My goal is to determine what is the center of a ring $R$ generated by differential operators $z_i \frac{\partial}{\partial z_j}$ for $i,j \in \{1,2\}$ with coefficients in polynomial ring $\mathbb{C}[z_1,z_2].$
I first write the definition of $R$ $$ R=\{p_1z_1\frac{\partial}{\partial z_2}+p_2z_2\frac{\partial}{\partial z_1}+p_3z_1\frac{\partial}{\partial z_2}+p_4z_2\frac{\partial}{\partial z_2}: p_i \in \mathbb{C}[z_1,z_2] \} .$$ Is this correct? Now using the defintion of center $$Z(R):=\{x\in R: xr=rx, \ \forall r\in R\}$$ I tried to determine $Z(R)$ purely by algebraic manipulation as below but this seemed very messy $$\Bigg(p_1z_1\frac{\partial}{\partial z_2}+p_2z_2\frac{\partial}{\partial z_1}+p_3z_1\frac{\partial}{\partial z_2}+p_4z_2\frac{\partial}{\partial z_2}\Bigg)\Bigg(q_1z_1\frac{\partial}{\partial z_2}+q_2z_2\frac{\partial}{\partial z_1}+q_3z_1\frac{\partial}{\partial z_2}+q_4z_2\frac{\partial}{\partial z_2}\Bigg)=\Bigg(q_1z_1\frac{\partial}{\partial z_2}+q_2z_2\frac{\partial}{\partial z_1}+q_3z_1\frac{\partial}{\partial z_2}+q_4z_2\frac{\partial}{\partial z_2}\Bigg)\Bigg(p_1z_1\frac{\partial}{\partial z_2}+p_2z_2\frac{\partial}{\partial z_1}+p_3z_1\frac{\partial}{\partial z_2}+p_4z_2\frac{\partial}{\partial z_2}\Bigg)$$ where $p_i \in \mathbb{C}[z_1,z_2]$ are fixed and $q_i \in \mathbb{C}[z_1,z_2]$ are arbitrary.
Is there a neat way to do this?
You're looking at the Weyl algebra $W_2$ which is the subalgebra of the endomorphisms of $\Bbb C[z_1,z_2]$ generated by $\partial_1,\partial_2,z_1,z_2$, where the $z_i$ act on polynomials by multiplication and the $\partial_i$ by differentiation. We have $[z_i,z_j]=[\partial_i,\partial_j]=0$ and $[\partial_i,z_j]=\delta_{ij}$. Here $[x,y]=xy-yx$, so $[x,y]=0$ iff $x,y$ commute. We will take advantage of the fact that $W_1$ admits a basis to show it has center $\Bbb C$. I hope my idea helps you generalize this for $W_2$.
One can show that $W_1$ is central, i.e. $Z(W_1)=\Bbb C$: the elements $z_1^i\partial_1^j$ form a basis$^1$ of $W_1$ as a $\Bbb C$-module, hence pick an element $p=\sum a_{ij}z_1^i\partial_1^j=\sum p_j(z_1)\partial_1^j$.
By the previous remark $p$ is central iff it commutes with $z_1$ and $\partial_1$, since a basis is in particular a set of generators.
You can check that $[\partial_1^j,z_1]=j \partial_1^{j-1}$ and $[\partial_1,z_1^j]=jz_1^{j-1}$ provided $j>0$. Since $[-,-]$ is linear in each of its coordinates, we get $$-[p,\partial_1]=\sum p_j'(z)\partial_1^j=0$$
and since we know the $z_1^i\partial_1^j$ are a basis, $p_j'(z)=0$ for each $j$. Since $\Bbb C$ has characteristic zero, $p_j\in\Bbb C$ for each $j$. Hence $p=\sum a_j \partial_1^j$. But now $[p,z_1]=\sum a_j j\partial_1^{j-1}=0$, so each $a_i,i\neq 0$ must be $0$ (again by zero char.) so $p$ is indeed a scalar.
Now $W_2$ is obtained by looking at $W_1$ as a subalgebra of the endomorphisms of $\Bbb C[z_1,z_2]$ and ajoining $z_2,\partial_2$. Can you mimic the above to generalize?
$1$. Proof Consider an element $\sum p_j(z_1)\partial_1^j$. Evaluating this endomorphism at $1$, we obtain that $p_0(z_1)=0$, since the derivatives kill the constant. Hence $p_0=0$. Now we evaluate at $z_1$, and we otain $p_1(z_1)=0$, since the $\partial_1^j,j>1$ kill $z_1$. And so on. This gives linear independence. The commutativity relations give they generate the algebra.
Add Using the given relations, you can show that each $W_n$ is simple central.