Let $X$ be a variety over algebraically closed field $k$, $\dim X > 0$, and let $H$ be ample Cartier divisor on X. Suppose $m < 0$. Show that $\mathcal{O}(mH)$ has no global sections.

My result so far is that if $X$ is a complete non-singular curve, then it follows from Riemann-Roch theorem that $H$ being ample implies that $\deg H > 0$ (since otherwise $kH$ wouldn't be very ample for large enough $k$), so $\deg mH < 0$ and Riemann-Roch again implies that $\mathcal{O}(mH)$ has no nonzero global sections.

Thus, we can conclude that every global section of $\mathcal{O}(mH)$ is zero over every complete non-singular $Y \subset X$. If we could show that the set of points $p \in X$ that belong to some complete non-singular curve is dense, the result would follow. I'm not sure if that's even true, though.

Also, is there a simple argument that avoids using Riemann-Roch here? Can we somehow relax the assumptions on X?

  • 2
    $\begingroup$ You must suppose $X$ is projective. But it is not necessary to assume $k$ is algebraically closed. $\endgroup$
    – Cantlog
    Oct 8, 2013 at 22:39
  • $\begingroup$ $X$ is projective by the fact that $H$ is ample (thus $kH$ is very ample for some large $k$, which automatically implies that $X$ is projective). $\endgroup$
    – xyzzyz
    Oct 8, 2013 at 22:55
  • $\begingroup$ I guess one way to avoid worrying about singularities is to pull back $H$ to the normalization of a curve $Y$. $\endgroup$
    – user64687
    Oct 9, 2013 at 8:03
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    $\begingroup$ @xyzzyz: Don't you only get that $X$ is quasi-projective? If you assume $X$ to be proper over $k$, then this implies projective, but not in general. $\endgroup$ Oct 9, 2013 at 9:54
  • 1
    $\begingroup$ For $X$ nonsingular and projective: A global section $s\in\mathcal L(X)$, the divisor of zeros $Z(s)$ is an effective divisor $D$ such that $\mathcal L \cong \mathcal O(D)$. This is II, Proposition 7.7 in Hartshorne. Hence, in your case, this would imply that $mH$ is effective. Then, $\mathcal O(-mH)$ is the ideal sheaf of $mH$. Since it is not globally generated, it can not be ample, but it is supposed to be because $H$ is. The assumptions are much stronger than yours, but maybe it helps. $\endgroup$ Oct 9, 2013 at 10:22

1 Answer 1


Suppose that $X$ is reduced, with positive dimensional connected components, and is projective over a field (which is not implied by the existence of an ample sheaf following standard definition of ample and of very ample). Then it is true that for any $n\ge 1$, $O_X(-nH)$ has no non-zero global sections.

As the restriction of an ample divisor to a connected component is ample, it is enough to work with connected and reduced $X$, in which case $H^0(X, O_X)$ is a field.

If $O_X(-nH)$ has a non-zero global section $s$, then for any positive integer $d$, $s^{\otimes d}$ is a non-zero global section of $O_X(-ndH)$. Now as you said, if $d$ is big enough, $ndH$ is very ample, so $ndH$ is linearly equivalent to an effective divisor $D$ (embedd $X$ into some projective space using $O_X(ndH)$, take the trace to $X$ of a hyperplane). We have $D\ne 0$ because otherwise $X$ would be affine (or if you prefer, a hyperplane in a projective space containing $X$ must meet $X$ when $\dim X\ge 1$). This implies that $O_X(-ndH)\cong O_X(-D)$ is a sheaf of ideals. It is then enough to show that $H^0(X, O_X(-D))=0$. But this follows immediately from the fact that $H^0(X, O_X)$ is a field and any global section of $O_X(-D)$ has a zero somewhere in $X$.

If $X$ is not reduced, with sheaf of nilpotent ideals $N$, the question is whether $N(-D)$ has non-zero global section. I think the answer is yes, this can happen. I don't have an explicit example, but it is not clear for me whether you consider non-reduced algebraic varieties...

  • $\begingroup$ Thank you! It's clear to me now. And no, I don't work with nonreduced varieties. $\endgroup$
    – xyzzyz
    Oct 10, 2013 at 0:40

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