Let $k$ be an algebraically closed field and let $p$ be the characteristic of the field. Let $f : X \to S$ be a projective morphism such that the fibres are generically reduced, pure of dimension $1$. We suppose $X$ and $S$ are both irreducible and quasi-projective varieties. For every $s \in S(k)$, does there exist a Zariski open neighbourhood $S'$ of $s$ such that the morphism $f$ factors through a finite morphism $g : X \to \mathbb{P}^1 \times S'$ and is such that for every $x \in S'(k)$, the degree of the morphism $g_x : X_x \to \mathbb{P}^1 \times \{x\}$ is co-prime to $p$. By this we mean that for every generic point $\eta' \in X_x$, the degree of the extension of function fields $k(\mathbb{P}^1) \hookrightarrow k(\eta')$ is coprime to $p$.
1 Answer
No, that is not always possible. For instance, if $p$ equals $2$, if $S$ is the dense, Zariski open subscheme of $\mathbb{P}^5$ parameterizing smooth plane conics in $\mathbb{P}^2_k$, and if $f:X\to S$ is the restriction to $S$ of the universal conic, then the index of $f$ equals $2$.
The simplest way to see this is to consider the universal conic $\overline{X} \subset \mathbb{P}^5_k \times_k \mathbb{P}^2_k$. The projection $g:\overline{X} \to \mathbb{P}^2_k$ is a projective subbundle of $\mathbb{P}^5_k\times_k \mathbb{P}^2_k$ of relative dimension $4$. In particular, the Abelian group $\text{CH}^1(\overline{X})$ is $$\text{CH}^0(\mathbb{P}^2)\cdot H \oplus \text{CH}^1(\mathbb{P}^2) = \mathbb{Z} H \oplus \mathbb{Z}h,$$ where $H$, resp. $h$, is the restriction to $\text{CH}^1(\overline{X})$ of the pullback of the universal hyperplane class in $\text{CH}^1(\mathbb{P}^5)$, resp. $\text{CH}^1(\mathbb{P}^2)$. Consider the proper pushforward homomorphism, $$f_*: \text{CH}^1(\overline{X}) \to \text{CH}^0(\mathbb{P}^5).$$ The summand $\mathbb{Z} H$ is in the kernel of $f_*$. Thus, the image of $f_*$, as a subgroup of $\text{CH}^0(\mathbb{P}^5) = \mathbb{Z}$, is generated by $f_*h$. Since the plane conics have degree $2$, $f_*h$ is $2$ times the generator of $\text{CH}^0(\mathbb{P}^5)$. Thus, every cycle class on $\overline{X}$ of relative dimension $0$ over $\mathbb{P}^5$ has degree divisible by $2$. For a cycle $Z$ in $X$ that has relative dimension $0$ over $S$, the closure $\overline{Z}$ in $\overline{X}$ is a cycle that has relative dimension $0$ over $\mathbb{P}^5$. Thus the index of $X$ over $S$ equals $2$.