# Resolution of the ideal of a scroll

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.

• 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 $r$!). Dec 17 '19 at 16:13
• @Kapil You're right. The degree of $L$ should be large sufficiently. Dec 17 '19 at 18:03