Does the image of a differential operator always contain an ideal?

Question

Let $\delta$ denote a non-zero complex algebraic differential operator in a single variable x. That is, it can be written as a sum $$ \delta = \sum_i f_i\partial_x^i$$ where there $f_i$ are complex polynomials in x.

Let $R=\mathbb{C}[x]$, and consider the image of $\delta$ as a map on R. As a subspace of $R$, does $Im(\delta)$ always contain an non-trivial ideal?

It does in every case I can think of where there is some trick I can use to understand the image better:

• When $\delta$ is a function.
• When $\delta$ is a constant coefficient differential operator.
• When $\delta$ has order 1.
• When $\delta$ is homogeneous for the Euler grading; that is, it takes monomials to monomials.

It seems like it should be related to the simpler fact that $\delta$ is zero if $\delta$ kills functions of unboundedly high degree, which can be shown from the Formal Continuity of differential operators.

Remark. For more than one variable, the above question is false. If $\delta=x\partial_x-y\partial_y$, then $\delta$ is homogeneous for the Euler bigrading (it takes monomials to monomials), but it kills all monomials of the form $x^iy^i$. Since any monomial ideal in $\mathbb{C}[x,y]$ must contain some monomial of this form, the image of this $\delta$ contains no ideal.

Answer 1

No. Let $\delta=x-\partial$ and $L=Im(\delta).$ I claim that $L$ does not contain any non-zero ideal of $\mathbb{C}[x].$ Indeed, $x^k\equiv (k-1)x^{k-2}\ (\mod L)$ and, by induction,

$$x^{2n+1}\equiv (2n)!!x\equiv 0(\mod L),\ x^{2n}\equiv (2n-1)!!\ (\mod L).$$

Thus $L$ contains all odd powers of $x$ and has codimension 1 in $R.$ Suppose that $L$ contains a principal ideal $(f)$. Let $f=f_0+f_1$ be the decomposition of $f$ into the even and odd parts ($f_0$ is the span of the even degree monomials of $f$). Then $x^{2N}f_0\in L$ and $x^{2N+1}f_1\in L$ for any $N\geq 0.$ At least one of $f_0$ and $xf_1$ is non-zero and has the form $g=\sum_n a_{n}x^{2n}.$ Then for any $N\geq 0,$

$$x^{2N}g=\sum_n a_nx^{2(n+N)}\equiv \sum_n (2n+2N)!!a_n\ (\mod L)\quad \text{ and }\quad \sum_n (2n+2N)!!a_n=0.$$

However, this is impossible: for sufficiently large $N,$ the term involving $a_n\ne 0$ with the largest $n$ clearly dominates the rest of the sum.