Suppose that $X$ is a smooth algebraic variety over an algebraically closed (uncountable if it helps) field of characteristic $p > 0$. Suppose that $L$ is a line bundle, probably ample or at least positive, and that $\delta \subseteq |H^0(X, L)|$ is a linear system.

It is well known that just because $\delta$ is base-point-free, it does not mean that a general member defines a smooth subscheme (it need not even be reduced, for example Frobenius pull-backs of linear systems, Remark 10.9.3 in Hartshorne's algebraic geometry).

However, let's suppose the following:

$f : X \to Y$ is a map, $M$ is a very ample line bundle on $Y$ and $L = f^* M$. Suppose further that $\delta$ is the pull-back of the complete linear system $|H^0(Y, M)|$. Are there any (separability?) conditions on the map $f$ which still guarantee that Bertini holds for $\delta$? I imagine this must be well known, but I don't know the right references.

In particular, I am looking for conditions weaker than etale (/ etale outside a finite set of points)? Say in the birational case, or the finite case?

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    $\begingroup$ I have nothing intelligent to say about this nice question, but I think $\sigma$ in the second pgph. should be $\delta$. $\endgroup$ – user5117 Jun 9 '11 at 7:24
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    $\begingroup$ Have you looked at Jouanolou's "Theoremes de Bertini et Applications"? I haven't looked at it recently, so don't remember the details, but I think that what he proves is close to the best possible. $\endgroup$ – naf Jun 9 '11 at 8:16
  • $\begingroup$ @Artie, thanks for the typo. @ulrich, I'll check that out, thanks! $\endgroup$ – Karl Schwede Jun 9 '11 at 12:31
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    $\begingroup$ Karl, Ulrich is right that Jouanolou has a strong version (I'm travelling but there should be a precise reference in my "Mirabolic Langlands/quantum Calogero-Moser" paper, though perhaps that's not the precise result from Jouanolou you need). I think it's along the lines of generically unramified maybe? $\endgroup$ – Thomas Nevins Jun 9 '11 at 21:39
  • $\begingroup$ Tom, thanks. I still have to procure my copy of Jounalou, but your paper at least states it if the morphism is unramified, and refers to Theorem 6.3 in that book. To me, unramified is a very strong condition. But perhaps one can weaken it somewhat (in some generic way as you suggest). $\endgroup$ – Karl Schwede Jun 10 '11 at 2:49

I think Corollary 4.3 of Spreafico's Axiomatic theory for transversality and Bertini type theorems does what you want. It says (in the case where the property is taken to be smoothness) that if $f:X\to \mathbb P^n$ is a finite type morphism from a smooth scheme $X$ over any infinite field, and if $f$ is residually separated (i.e. the induced extensions of residue fields are separable), then the pullback of a generic hyperplane is smooth.

  • $\begingroup$ Thanks, I hadn't seen that paper before. I had found the work of Cumino, Greco and Manaresi which Spreafico generalizes. $\endgroup$ – Karl Schwede Aug 23 '11 at 23:40

I thought it might be useful to post a statement of an answer. This is contained within Cumino-Greco-Manaresi and also Spreafico's paper which Anton mentioned earlier, but can't really fit in a comment and I think might be useful to someone else. I probably should have posted this earlier.

Indeed, in the context of my original question, if $f : X \to Y$ induces separable residue field extensions (or separably generated not necessarily algebraic residue field extensions), then everything is fine (say over an algebraically closed field). I should note that even really nice maps may fail to have separably generated residue field extensions.

Indeed, consider the projection map $\mathbb{A}^2 \to \mathbb{A}^1$ (if you want, this can be an open chart from the projection $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$) over a field $k$ of characteristic $p > 0$. This corresponds to the inclusion $k[x] \subseteq k[x,y]$. This does not have separable residue field extensions.

Consider the ideal $(x-y^p) \subseteq k[x,y]$. This pulls back to the zero ideal in $k[x]$ and induces the residue field extension $k(x) \subseteq k(x^{1/p})$ which is clearly non-separable.

This indicates to me that even birational maps can fail this property horribly (for example, if I blow up a curve on a 3-fold similar things can be arranged). Finite maps are much better behaved of course.


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