Let $x=a+bi$ be an arbitrary Gaussian integer and consider the qoutient ring $S := \frac{\mathbb{Z}[i]}{(x)}$. I know that the number of elements of $S$ is equal to $a^2 + b^2$. Is it true that $S$ is isomorphic to $\mathbb{Z}_{a^2+b^2}$ (as rings)?
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What are your thoughts on the problem? What have you tried and where did you get stuck? – Servaes Dec 03 '18 at 08:52
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3@Servaes I think the answer is negative because the ideal generated by 3 is maximal but $\mathbb{Z}_9$ is not a field. Am I correct !? – user390026 Dec 03 '18 at 08:57
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You are absolutely correct, and I like your thinking. – Servaes Dec 03 '18 at 08:58
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@Servaes Thanks. Is there a known characterization of maximal ideals of Gaussian integers ? – user390026 Dec 03 '18 at 08:59
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1Depending on your knowledge of ring theory; the Gaussian integers are a principal ideal domain, so its maximal ideals are precisely its nonzero prime ideals, which are the ideals generated by prime elements. The Gaussian primes the (normal) primes $p\equiv3\pmod{4}$, and the Gaussian integers $a+bi$ with $a^2+b^2=p$, with $p$ a (normal) prime. The wikipedia page gives a decent overview. – Servaes Dec 03 '18 at 09:05
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A useful observation in this context is that the inclusion $f:\ \Bbb{Z}\ \longrightarrow\ \Bbb{Z}[i]$ is a ring homomorphism, so for every prime ideal $I\subset\Bbb{Z}[i]$ also $f^{-1}(I)=I\cap\Bbb{Z}$ is a prime ideal, so $I$ contains a prime number $p$. Because all ideals in $\Bbb{Z}[i]$ are principal ideals there is some Gaussian integer $a+bi$ such that $I=(a+bi)$, and $p\in I$ implies that $a+bi$ divides $p$. This line of reasoning works for many rings of the form $\Bbb{Z}[\alpha]$. – Servaes Dec 03 '18 at 09:10
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Similar questions: 1, 2, 3. The claim is true if $\gcd(a,b) = 1$. – Viktor Vaughn Dec 03 '18 at 16:24
