Polynomials in one variable simple module here or not?

Question

Let $A$ be the $k$-algebra of differential operators with polynomial coefficients in one variable, where $k$ is a field of characteristic $0$. I know that $k[t]$ has a natural $A$-module structure.

Question. Is $k[t]$ a simple $A$-module or not?

Answer 1

The short answer is yes.

For a bit more detail, note that $A$ is the $k$-algebra generated by two elements $x$ and $d$ with the relation $dx-xd=1$ (here you should think of $d=\partial/\partial x$). Then, $k[t]$ is an $A$-module where $x$ acts as multiplication by $t$, and $d$ acts by $\partial/\partial t$.

To show that $k[t]$ is simple, suppose $V\subset k[t]$ is an $A$-submodule. We will show that either $V=0$ or $V=k[t]$. Indeed, if $V\neq 0$, then there exists some $$f(t)=a_nt^n+\cdots+a_0\in V$$ with $a_n\neq 0$. Note that $$\frac{1}{a_n}d^n.f(t)=1\in V$$ and now, for any $g(t)\in k[t]$, $g(x).1=g(t)\in V$. Hence $V=k[t]$ as required.