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Consider the quotient ring $\mathbb{Z}[i]/2\langle a + bi \rangle$. Then this ring is isomorphic to $\mathbb{Z}[x]/\langle x^2 + 1, 2(a + bx) \rangle$. Then $x = i$ and $x = -\frac{a}{b}$, such that $-\frac{a}{b} \in \mathbb{Z}$. Am I getting this right?

But how do I proceed from here? I've read some posts on similar problems, but I don't know/see something peculiar, which is omitted in all of those posts I've seen. I'd really appreciate if someone could explain the real mechanics behind this kind of determination.

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  • $\begingroup$ Why should $-\frac{a}{b} \in \mathbb{Z}$? Appart from that, if you want to know the answer for all $a,b \in \mathbb{Z}$, you are asking for all factor rings of $\mathbb{Z}[i]$, or equivalent: All simple $\mathbb{Z}[i]$ modules. I'm not sure if there is an easy answer to that, but for $\mathbb{Z}[i]$, the ring of Gaussian integers, many results can be found on the web. $\endgroup$ – Dirk Apr 12 '17 at 9:00
  • $\begingroup$ @Bemte $-\frac{a}{b}$ must be in $\mathbb{Z}$ because we are working with a ring over $\mathbb{Z}$. $\endgroup$ – sequence Apr 12 '17 at 9:11
  • $\begingroup$ Possibly related: math.stackexchange.com/questions/1868345 $\endgroup$ – Watson Apr 21 '17 at 19:50
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It is well known that $|\mathbb Z[i]/\langle a+bi\rangle|=N(a+bi)=a^2+b^2$.

This fact follows from the structure theorem of finitely generated abelian groups, which in particular states that the number of elements of the co-kernel of $$\mathbb Z[i] \xrightarrow{\cdot (a+bi)} \mathbb Z[i]$$ is equal to the determinant of that map (viewed as map of free $\mathbb Z$-modules). This determinant is equal to $N(a+bi)$.

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