# Quotient rings of Gaussian integers

I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following:

For any $\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1$, we have a ring isomorphism $$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.$$

Could someone show me an isomorphism between these rings to prove this?

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Look at the multiplication with a-bi – Ehsan M. Kermani Apr 26 '13 at 1:36
i don't see how you get something multiplicative from that? – pad Apr 26 '13 at 1:55
Duplicate of math.stackexchange.com/questions/23358/… – user26857 Aug 9 '13 at 8:34

$\rm\: (a,b)=1\stackrel{Bezout}\Rightarrow ak\!+\!bj=1.\:$ Let $\rm\ w = a\!+\!b\,{\it i}\,.\,\ \langle w\rangle \ni (a\!+\!b\,{\it i}\,)(j\!+\!k\,{\it i}\,)\, =\, aj\!-\!bk+{\it i}\, =:\, \color{#f0f}{ e+{\it i}}$

$\rm\quad \Bbb Z\stackrel{h}{\to}\, \Bbb Z[{\it i}\,]/\langle w\rangle\$ is $\rm\,\color{#0b0}{\bf onto,\ }$ by $\rm\ mod\,\ w:\,\ \color{#f0f}{{\it i}\,\equiv -e}\phantom{\dfrac{|}{|}}\!\!\Rightarrow\:c\!+\!d\,{\it i}\:\equiv\, c\!-\!d\,e\,\in\, \Bbb Z.\ \$ Let $\rm\ n = ww'$
$\rm\quad\!\begin{eqnarray}\rm m\in ker\ h &\iff&\rm w\mid m\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\!\! \dfrac{m}{w}\: =\: \dfrac{m\,w'}{ww'}=\,\dfrac{ma\!-\!mb\,{\it i}}n\:\in\: \Bbb Z[{\it i}\,]\\ &\iff&\rm n\mid ma,mb\!\iff\! \color{#c00}n\mid(ma,mb)=m(a,b)=m\end{eqnarray}$

$\rm\quad So \ \ \ \Bbb Z[{\it i}\,]/\langle w\rangle\, \color{#0b0}{\bf =\ Im\:h}\:\cong\: \Bbb Z/ker\:h \,=\, \Bbb Z/\color{#c00}n\,\Bbb Z\$ $\ \$ QED

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@YACP The above proof has never been posted here before (though I did post some special cases). – Math Gems Apr 26 '13 at 5:18
MathGems has achieved maximum StackExchange TeX level. Looking at his source code genuinely terrifies me. +1 – Alexander Gruber Apr 26 '13 at 5:27
Hopefully the code is not manually generated? – copper.hat Apr 26 '13 at 5:28
@Alex The TeX code is for machines to read, not humans! (it is generated by macros). Even I don't look at it. – Math Gems Apr 26 '13 at 5:30
+1: Nice. ${}{}{}$ – copper.hat Apr 26 '13 at 6:02

Hint: Define a map $\phi : \mathbb{Z}[i] \to {\mathbb Z}_{a^2 + b^2}$ by $\phi (x + yi) = x-(ab)^{-1}y$. Next, show that $\phi$ is surjective homomorphism and find its kernel. For your hint $\ker(\phi) = \langle a+bi\rangle$.

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There's going to be a problem here for sure if $\,ab=0\,$ ... – DonAntonio Apr 26 '13 at 3:06
@DonAntonio we have assumed that $gcd(a, b) = 1$ so we can assume without loss of generality that $a$ and $b$ are both positive. – srijan Apr 26 '13 at 3:49
You're right, Srijan. Thanks.+ 1 – DonAntonio Apr 26 '13 at 16:09
@DonAntonio thank you :) – srijan Apr 26 '13 at 17:58