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Lately I am studying the bend-and-break, and I follow the proof in the following note written by Olivier Debarre:

http://www.math.ens.fr/~debarre/M2.pdf

There is a detail that I just cannot go through in the top of page 92, the situation here is: there is a normal surface $S$ and a smooth curve $C$ with a flat morphism $\pi: S \to C$, and we have a fiber $F$ of $\pi$ in $S$ over $C$ which is not integral and has no embedded point. I already know that the genus of the fiber $F$ which is a curve in $S$ is $0$. But I just can't understand why this implies that $h^0(F, \mathcal O_F) = 1$.

Hope somebody can help me get through this, thanks!

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  • $\begingroup$ Holomorphic functions over connected, compact complex varieties are constant (Liouville theorem, essentially). $\endgroup$
    – Francesco Polizzi
    Commented Feb 13, 2015 at 14:50
  • $\begingroup$ But it is not necessarily over $\mathbb C$, it is over an algebraically closed field $k$. And I don't know if it is connected yet. Connectedness of this $F$ seems to be a result of $h^0=1$.... $\endgroup$
    – user50190
    Commented Feb 13, 2015 at 14:58
  • $\begingroup$ of characteristic 0, I guess. So you can use Lefschetz principle or something like that. Anyway, your fiber must be connected (I guess you forgot to say this), otherwise the result is obviously not true. $\endgroup$
    – Francesco Polizzi
    Commented Feb 13, 2015 at 14:59
  • $\begingroup$ I am so confused because I cannot understand what the condition of having no embedded point here is about. $\endgroup$
    – user50190
    Commented Feb 13, 2015 at 15:02
  • $\begingroup$ It seems that $F$ needs to be connected, but all I know is there is a dense open subset $U$ of $C$ such that every fiber of $\pi$ over points in $U$ is $\mathbb P^1$, maybe this gives connectedness of $F$? $\endgroup$
    – user50190
    Commented Feb 13, 2015 at 15:51

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I believe that you misinterpreted the top of page 92 of Debarre's notes.

At that stage Debarre works with a flat morphism $\pi: S\to \overline{T}$, where $S$ is a normal surface and the general fiber of $\pi$ is a smooth rational curve.

He then assumes that every fiber of $\pi$ is integral. Then flatness etc. implies that every fiber of $\pi$ is a smooth rational curve (middle of page 91) and hence $S$ is a ruled surface over $T$. This implies that the two sections $T_0$ and $T_\infty$ have both negative self-intersection and one easily deduces a contradiction from this (first paragraph of page 92).

This implies that $\pi$ has at least one fiber $F$ which is not integral, hence this fiber $F$ is either a multiple fiber of has several irreducible components, or a combination of this.

However, $\pi$ is proper and flat, and the generic fiber is connected. This implies that every fiber of $\pi$ is connected. (See e.g., Number of irreducible and connected components constant in flat families )

In particular, $F$ is a connected, but it may be nonreduced, and hence $h^0(\mathcal{O}_F)$ can be larger than one.

Using that $\pi$ is flat you get that every fiber is a connected curve with arithmetic genus $0$. In particular, every irreducible component of $F_{red}$ is a rational curve.

Debarre then consider $S''$, a resolution of singularities of $S$. In this way he introduces further rational curves, and obtains that every fiber of $S''\to T$ consists of rational curves only.

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