0

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 ?

inequal
  • 1,377

1 Answers1

2

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.

Mariano Suárez-Álvarez
  • 139,300
  • Sorry again but is that matrix unique, or do I have to choose an appropriate one, and how do you multiply with matrices from left or right ? – inequal Oct 05 '14 at 20:10
  • The matrix in the first paragraph is determined uniquely by $a$, as it is the matrix of the map $m_a$. You can choose too multiply on the left or on the right, but for each side you'll have a different matrix (although the two are related in a very simple way: which one?) – Mariano Suárez-Álvarez Oct 05 '14 at 20:42
  • yes one is transposed of the other, and so the determinant remains the same, but in your case, I think you consider the element $u$ as a row, so first the element then the matrix. – inequal Oct 05 '14 at 20:59
  • I hope that my solution is correct https://imagizer.imageshack.us/v2/518x691q90/673/EI1YIL.jpg, thanks a lot, but what is the name of the theorem in the grey box ? – inequal Oct 05 '14 at 21:44