Understanding the first Weyl algebra $A$ and $\operatorname{End}_A (V)$

Question

Let $\mathbb{F}$ be a field, $V=\mathbb{F}[x]$ the vector space of polynomials. Suppose we have the first Weyl algebra $A$, the $\mathbb{F}$-subalgebra of $\operatorname{End}_{\mathbb{F}}V$ generated by $z, d\in \operatorname{End}_{\mathbb{F}}V$ and defined by \begin{equation} z(f(x)) = xf(x), d(f(x))=f'(x) \end{equation} for all $f(x) \in V$. Let's suppose the characteristic of $\mathbb{F}$ is zero. Could someone help me understand the structure of $\operatorname{End}_A V$? I proved that $dz-zd=1$ and I suppose I would be useful in any of my two questions but I don't know how to proceed from there. I believe I should check an arbitrary $\phi \in \operatorname{End}_A (V)$ and its action over $1_A$, but don't know how that helps.

Answer 1

$\DeclareMathOperator{\End}{End}$

$\End_A(V)$ consists of just multiplications by scalars; that is, it is isomorphic to $F$. The $A$-module $V$ is simple, so Schur's Lemma tells us $\End_A(V)$ is a division ring, and the division ring we see hanging around is $F$.

To prove this, let $\phi\in \End_A(V)$. Then $\phi$ commutes with $d$, so $d(\phi(1))=\phi(d(1))=\phi(0)=0$. Thus $\phi(1)$ is a constant. Now use $z$ to show $\phi(v)=\phi(1)v$ for any $v\in V$.

One insight for working with the Weyl algebra is that $d$ lowers degree by $1$ (in characteristic $0$) and $z$ raises degree by $1$.