Let $X\subset \mathbb{P}^n\_k$ be a smooth projective variety, a point $p\in \mathbb{P}^{n,\vee}\_k$ gives rise to a hyperplane $H\_p\subset \mathbb{P}^n$, hence an intersection $X\_p:=H\_p\cap X$.


We say a line $L\subset\mathbb{P}\_k^{n,\vee}$ is a Lefschetz pencil if


(1) There exists $0,\infty \in L(k)$ such that $H\_0, H\_\infty, X$ intersect transversally.


(2) For $p\in L-\{s\_1,...,s\_r\}$, the section $X\_p$ is smooth, $H\_{s\_i}\cap X$ has exactly one quadratic ordinary singularity.


When $k$ is an algebraically closed field of $\mathrm{char}(k)=p$, do we know if Lefschetz pencils always exist? If not, can we relax the assumption by allowing $L$ be a smooth curve in $\mathbb{P}^{n,\vee}\_k$?


(From Theorem 3 [here](https://arxiv.org/pdf/0911.1470.pdf), Lefschetz pencils always exist over any field after $d$-uple embedding of $X$)



Does Lefschetz pencil always exist in char $p$?
I think you are asking whether a Lefschetz pencil exists without re-embedding. Then the answer is no. In cor. 3.5.0 of expose XVII of SGA 7, Katz gives a necessary and sufficient condition for a Lefschetz pencil to exist. In the case of a hypersurface $X=V(F)$ in $\mathbb{P}^n$ and $char k=p\not=2$, the condition amounts to the Gauss map $\phi:X\to (\mathbb{P}^n)^\vee$ defined by $\phi(x\_0,\ldots, x\_n) = (\partial F/\partial x\_i)$ being seperable.
But he shows this condition fails for $F= \sum x\_i^{p^n+1}$.


