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Let $X$ be a projective surface and $D$ be a divsor. Then I know $D$ correspond to a curve of $X$. My qeustion is simple. If $D$ is very ample, then the corrsponding curve of $D$ is irreducible? More generally, if $X$ be a projective variety of dimension $r$, then the corrsponding subvariety of a very ample divsor $D$ is irreducible??

Sang Cheol Lee
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    I'll let others who can be more enlightening post a proper answer, but ampleness ( and very ampleness ) is really a question about the corresponding line bundle, and divisors that are linearly equivalent give the same line bundle. In particular, take any embedded surface with a reducible hyperplane section and you can find a sort of counterexample to your claim. However, there are also Bertini theorems which very roughly say that your statement is true most of the time. – Callus - Reinstate Monica Mar 04 '14 at 07:00
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    Dear @Callus, you are too modest. Your comment summarises the situation perfectly: the statement as written is not true, but Bertini's theorem guarantees that the general member of a very ample linear system is smooth and irreducible. –  Mar 04 '14 at 09:13
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    A naive (but possibly helpful) observation: the sum of two ample divisors is ample. – Heitor Fontana Mar 04 '14 at 12:18

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No, very ample divisors needn't have irreducible support.
For example linearly embed $\mathbb P^r$ as $P\subset \mathbb P^{r+1}$ and consider three distinct hyperplanes $H,K,L\subset \mathbb P^{r+1}$ all different from $P$.
Then the divisor $1.(H\cap P)+1.(K\cap P)-1.(L\cap P)$ is very ample on $P\cong \mathbb P^r$ but its support has 3 irreducible components (which are 3 hyperplanes of $P$).

Edit
In line with Asal's judicious comment, let me confirm that the simplest ample divisor with reducible support on a complete surface is the sum $1.h+1.k$ of two lines $h,k \subset \mathbb P^2$.
The more complicated example I gave above might have the redeeming feature of being very ample but not effective i.e. it has a negative coefficient ($=-1$) in front of the component $L\cap P$.

Georges Elencwajg
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  • Dear @Georges, just out of curiosity, why take 3 hyperplanes, rather than say 2 whose intersections with $P$ are distinct? (And why the minus sign?) –  Mar 04 '14 at 08:43
  • Dear @Asal: this is because I first thought of a divisor on a quadric where things are a bit more complicated. Then I realized I could just take a projective space instead of a quadric but forgot to modify the structure of the divisor! But of course the example you suggest is simpler. – Georges Elencwajg Mar 04 '14 at 08:53
  • Dear @Georges, thanks for the clarification. (I suppose for the simplest possible example, one could just take 2 lines in $\mathbf P^2$...) –  Mar 04 '14 at 08:54
  • I was wondering: in the examples you guys suggested (e.g. the two lines in $\Bbb{P}^2$ for simplicity) we still have the line bundle irreducible up to linear equivalence (of course, not reduced though). What about an example 'genuinely' irreducible (i.e. not linearly equivalent to an irreducible divisor) ? – Heitor Fontana Mar 04 '14 at 12:25
  • Dear @HeitorFontana: I don't think I understand your question. What do you want an example of? –  Mar 04 '14 at 13:06
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    @HeitorFontana sorry if I misunderstood your question, but at least for smooth projective varieties this is exactly what Bertini forbids. Also, regarding your comment "of course, not reduced though", just want to clarify that the divisor class of two lines does have reduced and irreducible representative given by any conic. For an example of a singular scheme with a very ample line bundle that has no irreducible section... I'm not sure what the horizon of pathology you need to reach is, but certainly the doubled line is extreme enough. – Callus - Reinstate Monica Mar 04 '14 at 13:20