It is direct to check this is well-defined and gives an isomoprhism.

edited May 06 '21 at 11:32

Chris

2,763

answered Mar 18 '14 at 20:48

Ma Ming

7,482

2

do you look at (x-y) mod 2? – malxmusician212 Mar 18 '14 at 20:51
@MalcolmLazarow yes, that's what Ma means by $[x-y]$. – Ben Grossmann Mar 18 '14 at 20:53
1

why couldn't we look at (x-y) in Z/nZ, where n is not necessarily 2? – malxmusician212 Mar 18 '14 at 21:02
we also look at the kernel of this transformation = ${a+bi|a-b mod 2 \equiv 0 }$. 4+2i is in the kernel, and 4+2i/(1+i)=2, which is not equivalent to 0 in Z[i]/(1+i) – malxmusician212 Mar 18 '14 at 21:04
nevermind, yes it is as 2=(1+i)(1-i) – malxmusician212 Mar 18 '14 at 21:06

score 2 · Answer 3 · answered Mar 18 '14 at 21:03

I find it easiest to do $${\mathbb Z}[i]/(1+i) \cong {\mathbb Z}[x]/(x^2+1,1+x) \cong ({\mathbb Z}[x]/(x+1))/(x^2+1) \cong {\mathbb Z}/(2).$$ For the final step, note ${\mathbb Z}[x]/(x+1)$ is isomorphic to ${\mathbb Z}$, with $x$ mapping to $-1$. Under that isomorphism, the ideal $(x^2+1)$ of $Z[x]/(x+1)$ corresponds to the ideal $(2)$ of ${\mathbb Z}$.

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

3 Answers3

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