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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.

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    What's an "ann"?2011-10-26
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    $ann_r(S)=\{r \in R :rm=0, \forall m \in S\}$2011-10-26
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    And what is $\partial$? What is $A_1$?2011-10-26
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    Differential sign. $\partial^i= {\frac{d^i}{dx^i}}$2011-10-26
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    Yes, I know it's the differential sign. The question is, what is the context, or what does it mean here? You don't specify your context, you don't specify much of anything, so I don't know what $\partial$ means *for you here*. What is your $R$? What is $A_1$? You are finding the annihilator of what ring?2011-10-26
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    Presumably, $A_1$ is the first Weyl algebra, and the module $k[x]$ with its natural action is being considered, and the annihilator of $x$ is being sought. (simplicity: not everyone has a mind reading machine!)2011-10-26
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    Yes that is correct Mariano.2011-10-26
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    Please don't use titles that are entirely in $\TeX$...2011-10-26

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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.