Cardinality of a Quotient Ring

Question

Let $R=\mathbb Z[\xi]$, with $\xi=\frac{1+\sqrt{-19}}{2}$. What is the cardinality of $R/aR$, if $0\neq a\in R$ ?

Is the cardinality finite, and equal to the number of cosets ? So if $a$ is fix and is in $R$ then it must be of the form for example $s+t\xi$ with $s,t\in\mathbb Z$, so do I have to find another element in $R$ such that their product minimizes the number of cosets, and this number is then the cardinality ?

Answer 1

As an abelian group, $R$ is free with basis $\{1,\xi\}$. If $a=x+yi$ is in this ring, the function $m_a:u\in R\mapsto au$ is a morphism of abelian groups which, with respect to that basis, has matrix $A=\begin{pmatrix}x&x'\\y&y'\end{pmatrix}$, for some $x'$ and $y'$ that you can find. Now, a very useful result now is:

if $f:\mathbb Z^2\to\mathbb Z^2$ is a group homomorphism given by multiplication by a non-singular matrix $A\in M_2(\mathbb Z)$, then the cardinal of the quotient $\mathbb Z^2/A\mathbb Z^2$ is $|\det A|$.

This implies that the cardinal of $R/aR$ is precisely the absolute value of the determinant of our matrix.