2
$\begingroup$

Is there always a reducible curve (EDIT: with exactly two irreducible components intersecting in at least 2 points) in a complete linear system (EDIT: of dimension at least 2 with curves of genus at least 1) on a Del Pezzo surface?

$\endgroup$
7
$\begingroup$

No. Let $S$ be a Del Pezzo surface of degree 1, namely $K^2_S=1$. The anticanonical system $|-K_S|$ has dimension 1 and it contains no reducible curve, since $-K_S$ is ample and has self-intersection $1$.

EDIT: the system of lines in $\mathbb P^2$ shows that the answer to the edited question is also negative. Other counterexamples are given by the system of curves of $\mathbb P^1\times\mathbb P^1$ of bidegreee $(1,n)$, $n\ge 1$.

$\endgroup$
5
  • $\begingroup$ Ok, sorry, I meant linear systems of dimension at least 2. $\endgroup$
    – sqrt2sqrt2
    Oct 12 '13 at 9:47
  • 1
    $\begingroup$ If you are interested in a precise example of linear system, it may be possible to answer the question by using numerical considerations, but asking about the intersection of the components in general does not make much sense. $\endgroup$
    – rita
    Oct 12 '13 at 9:58
  • $\begingroup$ Ok. If I consider only the case of a reducible curve with exactly two irreducible components intersecting in at least 2 points, can I say that in every linear system of dimension at least 2 on a Del Pezzo there is always such a curve? $\endgroup$
    – sqrt2sqrt2
    Oct 12 '13 at 10:24
  • $\begingroup$ Thank you Rita for the answers. What about the case of genus at least 1 (and dimension of the linear system at least 2)? $\endgroup$
    – sqrt2sqrt2
    Oct 14 '13 at 13:32
  • 1
    $\begingroup$ My feeling is that the statement might be correct (I thought about it a bit couldn' find a counterexample), but I wouldn't know how to prove it. $\endgroup$
    – rita
    Oct 14 '13 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.