# Quotient rings of Gaussian integers [duplicate]

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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?

## marked as duplicate by Watson, Did, Vincent, Lee David Chung Lin, RebellosNov 26 '18 at 18:16

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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

## 3 Answers

$$\rm\overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$$ Let $$\rm\ w = a\!+\!b\,{\it i}.\$$ In $$\rm\, \langle w\rangle\,$$ is $$\,\rm(a\!+\!b\,{\it i}\,)(j\!+\!k\,{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{#d0f}{ e\,\ +\,\ {\it i}}}$$

$$\rm \smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/\langle w\rangle\$$ is $$\rm\,\color{#0a0}{onto,\ }$$ by $$\rm\bmod w\!:\,\ \color{#d0f}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\,{\it i}\:\equiv\, c\!-\!de\in \Bbb Z.\$$ Let $$\rm\ n = ww'$$
$$\!\rm\begin{eqnarray}\rm Note\ \ \color{#c00}{m\in ker\ h} &\iff&\rm w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{ww'}\!=\dfrac{ma\!-\!mb\,{\it i}}n\in \Bbb Z[{\it i}\,]\\ &\iff&\rm n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray}$$

$$\rm Thus \ \ \Bbb Z[{\it i}\,]/\langle w\rangle\! \color{#0a0}{= Im\:h}\,\cong\: \Bbb Z/\color{#c00}{ker\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\$$ by the First Isomorphism Theorem.

• @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
• Even uses mumbo-jumbo codes for the colors instead of \color{red}...! Damn, that's good. +1 – DonAntonio Apr 26 '13 at 16:14

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 $\ker(\phi) = \langle a+bi\rangle$.

• 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

$$\mathbb{Z}[i]/(a+ib)=\mathbb{Z}[i]/(a^2+b^2,a+ib)=\mathbb{Z}_{a^2+b^2}[i]/(a+ib) =\mathbb{Z}_{a^2+b^2}[x]/( x^2+1,a+bx)\\=\mathbb{Z}_{a^2+b^2}[x]/(a^2 x^2+a^2,a+bx) =\mathbb{Z}_{a^2+b^2}[x]/(-b^2 x^2+a^2,a+bx)\\ =\mathbb{Z}_{a^2+b^2}[x]/(a+bx) \simeq \mathbb{Z}_{a^2+b^2}$$

Where I used $a,b$ is invertible $\bmod a^2+b^2$

• Clearer: $\bmod n\! = a^2\!+b^2\!:\,\ (a/b)^2 \equiv -1\,\Rightarrow\ bx\!-\!a\mid x^2+1.\$ Iirc I posted a proof using this either here or sci.math or Ask an Algebraist. – Bill Dubuque Jul 1 '17 at 3:20