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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?

Bill Dubuque
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3 Answers3

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$\rm\overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ 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'}{\color{#90f}{ww'}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#90f}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.

Remark $ $ We tested divisibility by $\color{#90f}{\textit{rationalizing}}$ the denominator in order to reduce division by an algebraic irrational $\,\rm w\,$ to a simpler division by an integer (its norm $\,\rm \color{#90f}{ww'=n})$. This is a special case of the method of simpler multiples.

See also this method of rewriting the ideal as a module in Hermite normal form.

Bill Dubuque
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Math Gems
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    MathGems has achieved maximum StackExchange TeX level. Looking at his source code genuinely terrifies me. +1 – Alexander Gruber Apr 26 '13 at 05:27
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    Hopefully the code is not manually generated? – copper.hat Apr 26 '13 at 05:28
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    @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 05:30
  • Even uses mumbo-jumbo codes for the colors instead of \color{red}...! Damn, that's good. +1 – DonAntonio Apr 26 '13 at 16:14
  • @Don Alas, it is difficult to find colors that work good on all types of displays and, further, do not pose problems for color-challenged readers. If the colors do not render well or pose problems then please do let me know so that I can avoid using those color schemes. – Math Gems Apr 26 '13 at 16:21
  • I think all the colores you used look fine, @MathGems. I once tried "yellow" and it was unreadable. – DonAntonio Apr 26 '13 at 16:28
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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 $\ker(\phi) = \langle a+bi\rangle$.

user26857
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Srijan
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$$\mathbb{Z}[i]/(a+ib)=\mathbb{Z}[i]/(a^2+b^2,a+ib)=\mathbb{Z}_{a^2+b^2}[i]/(a+ib) =\mathbb{Z}_{a^2+b^2}[x]/( x^2+1,a+bx)\\=\mathbb{Z}_{a^2+b^2}[x]/(a^2 x^2+a^2,a+bx) =\mathbb{Z}_{a^2+b^2}[x]/(-b^2 x^2+a^2,a+bx)\\ =\mathbb{Z}_{a^2+b^2}[x]/(a+bx) \simeq \mathbb{Z}_{a^2+b^2}$$

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

reuns
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  • 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 01 '17 at 03:20
  • Proof is absolutely correct. Maybe if I you make some written comments (here and her I used the isomorphism theorem) then I think you could obtain more upvote. – vesszabo Aug 22 '22 at 14:15