# Testing ideal membership in the Weyl algebra: a simple example

In Example 1.1.4 of the book Grobner Deformations of Hypergeometric Differential Equations, 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.

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.