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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.

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

Let $\delta$ denote a 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 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.