4
$\begingroup$

Let $X$ be a projective variety over an imperfect (hence infinite and char(k)=p>0) field $k$. If the local rings of $X$ are all regular, then can we say that a general hyperplane section $H$ is also regular? If it helps, you can assume any combination of the following hypothesis on $k$: $k$ contains a perfect (infinite) subfield $k_0$ or $k$ is a $F$-finite field or $k$ is a differentially finite over an (infinite) perfect subfield $k_0$ i.e., $\Omega_{k/k_0}$ is a finite dimensional vector space over $k_0$, or perhaps even the most geometric scenario that $k$ is a finite dimensional function field over an algebraically closed field $k_0$.

These type of results are very useful for studying varieties over imperfect fields; in particular studying families in positive characteristic over algebraically closed fields.

Note that there is a theorem of Seidenberg which says that if $X$ is a normal projective variety of dimension at least $2$ over an infinite field $k$, then a general hyperplane section $H$ of $X$ is irreducible and normal.

$\endgroup$
  • 1
    $\begingroup$ If $k$ is finitely generated over a perfect subfield $k_0$, then every regular $k$-variety $X$ is a localisation of a smooth $k_0$-variety $\mathcal X$. You could try to use Bertini theorems on $\mathcal X$ (if $k_0$ is infinite, or Poonen–Bertini if $k_0$ is finite) to prove statements on $X$. $\endgroup$ – R. van Dobben de Bruyn Oct 5 '17 at 19:31
  • 1
    $\begingroup$ Have you looked at Spreafico's Axiomatic theory for transversality and Bertini type theorems? Spreafico works over arbitrary infinite fields, and shows that (in particular, see Cor. 4.3 and what follows) if $X$ is regular and $X \to \mathbf{P}^n$ induces separable field extensions of residue fields at all points, then the inverse image of a general hyperplane is regular. I think this condition is satisfied for closed embeddings. $\endgroup$ – Takumi Murayama Oct 5 '17 at 21:54
  • $\begingroup$ @ R. van Dobben de Bruyn, this is what I had in mind but would like to know if there is a more direct approach, since my filed $k$ not always a function field. $\endgroup$ – user80473 Oct 5 '17 at 22:08
  • $\begingroup$ @Takumi Murayama, thank you so much for the reference, I was not aware of this article. I will take a look. $\endgroup$ – user80473 Oct 5 '17 at 22:09
  • 1
    $\begingroup$ @Jason Starr, by general hyperplane $H$ I mean the following: Let $X\subset\mathbb{P}^n_k$. Then there exists a Zarsiki dense open set $U\subset{\mathbb{P}^n_k}^\vee$ such that $H$ corresponds to a $k$-rational point $[H]\in U(k)$. $\endgroup$ – user80473 Oct 6 '17 at 2:02
4
$\begingroup$

I needed to know the answer to this myself, so here is a good reference:

Hubert Flenner, Liam O’Carroll, and Wolfgang Vogel, Joins and intersections, Springer Monographs in Mathematics, Springer-Verlag, Berlin, 1999. MR 1724388 DOI 10.1007/978-3-662-03817-8

The relevant results are the following:

Theorem 3.4.10. Let $X$ be a variety over an infinite field $K$, and let $D$ be a Cartier divisor on $X$. Assume that $\Gamma \subseteq \lvert D \rvert$ is a finite-dimensional linear system that is not composed with a pencil and satisfies $\operatorname{codim} \operatorname{Bs}(\Gamma) \ge 2$. Then, a generic member of $\Gamma$ is irreducible.

Corollary 3.4.14. Let $X \subseteq \mathbf{P}^n_K$ be a projective scheme over an infinite field $K$ which is regular (resp. normal, reduced, satisfies $R_k$). Then, for a generic hyperplane $H \subseteq \mathbf{P}^n_K$, the intersection $X \cap H$ has the same property.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.