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.


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