Let us assume we are given a field $K$ and the quaternion algebra $Q=(a,b)_K$ with $a,b\in K^{\times}$.

$Q$ is generated as a $K$-vector space by elements $1,i,j,k$ with $k=ij$, $i^2=a$, $j^2=b$ and $ij=-ji$. We have $\dim_K(Q)=4$ and $Q$ is a noncommutative $K$-algebra. The algebra has a norm given by $$N(x_11+x_2i+x_3j+x_4k)=x_1^2-ax_2^2-bx_3^2+abx_4^2$$

Now let $I\subset Q$ be a left ideal such that $\dim_K(I)=2$.

  • Why do we have $I=Q\alpha$ with $\alpha=x_11+x_2i+x_3j\in Q$ and $N(\alpha)=x_1^2-ax_2^2-bx_3^2=0$?

  • Furthermore why is $Q(x_11+x_2i+x_3j)=Q(y_11+y_2i+y_3j)$ if and only if there exists $r\in K$ such that $(y_1,y_2,y_3)=r(x_1,x_2,x_3)$.

I get that we must have $N(\alpha)=0$ for otherwise $\alpha$ would be a unit and $I=Q$ which does not have dimension two. But why can we write every two dimensional ideal as a principal ideal generated by an element from $\mathrm{span}(1,i,j)$? How do we lose the $k$?

The only if part of the second question is clear but how to see that if $Q\alpha=Q\beta$ the we must have $\beta=r\alpha$?

All results together should show that the two-dimensional left ideals in $Q$ are classified by the conic $u^2-av^2-bw^2=0$ defined in the projective space $\mathbb{P}^2_k$ with homogeneous coordinates $[u:v:w]$.

  • $\begingroup$ Note that $N\alpha = 0$ for some $\alpha \ne 0$ means $Q$ is not a division algebra, so $Q \simeq M_2(K)$. So you can translate the problem to matrix algebras. Does this help? $\endgroup$
    – Kimball
    Dec 22, 2016 at 15:14
  • $\begingroup$ By the way, the simple modules describe here: math.stackexchange.com/a/2060075/11323 will be the ideals of dimension 2 in $M_2(K)$. $\endgroup$
    – Kimball
    Dec 23, 2016 at 16:45


You must log in to answer this question.

Browse other questions tagged .