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I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.

Proof: Notice that $$\langle 1 - i \rangle\\ \Rightarrow 1 = i\\ \Rightarrow 2 = 0.$$ Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus $$ \mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}. $$

This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$

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Yes that's correct. Be a bit careful about writing things like $1=i$ when you mean that their images are equivalent in the quotient. It's fine for simple examples but this things can really catch you out later on. –  tharris Apr 15 '13 at 23:41
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I'd advise you to answer your own question, perhaps make it a community wiki (if you want), and when you are able, mark it accepted. –  mixedmath Apr 15 '13 at 23:43
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You also need to prove that $1\not\equiv 0,\:$ i.e. that the quotient ring is not trivial. –  Math Gems Apr 15 '13 at 23:52
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This may be a nitpick, but you haven't actually shown that $0 + \langle 1-i \rangle \neq 1 + \langle 1-i \rangle$, so there's a little more work to do. –  Hurkyl Apr 15 '13 at 23:52
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@Eric You need to show $\rm\:1\not\equiv 0,\:$ i.e. $\rm\:1\not\in (1-{\it i}\,),\:$ i.e. $\rm\:1-{\it i}\,\nmid 1,\:$ i.e. $\rm\:1-{\it i}\ $ is not a unit. One easy way is to rationalize denominators, as in my answer. Or you can use norms. –  Math Gems Apr 16 '13 at 0:24

3 Answers 3

up vote 4 down vote accepted

Your answer is great, but I'd like to give a different view as well.

A standard first or second example of a Euclidean Domain are the Gaussian integers $\mathbb{Z}[i]$, so that in particular the Gaussian integers form a principal ideal domain. We also know that in PIDs, nonzero prime ideals are maximal. So if we were to show that $1 - i$ is a Gaussian prime, then $\langle 1 - i \rangle$ would be a prime ideal, and thus a maximal ideal. Thus quotienting by it would give a field.

So how do we show that $1 - i$ is prime? Well, compute its norm (from the Euclidean Domain norm, where $|x + iy| = x^2 + y^2$. It's norm is $2$. Norms are multiplicative, so if $1-i = ab$, then $2 = |a||b|$. But it's norm is also an integer, and $2$ is a prime (in the reals). Thus $1-i$ is a prime.

And so we have it.

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Your proof only shows that there are at most two elements. So you also have to check that these two elements differ, i.e. that $1-i$ is not a unit. But instead, you can also do it directly, without any elements at all:

$\mathbb{Z}[i]/(i-1)=\mathbb{Z}[x]/(x^2+1)/(x-1)=\mathbb{Z}/(1^2+1)=\mathbb{F}_2$.

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+1 Exactly how I would do it. –  user38268 Apr 16 '13 at 0:06
    
But it's $\mathbb{Z} [i]/(1-i)$ not $\mathbb{Z} [i]/(i-1)$. –  Milad Jun 13 at 14:07

One must also prove that the quotient ring is $\ne \{0\}.\:$ Below is a complete proof. $\rm\quad \Bbb Z\stackrel{h}{\to}\, \Bbb Z[{\it i}\,]/(1\!-\!{\it i}\,)\:$ is $\rm\,\color{#0b0}{\bf onto,\:}$ by $\rm\:mod\,\ 1\!-\!{\it i}\,:\ {\it i}\,\equiv 1\phantom{\dfrac{|}{|}}\!\!\!\Rightarrow\:a\!+\!b\,{\it i}\,\equiv a\!+\!b\in \Bbb Z\ $
$\rm\quad n\in ker\ h\iff 1\!-\!{\it i}\,\mid n\iff\phantom{\dfrac{|}{|_|}}\!\!\!\!\!\!\! \dfrac{n}{1\!-\!{\it i}}\, =\, \dfrac{n\,(1\!+\!{\it i}\,)}2\,\in\, \Bbb Z[{\it i}\,] \iff \color{#c00}2\mid n\ $
$\rm\quad So \ \ \ \Bbb Z[{\it i}\,]/(1\!-\!{\it i}\,)\, \color{#0b0}{\bf =\ Im\:h}\,\cong\, \Bbb Z/ker\:h \,=\, \Bbb Z/\color{#c00}2\,\Bbb Z\, =\, \Bbb F_2\ $ $\ \ $ QED

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