I was reading A. de Jong and J. Starr's paper "Low degree complete intersections are rationally simply connected", which can be found at http://www.math.stonybrook.edu/~jstarr/papers/nk1006g.pdf, and I came across with the following claim in the proof of Lemma 5.5, page 23: for a general $m$-uple of points on a smooth, projective variety $X$, every irreducible curve $C$ containing these points satisfy $h^{1}(C, N_{C/X}(-m)) = 0$. The authors do not say about this fact, they just use it. I would like to know how to prove it, or a reference where I can find the proof. I am thinking it is necessary $C$ to be smooth. May anyone help me with this? I thank a lot.

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    $\begingroup$ Every curve we are studying in that part of the paper is a genus $0$ curve. For a genus $0$ curve $C$, for every locally free sheaf $\mathcal{E}$ on $C$, if $\mathcal{E}$ is globally generated, then $h^1(C,\mathcal{E}(d))$ is zero for every $d\geq -1$. For a general $m$-tuple of points, for a genus $0$ curve $C$ that contains those points, $N_{C/X}(-m)$ is globally generated. $\endgroup$ Jul 8, 2015 at 15:03
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    $\begingroup$ Just to say one more word: for $m$ linearly general points in $\mathbb{P}^n$, every irreducible, projective curve $C$ in $\mathbb{P}^n$ containing the points and having $\mathcal{O}(1)$-degree $m-1$ is automatically genus $0$. This can be proved by induction on $m$. Projection away from one of the $m$ points gives a curve in $\mathbb{P}^{n-1}$ containing $m-1$ general points and with degree $m-2$. The base case is for $m=2$: every line has genus $0$. $\endgroup$ Jul 8, 2015 at 15:08
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    $\begingroup$ Okay, two more words. The induction argument above applies to arithmetic genus, not just geometric genus. An irreducible, projective (reduced) curve of arithmetic genus $0$ is automatically smooth of geometric genus $0$. So smoothness is automatic for the curves $C$ in that proof. At any rate, the OP is correct that de Jong and I should have spelled these things out for the reader. $\endgroup$ Jul 8, 2015 at 15:35
  • $\begingroup$ Dear @JasonStarr, thank you very much for the help. $\endgroup$
    – theStudent
    Jul 9, 2015 at 0:50


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