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Let $S$ be an algebraic smooth surface (over $\Bbb{C}$).

Suppose we have some fibration $p\colon S\rightarrow C$ onto a smooth curve. Let $f$ be a fiber, a curve, say.

Let $K$ be a canonical divisor on $S$ and assume $K\cdot f=0$. My question is: can we conclude that each $m$-canonical map $\varphi_{mK}\colon S\rightarrow\Bbb{P}^N$, when defined, will contract $f$ to a point ?

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If $|mK|$ is base point free and if $f$ is irreducible, then $\varphi_{mK}(f)$ is a point.

Otherwise, there exists $x$, $y\in f$ s.t. $\varphi_{mK}(x)\neq \varphi_{mK}(y)$. By the definition of $\varphi_{mK}$, it is equivalent to say that there exists $s\in H^0(S,\mathcal{O}_S(mK))$, such that $s(x)=0$ and $s(y)\neq 0$.

Let $D:=div(s)$, then $D$ is an effective divisor, linearly equivalent to $mK$, and satisfying $x\in \operatorname{Supp}(D)$ while $y\notin \operatorname{Supp}(D)$. Therefore $f$ is not a component of $D$ (since $y\notin \operatorname{Supp}(D)$), hence $D\cdot f\geq 1$ (since $D$, $f$ have no components in common and $x\in \operatorname{Supp}(D)\cap f$). So $mK\cdot f=D\cdot f\geq 1$, we get a contradiction.

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One slick way to explain why the answer is yes uses the projection formula. For our purposes the following formulation is enough: if $f: X \rightarrow Y$ is a morphism between smooth projective varieties, $L$ is a line bundle on $Y$, and $C$ is a curve on $X$, then $$f^*L \cdot C = L \cdot f_*C$$ (where $f_*$ denotes the push-forward of cycles, and $\cdot$ is intersection of cycles).

By definition of the morphism $\varphi_L: X \rightarrow \mathbf{P}^n$ associated to a basepoint-free line bundle on a variety $X$, we have $\varphi_L^* \mathcal{O}(1) =L$. So if $C$ is any curve with $L \cdot C =0$, the projection formula gives $(\varphi_L)_*C \cdot \mathcal{O}(1)=0$. But $\mathcal{O}(1)$ has strictly positive degree on any curve in $\mathbf{P}^n$, so the 1-cycle $(\varphi_L)_*(C)$ must be zero: that is, $C$ must be contracted by $\varphi$.

Now just apply this in your case with $L=mK$. (Note that for each integer $m$, the conditions $K \cdot C=0$ and $mK \cdot C=0$ are equivalent.)

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  • $\begingroup$ Thank you, this proof is more elegant, though I'll accept the one above, being more elementary. $\endgroup$ – Heitor Fontana May 22 '13 at 17:30

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