Assume that $P(x,y), Q(x,y) \in \mathbb{R}[x,y]$ are two polynomials. We define a linear map

$D$ on $\mathbb{R}[x,y]$ with $D(U)=PU_{x}+QU_{y}$. In fact $D$ is the differential operator corresponding to the vector field $P\partial_{x}+Q\partial_{y}$.

Are there polynomials $P$ and $Q$ such that the codimension of the range of $D$ is finite but different from $0$ and $1$?

**Note** This codimension is $0$, $1$ and $\infty$ for $\partial_{x}$, $x\partial_{x}+y\partial_{y}$ and $x\partial_{x}-y\partial_{y}$, respectively.

**Motivation** It can be easily shown that this codimension is an upper bound for the number of closed orbits of the polynomial vector field $P\partial_{x}+Q\partial_{y}$.

**However there is a Painful fact**: If an smooth vector field $X$ on $\mathbb{R}^{2}$ has a limit cycle which surrounds a nondegenerate singularity (or a singularity which has a local smooth first integral at a deleted neighborhood of the singularity and is **discontinuous** at the singularity), then the codimension of the range of the differential operator $D_{X}$ on $C^{\infty} (\mathbb{R}^{2})$ is infinite ! The reason is as follows.

Assume that a limit cycle $\gamma$ is attractor and surounds a source singularity at the origin.

Non degeneracy of the singularity help us to find an open set D around origin whose boundary $S$ is a smooth closed curve and the vector field is transverse to $S$ toward the exterior of $S$. So the flow of $X$ defines a smooth retraction $r:\bar{D}\setminus \{0\} \to S$. In fact $r(x)$ is the intersection point of the orbit of $x$ with $S$ This shows that there is an smooth real valued function $\phi$ on $\mathbb{R}^2 \setminus \{0\}$ such that locally around the origin we have $X. \phi =0$ and $P\circ \phi$ is discontinuous at origin for every polynomial $P(x)$. Such $\phi$ can be constructed locally by $\psi \circ r$ where $\psi:S \to \mathbb{R}$ is an arbitrary non constant smooth function.Now we choose an arbitrary smooth extension of this locally defined $\phi$ to whole punctured plane. This extension is denoted again by $\phi$. Obviously every nontrivial linear combination $\sum \lambda_i \phi^i$ is discontinuos at origin. Because its restriction to every small neighborhood of the origin has the same range(image) as the range of $\sum \lambda_i \psi^i: S \to \mathbb{R}$. However $X.\phi^i$ is a smooth function on whole plane, for every $i$ since it vanish locally around the origin.

Now for every $n$,the following set represents an independent subset of $C^{\infty}(\mathbb{R}^2)/Range(D_X)$

$$\{X.\phi, X.\phi^2, \ldots, X.\phi^n \}$$ To prove this, we strongly use the fact that we have at least one limit cycle.

More precisely: assume that $\sum_{i=1}^n \lambda_i (X.\phi^i)= X.F$ for some smooth function $F\in C^{\infty} (\mathbb{R}^2)$. Then $F-\sum \lambda_i \phi^i$ is a first integral(constant of motion) on the puntured plane. So it is constant on the basian of attraction of the limit cycle. A punctured neighborhood of the origin is contained in the basian of attraction of limit cycle. Thus locally around the origin,$F$ differs $\sum_{i=1}^n \lambda_i \phi^i$ by a constant(In a deleted neighborhood). This is a contradiction because $F$ is smooth at origin but $\sum \lambda_i \phi^i$ is discontinuous at origin.

**So unfortunately we encounter with the following two inconsistent and opposite but true results:**

1)The codimension of the range of $D_X$ is an upper bound for the number of limit cycles.

2.If the codimension is finite then there is no any limit cycle.

So we should choose an appropriate function algebra, different from $C^{\infty} (\mathbb{R}^{2})$, which is invariant under the differential operator corresponding to an algebraic vector field.

#Added:

According to the answer of Loic Teyssier we put $X=x^k\partial_x+y\partial_y$. Then $X$ can act on various spaces $C^{\infty}(\mathbb{R}^2),\; C^{\omega}(\mathbb{R}^2)$ or the space of holomorphic functions from $\mathbb{C}^2$ to $\mathbb{C}$. What can be said about the dimension of the cokernels in each of these actions?