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It is known that $\mathbb{Z}[i]$ is a PID and that $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ is finite for all $(a,b) \in \mathbb{Z}^2\backslash \{(0,0)\}$.

My question :

Is there any result on the cardinal of $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ ?

user26857
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Tulip
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2 Answers2

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Here's a geometric solution: In the complex plane, consider the parallelogram with vertices at $0, a+bi, -b+ai$ and $a-b+(a+b)i$. Show that every point in $\mathbb{Z}[i]$ is congruent to some lattice point in this parallelogram modulo $a+bi$. Now count the number of lattice points that are in the parallelogram using Pick's Formula.

ArtW
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$(a+bi)\Bbb Z$ is the additive subgroup generated by $(a+bi)$ and $i(a+bi) = (-b+ai)$.

So, as a group, $\Bbb Z[i]/(a+ib) \cong \Bbb Z^2/\langle(a,b) ; (-b,a)\rangle$ and its cardinality (up to sign) is $\begin{vmatrix}a & -b \\ b & a\end{vmatrix} = a^2+b^2$.

mercio
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    More generally, if $L$ is a subgroup of $\mathbf Z^n$ with rank $n$ then the quotient group $\mathbf Z^n/L$ has order $|\det(A)|$ where $A$ is any $n \times n$ integral matrix such that $L = A(\mathbf Z^n)$. – KCd Jun 11 '16 at 12:45