# Modules over quantum complete intersections

Let $a_i \geq 2$ be natural numbers and $q_{ij}$ field elements of the field $k$ for $i>j$. A quantum complete intersection is the algebra $A:=k<x_1,...,x_n>/(x_i^{a_i},x_i x_j - q_{ij} x_j x_i)$. This is a local finite dimensional Frobenius algebra of dimension $a_1 a_2 ... a_n$.

Let $b_i$ be natural numbers with $0 \leq b_i \leq a_i -1$ and $c_i$ field elments. Define the vectors $b=(b_1,...,b_n)$ and $c=(c_1,...,c_n)$.

Define the modules (right ideals) by $M_c^b:=(c_1 x_1^{b_1}+...+c_n x_n^{b_n}) A$. So for example $M_c^b$ is equal to the algebra when choosing $b_i=0$ and $c_i=1$.

Specialising some parameters(I only looked at the at most three variables case), I could do some interesting contstructions but looking at the general case the computations seem to become a mess.

But maybe there are nice methods to answer the following questions:

1. What are those right ideals with dimension equal to half the dimension of the algebra? I think those build an especially nice subcategory, which is closed under syzygies. Reason: There is then a short exact sequence $0 \rightarrow U \rightarrow A \rightarrow$M_c^b$\rightarrow 0$ with $U$ also having dimension equal to half the dimension of $A$. $U$ then should be equal to $uA$, when $u$ is an element such that $(c_1 x_1^{b_1}+...+c_n x_n^{b_n}) u=0$.

2. What is the dimension of $Hom_A(M_c^b, M_{c'}^{b'})$? This should also answer when two such right modules are isomorphic. My guess is that they are isomorphic iff $b=b'$ and $c$ and $c'$ are equal in the respective projective space. at least when all $b_i \geq 1$. The special case $M_c^b=A$ would give the general dimension of $M_{c'}^{b'}$.

3. When is $M_c^b$ a two-sided ideal?

Answers to 2. and 3. would also be interesting restricted to the right ideals with dimension equal to half the dimension of the algebra.

1. (maybe very hard question) When is $A$ a Hopf algebra?