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I'm studying the proof of Bertini's theorem on "Principles of Algebraic Geometry" by Griffiths and Harris (page 137). The statement is as follows:

The generic element of a linear system is smooth away from the base locus of the system.

The authors claim that it suffices to prove Bertini for a pencil (that is, a linear system of dimension one). Unfortunately, I am not able to understand how this reduction to pencils is possible. I am quite a newcomer in Algebraic Geometry, so I'm looking for a very detailed explanation. It is probably a triviality, nonetheless I couldn't manage to find a proof, either by myself, or by asking my colleagues for help.

Thank you in advance!

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  • $\begingroup$ This is just a quick guess, but if we take a generic element of an n-dimensional system, it is contained in a plethora of pencils contained in this n-dimensional system. So it is a generic element of a pencil in quite a lot of ways. $\endgroup$
    – Joachim
    Jul 1, 2013 at 16:19
  • $\begingroup$ One probably should find a way to deal with the fact that a pencil contained in the n-dimensional system can have a bigger base locus, but this should be doable.. Probably by considering all the different pencils your generic element is contained in.. But really someone else should give a complete answer... $\endgroup$
    – Joachim
    Jul 1, 2013 at 16:21
  • $\begingroup$ A problem I encountered is about the base locuses: the base locus of a linear system of dimension n > 1 is smaller than the base locus of a pencil contained in it. Even if we know that a generic element of any "subpencil" is smooth away from the base locus of that pencil, that does not allow us to conclude without any other argument, I think. EDIT: You've written about this problem as I was writing about it! Yes, of course it is "doable", but I really can't understand how! $\endgroup$ Jul 1, 2013 at 16:24
  • $\begingroup$ But isn't the intersection of the base locus of all the pencils that pass through a fixed element equal to the base locus of the entire system? So if a generic element is smooth except possibly in the base locus of a pencil that contains it, and this holds for all such pencils, it must be smooh except for the base locus of the entire system? Not sure if this works but it seems to.. $\endgroup$
    – Joachim
    Jul 2, 2013 at 1:10
  • $\begingroup$ Since the proof on G&H is likely to be false, does the OP know where one can find a complete proof of this theorem (in the same, i.e., complex analytic context)? $\endgroup$
    – Colescu
    Jul 8, 2020 at 11:56

1 Answer 1

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I asked myself this question some time ago, and I asked for help from several people. I will now try to reconstruct all the story, even though...

I am sorry, but I do not know whether the reduction to pencils is actually possible; I am trying to outline why in my opinion it is not possible. In summary, here is the reason why I think it's not possible: The Bertini condition is not (visibly) open.

Here is the strategy: I want to give a (false!) proof of the fact that the reduction is possible, and I will mark with a $(\star)$ the point where, won by tempation, I will be deducing something that is likely to be false.

So, let's get started: we want to show that

$$ \textrm{Bertini true for a generic pencil} \overset{(\bullet)}{\Longrightarrow}\textrm{Bertini true for the whole system.} $$

Notation: $X$ is our variety, and we have a linear system $\psi:X\dashrightarrow \mathbb P^r$ of dimension $r>1$. We will call "good" a divisor with no singular points outside the base locus of the linear system it is regarded into; it will be called "bad" otherwise. Let us introduce the Grassmannian $G=\mathbb G(1,r)$, the space of lines in $\mathbb P^r$. We denote by $p:\mathbb P^r\times G\to \mathbb P^r$ and $q:\mathbb P^r\times G\to G$ the projections.

Proof of $(\bullet)$. Let us assume that Bertini holds for a generic pencil $\ell\subset \mathbb P^r$ contained in the big system: this means that for $\lambda$ generic in $\ell$ the divisor $D_\lambda$ is good with respect to $\ell$. We want to find a dense open subset of $\mathbb P^r$ parameterizing good divisors for $\mathbb P^r$. To do so, let us describe the "bad" points of $\mathbb P^r$. There are two possibilities:

  1. The point $\lambda\in\mathbb P^r$ is on a "bad" line, that is, a line for which Bertini does not hold: there is no dense open subset of this line parameterizing good divisors for this pencil. Since by hypothesis we deal with a generic pencil, $$\textrm{the set of these lines is a proper closed subset } B\subset G.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\star)$$
  2. Let us call $U=G\setminus B$ the open set of "good" lines. If $[\ell]\in U$ is one of them, it might contain some bad points $P_1(\ell),\dots,P_{n_\ell}(\ell)$. However, these points are in a finite number, because of our assumption (Bertini): the generic $\lambda\in\ell$ is good. So we have a bad locus $$ C=\{(P_i(\ell),[\ell])\,|\,1\leq i\leq n_\ell\,\textrm{ and }\,[\ell]\in U\}\subset\mathcal L, $$ where $\mathcal L\subset \mathbb P^r\times G$ is the universal line. As the dependence of the points $P_i(\ell)$ on the line $\ell$ is algebraic (it is given by the equation of $\ell$), we conclude that $C$ is a closed subset of $q^{-1}(U)$.

Summary: $q^{-1}(B)$ consists exactly of the bad points on $\mathcal L$ which lie on some bad line (we get these through point 1); the only other possibility for being bad is to lie on $C$ (we get these through point 2).

So $q^{-1}(B)\cup C$ is the bad locus inside $\mathcal L$; note that it is closed. The bad points of $\mathbb P^r$ which we are interested in are exactly the points in $p(q^{-1}(B)\cup C)$. But $p$ is a proper morphism, hence closed, so the bad points form a closed subset of $\mathbb P^r$. Hence the good points form an open, thus dense, subset of $\mathbb P^r$, which means that Bertini holds for the $r$-dimensional linear system. $\,\,\,\,\,\square$

Yes, but to assert $(\star)$ is to say that the Bertini condition is open, and this is by no means evident.

$$\ast\,\,\,\,\,\,\,\,\ast\,\,\,\,\,\,\,\,\ast$$

Another approach (in fact, just a translation): try to show $(\bullet)$ by contrast. Let us give a name, say $V$, to the good locus, i.e. the locus of hyperplanes giving a smooth transversal intersection with our variety $X$: $$ V:=\{\textrm{hyperplanes }H\,\textrm{such that}\,|\,\psi^{-1}(H)\,\textrm{is smooth of dimension}\, \dim X-1\}. $$ Then, to say that Bertini is false for $\mathbb P^r$ is to say that $V$ has empty interior (i.e. there is no open subset of $\mathbb P^r$ parametrizing good divisors. I cannot see why this should imply that for a generic line $\ell\subset \mathbb P^{r}$ the corresponding $V_\ell$ has empty interior as well.

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  • $\begingroup$ Then, should I conclude that Griffiths and Harris give a wrong proof? They argue as follows (I quote the exact words): If the generic element of a linear system is singular away from the base locus of the system, then the same will be true for a generic pencil contained in the system; thus it suffices to prove Bertini for a pencil. I don't know how the negation of "the generic element is smooth" could be "the generic element is singular", actually! I find this argument quite unclear. $\endgroup$ Jul 2, 2013 at 6:13
  • $\begingroup$ You are definitely right, what you quote is not the negation of Bertini. The negation is what I wrote in the last paragraph. I do not know if their proof is wrong. If the reduction is possible, it is not obvious at all. However I think it is not possible. $\endgroup$
    – Brenin
    Jul 2, 2013 at 6:48
  • $\begingroup$ @Brenin What do you mean by a "generic pencil"? $\endgroup$
    – Roxana
    May 20, 2021 at 10:57

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