I don't understand the proof of $\mathrm{ann}_1(x) = A_1\partial^2 + A_1(x\partial - 1),$ where $\mathrm{ann}_r(S)=\{r \in R :rm=0, \forall m \in S\}$.
$A_1$ is a left module. The ring is $A_1=\{\sum_{i=1}^n f_i(x) \partial^i: f_i(x) \in \mathbb{C}[x], n \in \mathbb{N}\}$
It says take $a=\sum_{i=1}^n f_i(x) \partial^i$ then it written in this form $a=\beta \partial^2 + f(x)\partial + g(x)$, where $\beta \in A_{1}$ however in the notes this rearrangement is made.
$a=\beta \partial^2 + f_1(x)(x \partial -1) + \lambda \partial +g_1(x)$ where $\lambda \in \mathbb{C}$ and $f_1(x),g_1(x) \in \mathbb{C}[x]$.
Don't understand the rearrangement. Please can you explain it. I need to understand it to do the same for $x^2$ i.e. find the annihilator of that.