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.



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


