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Suppose $S\subset \mathbb{P}^n$ is a smooth del Pezzo surface and $C$ is an irreducible smooth curve (you can make it rational if it simplifies the setting) such that $\mathcal{L}=\vert -K_S-C\vert $ is non-empty. To simplify you may consider $-K_S$ is very ample and let me deal with the degree $1$ and $2$ cases. Moreover suppose that $h^0(S,\mathcal{O}(\mathcal{L}))\geq 2$ (i.e. $\mathcal{L}$ is at least a pencil).

It was my understanding that by Bertini's theorem one could choose a general member $L\in\mathcal{L}$ such that $L$ is smooth (and reduced and connected). I have been told this is wrong and after going to Hartshorne (and Wikipedia and some expository paper by Kleiman that Francesco added to the comments) I am also of the opinion that it may actually be wrong, but that $L$ must be irreducible away of the base locus of $\mathcal{L}$.

However I am unable of providing a proof nor a counter-example. Does someone have an insight on this? I also suspect the base locus of $\mathcal{L}$ may actually be empty.

Edit: Originally $H$ was a hyperplane section. The question is actually motivated by 'the' hyperplane section so I have rephrased it to meet this point. Apologies for the confusion.

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You need some conditions on the base locus. What if $S$ is the blow-up of $\mathbb PP^2$ at a point, take curves $H,C$ with classes $11L-E$, and $10L-3E$ respectively? These are both ample (and thus indeed contain smooth members). Here $\mathbb L=H-C=L+2E$ satisfies your condition of $h^0(\mathcal L)\ge 2$, but every element of $|\mathcal L|$ is reducible and is non-reduced. –  J.C. Ottem Oct 11 '12 at 18:13
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By the way, the Extended Bertini Theorem(s) quoted in my answer can be also found in Kleiman's paper arxiv.org/PS_cache/alg-geom/pdf/9704/9704018v1.pdf –  Francesco Polizzi Oct 11 '12 at 21:12
    
@Francesco that's the article I mentioned. @JC Ottem: You are right. I have rephrased this to choose the ample divisor I was thinking all the time on. I think in this case it might be true, but I am not sure how to prove it. –  Jesus Martinez Garcia Oct 12 '12 at 11:16
    
Update: For the problem I was trying to apply this on, this is not longer necessary (I solved the problem differently). However, it is something I could have applied on other occasions in many other settings (within del Pezzo surfaces, of course). I hope it gets an answer (by someone not necessarily in MO) at some point. –  Jesus Martinez Garcia Oct 12 '12 at 17:05

2 Answers 2

I think you are right in the special case of a Del Pezzo surface $S$. Here's an idea of proof.

a) we may assume that $K^2_S>1$.

Proof: since $K_S$ is ample, if $K^2_S=1$ then every curve of $|K_S|$ is irreducible.

b) if $|F|$ is an irreducible pencil, then $F^2=0$, $FK_S=-2$, i.e. the general $F$ is a smooth projective curve.

Proof: by Riemann-Roch, we have $2=h^0(F)=1+ (F^2-FK_S)/2$, namely $F^2-FK_S=2$. Since $F^2\ge 0$ and $-FK_S>0$, there are two possibilities:
$F^2=0, K_SF=-2$ or $F^2=1$ and $FK_S=-1$. The second possibility contradicts the index theorem, since $K^2_S>1$ by a).

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Now assume for contradiction that $-K_S=C+rF+Z$, where $C$ is smooth irreducible, $r>1$ is an integer, $|F|$ is an irreducible pencil and $Z$ is an effective divisor such that $|K_S-C|=Z+r|F|$. By b) we have $F^2=0$, $K_SF=-2$.

c) $S$ is not ${\mathbb P}^1\times {\mathbb P}^1$ or ${\mathbb P}^2$.

Proof: ${\mathbb P}^2$ has no free pencil $|F|$; if $S={\mathbb P}^1\times {\mathbb P}^1$ the only possible $|F|$ are the two rulings of the product.

d) $K^2_S\ge 5$

Proof: We have $K^2_S\ge -K_S(C+rF)\ge -K_SC+4\ge 5$, since $-K_S$ is ample.

e) $r\le 3$ and $K^2_S\ge 6$.

Proof: Assume that $S$ is the blow up of ${\mathbb P}^2$ at points $P_1,\dots P_k$, denote by $e_1,\dots e_k$ the corresponding exceptional curves and write $-K_S=3L-(e_1+\dots e_k)$, where $L$ is the class of a line in ${\mathbb P}^2$. Notice that $k\le 4$ by d). The image of $|F|$ in ${\mathbb P}^2$ is either the pencil of lines through, say, $P_1$ or the pencil of conics through, say $P_1, \dots P_4$. Since $-K_S-2F$ is effective, the second case cannot occur. If $k=4$, using a Cremona transformation the two cases can be switched, so $k=4$ (i.e., $K^2_S=5$) cannot occur either. In addition we have $r\le 3$, since $-K_S-rF$ is effective.

f) End of proof: the proof can be completed by enumeration, since the only possibility is that $S$ is the blow up of ${\mathbb P}^2$ at 1, 2 or 3 points and $|F|$ is the pencil of lines through one of the points.

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The point here is the following result, that you can find in Zariski's book "Algebraic Surfaces", page 26. Zariski calls it "Extended Theorem of Bertini".

Theorem (Extended Bertini)

(1) The general curve of an irreducible linear system cannot have multiple points outside the base locus of the system.

(2) A reducible linear system, without fixed components, is necessarily composed with the curves of a pencil.

Here "reducible" [resp. "irreducible"] means that the general curve of the system is reducible [resp. irreducible].

Now, let us write $\mathcal{L}=Z+\mathcal{M}$, where $Z$ is the fixed part and $\mathcal{M}$ is the moving part. Then by Extended Bertini it follows that the general element $M \in \mathcal{M}$ is necessarily irreducible, unless $\mathcal{M}$ is composed with a pencil.

The last situation can occur. For instance let $S=\mathbb{P}^1 \times \mathbb{P}^1$, whose natural pencils are denoted by $|F_1|$ and $|F_2|$, and take $H=F_1+2F_2$ and $C \in |F_1|$. Then $H$ is very ample but $\mathcal{L}=|H-C|=|2F_2|$, which is without fixed part and composed with the pencil $|F_2|$. In fact, any element of $|2F_2|$ is the union of two curves in the pencil $|F_2|$, in particular it is not irreducible.

Remark. The situation described in J. C. Ottem's comment is slightly different. In that example, indeed, we have a fixed part $Z=2E$; the moving part, however, is irreducible.

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Thank you, Francesco. This is the kind of thing I was thinking on. However your counter-example is not exactly what I was after: $H$ is not a hyperplane section, but rather $F_1+F_2$ is. So in some way, if $B$ is a hyperplane section, you are taking $H=2B-F_1$. Then it is clear you can extract a second $F_1$. My point was to start with a more 'canonical' choice of $H$. The question said 'hyperplane section', but what I actually really want is $H=-K_S$, which for Fano in dimension 2 is either very ample (deg >2), or has Fano index 1 (deg <1). I'll rephrase the question to fit that. –  Jesus Martinez Garcia Oct 12 '12 at 10:26
    
In the meanwhile, though, I think if you take in your setting $H=\frac{1}{2}-K$ or even $H=-K$ what I said seems to be true, namely: any general element of $\vert H-C\vert$ is irreducible. I quickly went through Kleiman's article yesterday night, which made me think about proving the following: If $\vert H-C \vert $ is base-point free, then the general member is irreducible. This is not too hard using the base-point free theorem (which probably wasn't available to Zariski). It still doesn't avoid (2) of Extended Bertini. Finally, what does it mean 'composed with the curves of a pencil'? –  Jesus Martinez Garcia Oct 12 '12 at 10:41
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Well, in my example $H$ is a hyperplane section if you embed $\mathbb{P}^1 \times \mathbb{P}^1$ in $\mathbb{P}^5$ by using the linear system $|F_1+2F_2|$. Often, "hyperplane section" is just used instead of "very ample divisor". But you are right, I will correct the answer. Of course, in this case as you noticed we do not have $H=K_S$. –  Francesco Polizzi Oct 12 '12 at 11:18
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'The linear system $\mathcal{L}$ is composed with the curves of a pencil' means exactly what one expects: i.e., there exists a rational pencil $|P|$ such that any curve of $\mathcal{L}$ is sum of curves of $|P|$. In particular, any curve of $\mathcal{L}$ is reducible. For instance, the linear system $|2F_1|$ on $\mathbb{P}^1 \times \mathbb{P}^1$ is composed with the pencil $|F_1|$. –  Francesco Polizzi Oct 12 '12 at 11:22
    
Thank you for the clarifications. I think it may be true (i.e. there is a smooth irreducible section in $\vert -K-C\vert$ when the linear system is at least a pencil on a del Pezzo) but I'm still struggling to prove it though. I'll let you know if I make progress. I suspect it is probably not true in higher dimensions, though. –  Jesus Martinez Garcia Oct 12 '12 at 12:36

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