Bertini's Theorem small print - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:54:59Z http://mathoverflow.net/feeds/question/109400 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109400/bertinis-theorem-small-print Bertini's Theorem small print Jesus Martinez Garcia 2012-10-11T17:58:02Z 2012-10-21T21:09:56Z <p>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).</p> <p>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 $\mathcal{L}$.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/109400/bertinis-theorem-small-print/109404#109404 Answer by Francesco Polizzi for Bertini's Theorem small print Francesco Polizzi 2012-10-11T18:23:24Z 2012-10-12T11:26:03Z <p>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".</p> <blockquote> <p><strong>Theorem (Extended Bertini)</strong></p> <p><strong>(1)</strong> The general curve of an irreducible linear system cannot have multiple points outside the base locus of the system.</p> <p><strong>(2)</strong> A reducible linear system, without fixed components, is necessarily composed with the curves of a pencil.</p> </blockquote> <p>Here "reducible" [resp. "irreducible"] means that the general curve of the system is reducible [resp. irreducible].</p> <p>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, <strong>unless $\mathcal{M}$ is composed with a pencil</strong>.</p> <p>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 <strong>not</strong> irreducible. </p> <p><strong>Remark.</strong> 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. </p> http://mathoverflow.net/questions/109400/bertinis-theorem-small-print/110271#110271 Answer by rita for Bertini's Theorem small print rita 2012-10-21T21:09:56Z 2012-10-21T21:09:56Z <p>I think you are right in the special case of a Del Pezzo surface $S$. Here's an idea of proof.</p> <p>a) we may assume that $K^2_S>1$. </p> <p>Proof: since $K_S$ is ample, if $K^2_S=1$ then every curve of $|K_S|$ is irreducible.</p> <p>b) if $|F|$ is an irreducible pencil, then $F^2=0$, $FK_S=-2$, i.e. the general $F$ is a smooth projective curve. </p> <p>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:<br> $F^2=0, K_SF=-2$ or $F^2=1$ and $FK_S=-1$. The second possibility contradicts the index theorem, since $K^2_S>1$ by a).</p> <p>======</p> <p>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 $F^2=0$, $K_SF=-2$.</p> <p>c) $S$ is not ${\mathbb P}^1\times {\mathbb P}^1$ or ${\mathbb P}^2$.</p> <p>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. </p> <p>d) $K^2_S\ge 5$</p> <p>Proof: We have $K^2_S\ge -K_S(C+rF)\ge -K_SC+4\ge 5$, since $-K_S$ is ample.</p> <p>e) $r\le 3$ and $K^2_S\ge 6$.</p> <p>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.</p> <p>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.</p>