Suppose that $X$ is an algebraic variety over $\mathbb C$, *not necessarily smooth*. Is it still true that each $\mathcal D\_X$-module ($\mathcal D\_X$ is of course the sheaf of differential operators) that is coherent as an $\mathcal O\_X$-vodule must be locally free as an $\mathcal O\_X$-module?


Thank you in advance,  

Serge



D-module that is coherent as O-module
[Edited to correct errors pointed out by David Ben-Zvi and to answer a query by serge\_I. These corrections reduce this ``answer'' to the status of a comment.] 


(1) (Over $\mathbb C$) Under strong assumptions on the singularities of $X$ (namely, that $X$ should be cuspidal, Ben-Zvi and Nevins, arXiv 0212094v3), if $\mathcal E$ is coherent on $X$, then giving it a structure as a $\mathcal D\_X$-module means giving an isomorphism $\phi:p\_1^\*\mathcal E\to p\_2^\*\mathcal E$, where $\mathcal X\_1$ is the completion of $X\times X$ along the diagonal and $p\_1,p\_2:\mathcal X\_1\to X$
are the projections. (There is also the cocycle condition,
$p\_{31}^\*\phi=p\_{32}^\*\phi\circ p\_{21}^\*\phi,$
where $p\_{ij}:\mathcal X\_2\to\mathcal X\_1$ are the projections from the completion
$\mathcal X\_2$ of $X\times X\times X$ along the diagonal, but we don't need this here.)


Since $X$ is noetherian, there is a unique flattening stratification $X=\coprod X\_i$ associated to $\mathcal E$: each $X\_i$ is locally closed in $X$ and for any $f:Y\to X$, $f^\*\mathcal E$ is locally free if and only if $f$ factors through some $X\_i$. The existence
of $\phi$ shows that $p\_1^\*\mathcal E$ and $p\_2^\*\mathcal E$ have the same flattening stratification, while $\{p\_1^{-1}(X\_i)\}\_i$ is the flattening stratification for $p\_1^\*\mathcal E$
and $\{p\_2^{-1}(X\_i)\}\_i$ is the flattening stratification for $p\_2^\*\mathcal E$.
Since $X$ is irreducible, there is a unique stratum $X\_0$ of maximal dimension, so that
$p\_1^{-1}(X\_0)=p\_2^{-1}(X\_0)$. Now think in terms of a tubular neighborhood of the diagonal in
$X\times X$ to see that this forces $X\_0=X$, so that $\mathcal E$ is locally free provided that $X$ is cuspidal.


(2) (In char. $p$, or over any base) If $X$ is smooth and $\mathcal D\_X$ is taken to be the full ring of differential operators rather than the subring generated by those operators of order at most $1$, then the same argument applies.


