Let $X$ be a smooth variety over a characteristic zero field $k$. It seems to be well-known that


"A coherent $\mathcal{D}\_X$-module is holonomic if and only if 'every element is annihilated by a nonzero differential operator'."


I know that, when $X=\mathbb{A}^1$, this is true in the sense that "a finitely generated $k[x]\langle\partial\rangle$-module $M$ is holonomic if and only if, for every $m\in M$, there exists $P\in k[x]\langle\partial\rangle\setminus 0$ such that $Pm=0$."


**Is a similar statement true in more generality? (For all $X$? For affine $X$?) If so, why?**



Holonomic = annihilated by some differential operator
"A coherent $D\_X$-module is holonomic if and only if 'every element is annihilated by a nonzero differential operator'."


This is very much not true in dim>1.


For example, let $D=k[x\_1,x\_2]\langle \partial\_1,\partial\_2\rangle$ be the ring of differential operators on $\mathbb A^2$, and consider the left $D$-module


$$M= D/D\partial\_1.$$


Then every element is annihilated by a suitably high power of $\partial\_1$, but $M$ is not holonomic (its characteristic variety is the hyperplane in $T^\ast\mathbb A^2= \operatorname{Spec} k[x\_1,x\_2,\xi\_1,\xi\_2]$ cut out by the equation $\xi\_1=0$).


EDIT: In fact, we can make a stronger statement. Let $M$ be a $D$-module, and suppose we can find $m\in M$ which is not annihilated by any non-zero differential operator. Then the homomorphism $D\to M$ given by $P \mapsto Pm$ is injective. It follows that $\mathit{SS}(D) \subseteq \mathit{SS}(M)$ (the characteristic variety operation takes short exact sequences to unions). Thus $\mathit{SS}(M)$ is the entire cotangent bundle in this case.


Put another way, any $D$-module whose characteristic variety is a proper subset of the cotangent bundle will satisfy the property that each element is annihilated by a non-zero differential operator.


In dimension 1, every proper closed coisotropic subset of the cotangent bundle is lagrangian, so every such $D$-module is holonomic. But this is definitely not the case in higher dimensions.


