Proving the first Weyl Algebra is a domain and $\mathrm{Z}(A_1)=\Bbb C$ We define the first Weyl Algebra $A_1$ as the ring of all linear differential operators with polynomial coefficients. That is,
$$A_1=A_1(\Bbb C):=\{a_n(X) \delta^n+\dotsb+a_1(X)\delta+a_0(X):a_i(X)\in\Bbb C[X],\in \Bbb N \}\subseteq \mathrm{End}_\Bbb C(\Bbb C[X]),$$
where $\delta:=\frac{d}{dX}$.
We assume that we know the Leibniz's Rule
$$\delta^n \circ x^m= x^m \circ \delta^n + \lambda_{m-1}x^{m-1} \delta^{m-1}+\lambda_{m-2}x^{m-2} \delta^{m-2}+\dotsb$$ we want to prove the following properties:


*

*$A_1$ is a domain.

*The center of $A_1$ is $\mathrm Z (A_1)=\Bbb C$.


My attempt. 1. We take $\alpha:=\sum_{i=1}^{n}a_i(X)\delta^i,\beta:=\sum_{j=1}^{m}b_j(X)\delta^j \in A_1$, which are not zero. Thus, we can assume that $a_n(X),b_m(X)\neq 0$. Then, if 
$$\alpha \circ \beta =0 \iff a_n(X)b_m(X)\delta^{n+m}+\text{lower order terms}=0.$$
Since the set $\{x^i\circ \delta^j:0\leq i,j \leq \infty\}$ is a basis, we obtain that $a_n(X),b_m(X)=0$, so we have a contradiction.


*As we know, 
$\mathrm{Z}(A_1)=\{\alpha \in A_1: \alpha \circ \rho=\rho \circ \alpha,\ \forall \rho \in A_1\}$. We want to show that $\mathrm{Z}(A_1)=\Bbb C$. Let's take a non-zero element $\alpha:=\sum_{i=0}^{n}a_i(X)\delta^i$ with $a_n(X)\neq 0$. My idea is to show that if $n\geq 1$, then $\alpha \circ x - x\circ \alpha \neq 0$. But I don't know how to apply it. 


Questions. 1) Is 1. correct?
2) What about 2.? Could you please write an elaborate proof? I couldn't find much stuff about this notion.
Thank you.
 A: (1) Looks correct. I suppose you could be more explicit about how you got the leading term you got, depending on the reader. (Also pet peeve of mine: the letter $m$ comes before $n$ in the alphabet.)
(2) Suppose $\alpha$ commutes with $\delta$. Subtract out all leading terms with constant coefficients until you get an expression $\overline{\alpha}=a_k(x)\delta^k+a_{k-1}(x)\delta^{k-1}+\cdots$. Suppose $k\ge1$. Then writing $\delta\overline{\alpha}=\overline{\alpha}\delta$ gives
$$ a_k(x)\delta^{k+1}+[\delta a_k(x)+a_{k-1}(x)]\delta^k+\cdots=a_k(x)\delta^{k+1}+a_{k-1}(x)\delta^k+\cdots $$
Equating coefficients of $\delta^k$ gives $\delta a_k(x)=0$, so $a_k(x)$ is constant, a contradiction. Therefore $k=0$, in which case $\delta\overline{\alpha}=\overline{\alpha}\delta$ implies $a_0(x)$ is constant too. Thus, $\alpha$ is purely a polynomial in $\delta$ with constant coefficients.
Writing $\alpha=\sum_j a_j\delta^j$ and $x^{\ell}\alpha=\alpha x^{\ell}$ I'll let you figure out why $\alpha$ can't contain any $\delta$s. (Pick $\ell$.)
