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Assume that $C$ is a smooth projective curve and $p:C\to\mathbb{P}^1$ is a degree $k$ branched cover. Let $L$ be a very ample line bundle on $C$ with very large degree, defining an embedding $C\hookrightarrow \mathbb{P}^r$, where $r$ is the dimension of $L$. Consider the scroll $S\subset\mathbb{P}^r$ spanned by the fibers of $p$. I want to ask: first, how $S$ contains $C$; and second, why does the syzygy of $I_{S/\mathbb{P}^r}$ contains a linear strand of length $r-k$ and why therefore it implies that the Koszul module $K_{p,1}(C,L)$ is nonzero for $1\leq p\leq r-k$? I'm a little familiar with curve theory but not with surfaces and their syzygies. Any help is appreciated.

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  • $\begingroup$ Perhaps you are assuming that the fibre of $p$ is contained in some hyperplane section of $C$ in $\mathbb{P}^r$. This is not automatic (for example $k$ could be much larger than $r$!). $\endgroup$
    – Kapil
    Dec 17 '19 at 16:13
  • $\begingroup$ @Kapil You're right. The degree of $L$ should be large sufficiently. $\endgroup$
    – Li Li
    Dec 17 '19 at 18:03

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