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I was trying to prove that $\mathbb Z[i]/(a+ib)$ ($a$, $b$ are integers, not both zero) has $a^2+b^2$ number of elements. I know it's already proven on this site but I was trying to do something different.

It's easily done when either $a=0$ or $b=0$. Again, it's easy when $\gcd(a,b)= 1$ using the canonical homomorphism $\mathbb Z\to\mathbb Z[i]/(a+ib)$.

I was wondering if it holds that if $\gcd(a,b)=d$ then $$\mathbb Z[i]/(a+ib)\\\cong \mathbb Z[i]/(d)\times \mathbb Z[i]/\left(\frac ad +i\frac bd\right).$$

I think $(a+ib)=(d)\cap\left(\frac ad +i\frac bd\right)$. But will $(d)+\left(\frac ad +i\frac bd\right)=\mathbb Z[i]$ hold? I would want it to hold so that I can use the Chinese Remainder.

Nothing special
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  • What about the answer with the submodule here? This seems to be even easier. – Dietrich Burde Nov 07 '24 at 16:11
  • @DietrichBurde I have not studied any module theory yet. I have knowledge of one semester of ring theory. – Nothing special Nov 07 '24 at 16:18
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    Then Pick's theorem from the first answer is the best idea, without using abstract algebra. You idea with factoring $a+bi$ into irreducible certainly looks more complicated. – Dietrich Burde Nov 07 '24 at 18:01
  • @DietrichBurde I just wanted to see a fully ring theoretic proof using isomorphisms and stuff like that. Thank you, I will look into Pick's theorem and also, study modules sometimes soon. – Nothing special Nov 07 '24 at 18:09

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The answer is no. $2+6i=2(1+3i)$, but $2$ and $1+3i$ are not coprime since $2=-i(1+i)^2$ and $1+3i=(1+i)(2+i)$.

Hence if $z_1=1+i$ and $z_2=2+i$, we have $\mathbb{Z}[i]/(2+6i)\simeq \mathbb{Z}[i]/(1+i)^3\times \mathbb{Z}[i]/(2+i)\simeq \mathbb{F}_2[X]/(X^3)\times\mathbb{F}_5$, while $\mathbb{Z}[i]/(2)\times \mathbb{Z}[i]/(1+3i)\simeq \mathbb{F}_2[X]/(X^2)\times\mathbb{F}_2\times\mathbb{F}_5$.

These two rings are not isomorphic.

GreginGre
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  • This gives me an idea. I can fix my argument in this way: Factor $a+ib$ into irreducibles $d(a_1+ib_1)\ldots(a_n+ib_n)$ then we shall have $\gcd(a_k, b_k)=1$. – Nothing special Nov 07 '24 at 17:59