Let $R$ be the ring $\mathbb{C}\langle x,y\rangle/(xy-yx-1)$, a quotient of free associative algebra on two generators.
(a) Show every nonzero $R$-module has infinite dimension as a complex vector space.
(b) Let $M$ be an $R$-module with a nonzero element $z$ such that $xz=0$. Show that $z, yz, y^2z, \ldots$ are $\mathbb{C}$-linearly independent in $M$.
I would love some good references to read about this stuff, because I couldn't find much. I'm probably not looking in the right places to be honest (not even sure what to tag this as).
My gut says something along the lines of modules over this ring have to somehow contain the ring, and the ring, although quotiented, still has an infinite basis over $\mathbb{C}$ as there does not appear to be a relation between $x, x^2, x^3, \ldots$. But again, I might be way off base here.
(a) $A$ is the ring $C\langle u,v\rangle/(uv-vu-1)$ which means that $[u,v]=1$. Consider some nonempty $A$-module. That is, $M$ is an abelian group under addition and we also have the operation $A\times M \rightarrow M$ satisfying all the module properties. Note that $C\subset A$, and thus we have the operation $C \times M\rightarrow M$. Thus $M$ can be thought of as a $C$ module, i.e. a complex vector space. From linear algebra we know that a nonempty vector space has a nonempty basis $B$. For sake of contradiction assume that the basis is finite $|B|=n >0$. Consider the operation on the vector space by elements $u, v \in A$.By the module properties we can see these act as linear operators, and thus we can represent them with finite matrices $X, Y \in M_n(C)$. However we have that these matrices satisfy $XY-YX=I_n$, and computing the trace (which we must be able to do in finite dimension) of both sides gives $0=Tr(XY)-Tr(YX)=Tr(XY-YX) = Tr(I_n)=n$, a contradiction. Thus we either have that the module is indeed empty or is infinite dimensional with undefined trace on the operators $u,v$.
(b) Let $y\in M$, a nonzero element in the nonempty $A$-module. We want to show that the elements $y, vy, v^2y, \ldots$ are $C$-linearly independent. This is equivalent to showing that for all $n>0$ that $c_0y+c_1vy+c_2v^y+\ldots c_nv^ny=0$ implies that all $c_i=0$ where the $c_i\in C$. To see this we consider operating on the sum by $u$. First we will want to show that $uv^ky=kv^{k-1}y$ for all positive integers $k$. The base case is simply $uvy=(1+vu)y=y+vuy=y+0=y$. Assume the hypothesis is true up to $n-1$ i.e. $uv^{n-1}y=(n-1)v^{n-2}y$. The inductive step is simply $uv^ny=uvv^{n-1}y=(1+vu)v^{n-1}y=v^{n-1}y+vuv^{n-1}y = v^{n-1}y+v(n-1)v^{n-2}y= nv^{n-1}y.$ Thus we can think of $u$ as the differential operator on the variable $v$. Repeated application of $u$ on $c_0y+c_1vy+c_2v^y+\ldots c_nv^ny=0$ gives $n!c_ny = 0$ which implies $c_n=0$ as neither $n!$ nor $y$ are zero. Following in this way we get that $c_{n-1}=0, \ldots c_1=0$. Thus all the $c_i=0$ and so the set $\{v^ky\}$ is $C$ linearly independent.