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.

Remarkdeal with monomial ideals such as $(x^2y)$ ? For what i does this ideal contain $x^iy^i$? Or, for that matter, in $\mathbb C[x,y,z]$, for what $i$ does $(x^2y,x^2z)$ contain some $x^iy^iz^i$? I'm not sure, either, how the exclusion of all monomial ideals per force excludes all ideals? – Mikael Vejdemo-Johansson Jun 14 '10 at 5:23