0
$\begingroup$

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.

  • 0
    Please don't use titles that are entirely in $\TeX$...2011-10-26

1 Answers 1

1

If $a=\beta \partial^2 + f(x)\partial + g(x),$ then also $a=\beta \partial^2 + f_1(x)(x \partial -1) + \lambda \partial +g_1(x)$ with $f_1(x)=\frac{f(x)-f(0)}{x},$ $\lambda=f(0)$ and $g_1=g(x)+f_1(x).$

You could have found this simply by expanding in the second expression for $a$, and comparing with the first one.