# Lifting vector fields to its resolution in char $p$

In this paper 4.7, the authors showed that on a normal variety $X$, if there is a tangent vector field on its smooth locus, then it can be lifted as a logarithmic tangent vector field on a log resolution $\tilde{X}$, with logarithmic poles along the exceptional locus. Roughly speaking, the lifting of the tangent vector field automatically doesn't have deep poles.

Unfortunately, the proof there was only in characteristic 0. My question is

If we assume that resolution of singularity (any reasonable version you want) holds in char $p$, can we prove such a result in char $p$ as well?

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Dear CX, Taking a short break from revising: I think this fails in positive characteristic. Let $k$ be a field of characteristic $p>0$. Let $m>1$ be an integer.
Let $\mathbb{A}^6_k$ have coordinates $x_0,x_1,x_2,y_0,y_1,y_2$. Consider the affine hypersurface $X$ with defining equation $f = x_0^py_0+x_1^py_1 - x_2^{pm}y_2$. By the Jacobian criterion, the singular locus is contained in the closed subset $Z(x_0^p,x_1^p,x_2^{pm})$, which has codimension $2$ in $X$. Since $X$ is Cohen-Macaulay and regular in codimension $1$, $X$ is normal by Serre's criterion. Let $R$ denote the coordinate ring of $X$, i.e., $R=k[x_i,y_j]/\langle f \rangle$. By direct computation, the sheaf of relative differentials is the $R$-module with presentation $\Omega_{R/k} = R\{dx_i,dy_j\}/\langle x_0^pdy_0 + x_1^pdy_1 - x_2^{pm}dy_2\rangle$. In particular, consider the derivation $$\theta:\Omega_{R/k} \to R, \ dy_j \mapsto 0, dx_0 \mapsto 1, dx_1 \mapsto 0, dx_2\mapsto 0.$$ Consider the (partial) resolution $\nu:\tilde{X} \to X$ where $\tilde{X}$ is the affine $5$-space with coordinates $u_0,u_1,x_2,y_0,y_1$, and where $\nu(u_0,u_1,x_2,y_0,y_1)$ is $$(x_0,x_1,x_2,y_0,y_1,y_2) = (u_0x_2^m,u_1x_2^m,x_2,y_0,y_1, u_0^py_0+u_1^py_1).$$ The exceptional set is the zero set of $x_2$. On the complement of this open set, $\theta$ is the derivation $(1/x_2)^m \partial/\partial u_0$. This derivation has a pole of order $m$ along the exceptional set. Thus, this derivation does not have log poles on the exceptional set.
$\textbf{Edit.}$ Changed pole order $2$ to arbitrary pole order $m$.