Lie algebra preserving ideal of functions

Let $$G$$ be an algebraic group acting on an affine variety $$X=\operatorname{Spec}A$$ (all over $$\mathbb{C}$$). This gives an action of $$G$$ on the $$\mathbb{C}$$-algebra $$A$$, and an action of the Lie algebra $$\mathfrak{g}$$ of $$G$$ on $$A$$ by derivations.

If the action is not transitive, then $$G$$ will preserve some nontrivial ideal of $$A$$ (namely take an ideal of a closed $$G$$-orbit). In particular, $$\mathfrak{g}$$ will preserve this ideal.

My question is whether this remains true if we only have the lie algebra and no group action. In particular, suppose that $$\mathfrak{g}$$ is a finite-dimensional Lie subalgebra of $$\operatorname{Der}_{\mathbb{C}}(A)$$, and suppose that it does not act 'transitively' on $$X$$, i.e. for some closed point $$x\in X(\mathbb{C})$$ the natural map $$\mathfrak{g}\to T_xX$$ is not surjective, where $$T_xX$$ is the tangent space of $$X$$ at $$x$$. Then, must there exist a non-trivial ideal $$I$$ of $$A$$ which is preserved by $$\mathfrak{g}$$? Feel free to assume $$X$$ is smooth, say. Note I am not assuming the action of $$\mathfrak{g}$$ on $$A$$ is integrable, else we could use the statement about group actions stated initially.

• What does it mean to say action of $\mathfrak{g}$ is integrable?.. I am not doubting anything I am simply asking because I don’t know :) – Praphulla Koushik Dec 31 '18 at 3:25
• I mean that $A$ is not a union of finite-dimensional $\mathfrak{g}$-modules. Equivalently, the action of $\mathfrak{g}$ on $A$ does not come from the action of an algebraic group $G$ acting on $A$. – freeRmodule Dec 31 '18 at 4:33
• Ok.. understood.. – Praphulla Koushik Dec 31 '18 at 5:01

A counterexample is $$X=\mathbb{A}^2\backslash\{y=0\}$$, $$A=\mathbb{C}[x,y,y^{-1}]$$, and $$\mathfrak{g}$$ the span of the derivation $$D(x)=1$$, $$D(y)=y$$. Now $$\mathfrak{g}$$ is $$1$$-dimensional and $$X$$ is $$2$$-dimensional, so the map $$\mathfrak{g}\to T_x X$$ is not surjective for any $$x\in X(\mathbb{C})$$. The integral curves of (the vector field associated with) $$D$$ are $$y=ce^x$$ for $$c\in\mathbb{C}^\times$$, and the intuition is that because each of these curves is Zariski dense, there cannot be a non-trivial ideal preserved by $$D$$. I'm a little confused about how to formalize this argument, especially because an ideal might not be reduced (maybe someone else sees how to do this). I will argue directly that there is no non-trivial ideal $$I\subset A$$ with $$D(I)\subset I$$.
Suppose $$I\neq (0)\subset A$$ satisfies $$D(I)\subset I$$. Let $$f(x,y)=\sum_{n=0}^m f_n(x) y^m\in\mathbb{C}[x,y], \,\,f_m\neq 0$$ be an element of $$I\cap \mathbb{C}[x,y]$$ for which $$m$$ is minimal, and for which $$\deg(f_m(x))$$ is minimal (among those elements with minimal $$m$$). Then $$mf-D(f) = \sum_{n=0}^m \big((m-n)f_n - f_n'\big)y^n\in I.$$ The coefficient of $$y^m$$ in the above element is $$-f_n'$$, which has degree less than that of $$f_n$$, so by the construction of $$f$$ we must have $$mf-D(f) = 0$$. Looking at the coefficient of $$y^0$$, we see $$m f_0= f_0'$$. This can only happen if either $$f_0=0$$, or if $$m=0$$ and $$f_0$$ is constant. If $$f_0=0$$, then $$y^{-1} f\in I\cap\mathbb{C}[x,y]$$ contradicts the minimality of $$f$$. If $$m=0$$ and $$f_0$$ is constant, then $$f$$ is a constant, and $$I=A$$.