I always believed the following statement: if $X$ is a smooth variety over an algebraically closed field of positive characteristic, assuming we know that the general member of a base point free linear system $|L|$ is reduced, then indeed a general member is smooth.

However, I realize this is not obvious, though all the examples I know which fail this Bertini theorem has non-reduced fibers.

So I was wondering whether indeed this statement is true. Or counterexamples are known.

[REEDIT:] After Laurent Moret-Bailly's nice counterexample. I was then wondering whether a counterexample exists when the linear system induces a birational morphism to its image. The assumption now is closer to the one in Bertini's theorem which says the statement is true when we assume the morphism is an embedding.

This sort of counterexamples will always help birational geometers to think what's going on in characteristic $p$.

  • $\begingroup$ you might want to have a look at Jouanolou's book 'Theoremes de Bertini et applications' $\endgroup$ – Christian Liedtke Jan 21 '12 at 17:15
  • $\begingroup$ I know this book which is not currently in our library though. But I'm not sure kind of examples exist there, do they? $\endgroup$ – CYXU Jan 31 '12 at 1:32

In characteristic 3, consider the surface $V\subset \mathbb{P}^2\times\mathbb{A}^1$ with equation $y^2z=x^3-tz^3$ (where $t$ is the coordinate on $\mathbb{A}^1$). It is easily seen to be smooth. The general fiber of the projection on $\mathbb{A}^1$ is a plane cubic which is reduced but not smooth. Taking a suitable projective completion $V\hookrightarrow X$ you get a morphism $X\to\mathbb{P}^1$ which is a counterexample.

  • $\begingroup$ Dear Laurent: is this an example of a quasi-elliptic surface? $\endgroup$ – user5117 Jan 20 '12 at 14:20
  • $\begingroup$ Dear Laurent Moret-Bailly: Thank you very much for this nice example! I was also wondering whether we can move one more step to ask when the linear system induces a birational morphism, whether the counterexample still exists. $\endgroup$ – CYXU Jan 20 '12 at 14:55
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    $\begingroup$ @Artie: yes, I think that's how they are called. $\endgroup$ – Laurent Moret-Bailly Jan 20 '12 at 15:07

[Added at the end: when I started this response I thought I had a complete answer, but as I started typing I realized that I don't. Yet, I think that this might still be a useful observation, so I will post it.]

The only place where $\mathrm{char}=0$ is used is to conclude that the family $f:W\to G$ of hyperplane sections of $X$ mapping to $G=\mathrm{PGL}_n$ is generically smooth. If you know that the general fiber is reduced, then it implies that the general fiber is generically smooth.

In order to get Bertini, we need that $f$ is generically smooth on $G$. In characteristic $0$ one can prove this by applying the above observation, but not for that family!

Define $W_0=\{\text{closed points }x\in W \ |\ T_{W,x}\to T_{G,f(x)}\text{ is not surjective} \}$, let $W_1=\overline W_0$ with the reduced induced scheme structure and consider the induced $f_1:W_1\to G$. If $f_1$ is not surjective, we are done. If it is, replace $W_1$ with an irreducible component on which $f_1$ is surjective. Now it follows that this $f_1$ cannot be generically smooth, because its tangent map is not surjective anywhere.

In other words, in order to get this more general Bertini you would have to know a sort of second order reducedness: You need that if you take the family of singular loci of all the hyperplane sections of $X$, take the reduced induced scheme structure on the family, then the general member of that family will be reduced.

I suppose the next obvious question is if someone can show an example that this does not hold automatically if your assumption holds or prove that such an example does not exist, so your desired statement is true. I am leaning towards the first possibility, but I don't know such an example off the top of my head.

  • $\begingroup$ To S\'andor: Thanks for your remark! Yes, my question is whether such examples exist. $\endgroup$ – CYXU Jan 20 '12 at 4:53

Since there's already an example posted, I'll point out that if your linear system gives a map $X \to \mathbb{P^n}$ with separable residue field extensions (at ALL points), then you are ok

EDIT: (of course, this is a very rare condition as pointed out in the comments below...):

Also see the question: of mine. That question has some links to some positive statements.

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    $\begingroup$ I don't see how this can happen except in trivial cases (e.g. the map is constant or finite). $\endgroup$ – Laurent Moret-Bailly Jan 20 '12 at 8:49
  • $\begingroup$ Indeed, this is a very rare condition with very little hope except in the finite case (perhaps one can also exclude certain loci where this happens as well). $\endgroup$ – Karl Schwede Jan 20 '12 at 11:39

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