negative multiple of ample line bundle has no global section

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?

Solution 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...