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Let $X$ be a projective, normal variety over complex field with canonical singularities. Suppose $|D|$ is a basepoint free linear system, then is it true that the generic elements in $|D|$ are irreducible?

Besides, I noticed that something might related to "free linear system" (see Mori, Kollár "Birational geometry of algebraic varieties" Page 158, Lemma 5.17). Is "free linear system" the same as basepoint free linear system?

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2 Answers 2

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No, the image of the associated morphism, $$\phi_{|D|}:X\to \mathbb{P}^n,$$ might be a curve. If the image has dimension $\geq 2$, then the general member is irreducible. Look up "Bertini theorems".

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  • $\begingroup$ Thank you so much! Which "Bertini Theorem" did you refer to? I looked up the Bertini theorem in "singularity in pairs" but was unable to find the similar statement... Besides, may I ask one more question? Is "free linear system" (as in Mori, Kollár "Birational geometry of algebraic varieties" Page 158, Lemma 5.17) the same as "basepoint free linear system"? Thank you again! I appreciate it very much! $\endgroup$
    – Li Yutong
    Mar 2, 2014 at 15:17
  • $\begingroup$ @LiYutong: Theorem 5.3, p. 25, of Steven Kleiman's beautiful historical article about Bertini's theorems, Bertini and His Two Fundamental Theorems. $\endgroup$ Mar 2, 2014 at 15:18
  • $\begingroup$ Great, it is so nice of you to point out this to me! $\endgroup$
    – Li Yutong
    Mar 2, 2014 at 15:21
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Just to illustrate this on a simple example: take on $\mathbb{P}^1\times \mathbb{P}^1$ the divisor $D$ equivalent to $2f$, where $f$ is the fibre of one projection. Then $|D|$ is base-point free but any member is the union of two fibres (computing the intersection with $f$ we get $0$ so it is vertical), so is not irreducible.

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  • $\begingroup$ I see! Thank you very much for this simply example! $\endgroup$
    – Li Yutong
    Mar 2, 2014 at 15:18

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