In *Example 1.1.4* of the book [*Grobner Deformations of Hypergeometric Differential Equations*](https://www.springer.com/gp/book/9783540660651), it is stated without proof that


$$\partial^2 \in D\cdot \langle x\partial^4, x^3\partial^2 \rangle \tag{$\star$}$$


where $D$ denotes the Weyl algebra over $\mathbb{k}[x]$, and $ D\cdot \langle x\partial^4, x^3\partial^2 \rangle$ denotes the left-ideal generated by the operators $x\partial^4,x^3\partial^2$. Since *Example 1.1.4* is at the very beginning of the book, and no proof is given by the authors, I'm presuming that there a simple way to verify this (i.e. without using the machinery of the main text). 


**My current line of attack**:


So far, this is my thinking: since $D$ is a domain, the equation $(\star)$ is equivalent to solving the following linear equation in the non-commutative ring $D$:


$$1=D\_1x\partial^2 +D\_2 x^3. ~~~~~~~~~~~~~~~~$$


where $D\_1,D\_2\in D$ are unknowns. I am able to prove (by brute force calculation) that $\text{ord}\,D\_2\geq 1$. Obviously, $\text{ord}\, D\_2 = \text{ord}\, D\_1+2$. The brute-force calculation in the general case quickly spirals out of control.



Testing ideal membership in the Weyl algebra: a simple example
Following my nose gave the following argument. Writing $I$ be the left ideal generated by $x\partial^2$ and $x^3$ and using $\cdot$ to stress multiplication we get 


$$ x^2 \cdot x\partial^2 - \partial^2\cdot x^3 = [ x^3, \partial^2] = -6x^2\partial - 6x\in I$$


So $$\frac{1}{6}x\partial\cdot (-6x^2\partial - 6x) + x^2\cdot x\partial^2 = x[x^2,\partial]\partial - x^2\partial + x[x,\partial] = -3x^2\partial - x\in I$$


Taking a suitable $\mathbb{k}$-linear combinations of these gives $x\in I$ and $x^2\partial\in I$. 


Then $$\partial^2\cdot x - 1\cdot x\partial^2 = [\partial^2,x]=2x\partial \in I.$$ Since also $2\partial x\in I$ we conclude $$\partial x- x\partial=[\partial,x]=1\in I$$ as required.


The general strategy at each step is to take two elements in the left ideal with the same principal symbol with respect to the filtration with $F\_0=\mathbb{k}$, $F\_1=\mathbb{k}\cdot\{1,x,\partial\}$ and $F\_n=F\_1^n$ and then compute their difference which will live in a lower filtered part. I haven't read the book but I'd imagine that this idea is at the heart of computations throughout.


