An question about the form of a primitive ideal of the bicyclic monoid algebra

Question

I am reading the paper 'Plactic algebra of rank 3'. You can visit http://link.springer.com/article/10.1007/s00233-011-9337-3.

My question is why the primitive ideal $P$ has the form $(p-\alpha,q-\beta)$?

Answer 1

$K[x,x^{-1}]$ is a commutative PID, and the primitive ideals are necessarily the maximal ideals. Therefore the image in $K[x,x^{-1}]$ (in which $p$ is being mapped to $x$ and $q$ is being mapped to $x^{-1}$) is generated by a single element $x-\alpha$. Of course $\alpha\in K$, and it is impossible for $\alpha$ to be zero since $x$ is a unit and does not generate a proper ideal.

The preimage of $(x-\alpha)$ in $K[B]$ is the ideal $(p-\alpha, 1-qp)$ in $K[B]$.

Then it is an easy matter to compute that

$$ -\alpha^{-1}q(p-\alpha)-\alpha^{-1}(1-qp)=q-\alpha^{-1}\in (p-\alpha, 1-qp) $$

So it's clear that $(p-\alpha, q-\alpha^{-1})$ is at least a subset. By modifying the string of equalities slightly:

$$ -\alpha^{-1}q(p-\alpha)-(q-\alpha^{-1})=\alpha^{-1}(1-qp)\,;\text{ and }\\ -q(p-\alpha)-\alpha(q-\alpha^{-1})=1-qp $$

So in fact $(p-\alpha, q-\alpha^{-1})=(p-\alpha, 1-qp)=P$