Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for an example of computation of the isomorphism classes of $n$-point modules over a non-commutative generated graded algebra (assuming all good properties such as Noetherian property). Recall that $n$-point module $M$ over graded $k$-algebra $A$ is defined to be a graded module generated in degree $0$ with Hilbert polynomial $H_M(t)=\frac{n}{1-t}$, i.e. $\dim_{k}M_d=n$. Of course, when $A$ is commutative, the isomorphism classes of $n$-point modules form the Hilbert scheme of $n$ points over $\mathrm{Proj}(A)$.

I have seen some computation of the isomorphism classes of $1$-point modules but not 2- or higher over strictly non-commutative algebra. Could anyone kindly give me a reference for non-trivial computation? Or show me some simple computation?

Thank you very much in advance.

share|improve this question

1 Answer 1

You could look at the paper "Linear Modules over Sklyanin Algebras" by Joanna Staniszkis. In this paper she calculates all the linear modules over a Sklyanin algebra on any number of generators. She also constructs all the possible short exact sequences involving linear modules.

The method is very geometric. Denote the $n$-dimensional Sklyanin algebra by $A_n(E,\tau)$ where $E$ is the associated (smooth) elliptic curve and $\tau$ a point on it. The $d$-dimensional linear modules correspond either to $d$-secants (so hyperplanes in projective space that intersect the elliptic in $d+1$ points) or so-called $d$-singular planes.

To see this coincides with the computations in previous papers for the smaller linear modules, one can see from Corollary 3.5(a) loc. cit. that the point modules for $n=4$ correspond to points on the elliptic curve and 4 singular points, which are the singular loci of the singular quadrics containing $E$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.