Let $A(k)$ be a $k$-algebra with $2$ generators $x$, $y$, and $1$ defining relation:$$yx - xy - 1= 0.$$Define an $A(k)$-action on the vector space $M := k[t]$ as follows:$$y \text{ acts as }{d\over{dt}},\text{ resp. }x\text{ acts as multiplication by }t.$$I have a few questions.
- How do I see that the above action makes $M$ an $A(k)$-module and that this module is simple if $\text{char}\, k = 0$?
- Let $\text{char}\,k > 0$. Is the $A(k)$-module $M$ necessarily simple? What is the construction of a simple $A(k)$-module?
In order to make a $k$-vector space $M$ into a module over the $k$-algebra with the presentation $\langle x, y\;|\;yx-xy=1\rangle$ it suffices to assign to each generator in the presentation a $k$-space endomorphism of $M$ in such a way that the defining relation holds for the assigned endomorphisms. In this example you are assigning to $y$ the endomorphism $f\mapsto f'$ for $f=f(t)\in M=k[t]$ and you are assigning to $x$ the endomorphism $f\mapsto tf$. To check that the relation is satisfied you must verify that $(yx-xy-1)f = 0f = 0$ for every $f\in k[t]$. That is, you must verify that $(tf)'-t(f')-f = 0$ for every $f\in k[t]$, which may be written suggestively as $(tf)' = tf'+1f$.
To check simplicity, you must show that if $f\in M$ is not $0$, then $M=\langle f\rangle$. For this, choose any $f\neq 0$ and multiply by $y$ repeatedly until you get a nonzero constant. You may scale it to show that $1\in\langle f\rangle$. Then multiply by $x$ repeatedly to show $t^m\in\langle f\rangle$. Now it is not hard to see why $\langle f\rangle =k[t]=M$.
This module $M=k[t]$ is not simple in characteristic $p>0$, since $x^pM=t^pk[t]$ is a proper submodule. You can show that $M/x^pM$ is simple by an argument like the one in the previous paragraph.