# Duality of an Ample invertible sheaf has no non-trivial global section?

This is EXERCISE 7.1 of Chapter III in Hartshorne's AG:

[Q] Let $$X$$ be an integral projective scheme of dimension $$\geq1$$ over a field $$k$$, and let $$\mathscr{L}$$ be an ample invertible sheaf on $$X$$. Then $$H^0(X,\mathscr{L}^{-1})=0$$.

Actually if $$s\in H^0(X,\mathscr{L}^{-1})\neq0$$, as $$H^0(X,\mathscr{L}^{-1})=\hom(\mathscr{O}_X,\mathscr{L}^{-1})$$, $$s$$ can be seen as a non-trivial map $$s:\mathscr{O}_X\to\mathscr{L}^{-1}$$. As $$X$$ integral, this map need be injective, so we get an inclusion $$s:\mathscr{L}^n\to\mathscr{O}_X$$ for all $$n>0$$. So induce an injection $$\Gamma(X,\mathscr{L}^n)\to\Gamma(X,\mathscr{O}_X)$$!

Now as $$\mathscr{L}$$ be an ample invertible sheaf, $$\dim_k(\Gamma(X,\mathscr{L}^n))$$ should be very big and $$\dim_k(\Gamma(X,\mathscr{O}_X))$$ should be not so big. So we get the contradiction.

(I) For $$\Gamma(X,\mathscr{O}_X)$$:

Here $$k$$ may not algebraic closed and $$X$$ may not geometrically reduced, so we can not get $$\Gamma(X,\mathscr{O}_X)=k$$, in fact this can be very large! (see: https://stacks.math.columbia.edu/tag/0BUG)

We go through that proof, $$t\in\Gamma(X,\mathscr{O}_X)=\hom(X,\mathbb{A}^1)$$. And we consider $$X\to\mathbb{A}^1\to\mathbb{P}^1$$, use $$X$$ is proper, we can get that $$X$$ maps to a single closed point of $$\mathbb{A}^1$$. So $$\Gamma(X,\mathscr{O}_X)$$ are all closed points in $$\mathbb{A}^1$$!

(II) For $$\Gamma(X,\mathscr{L}^n)$$:

As $$\mathscr{L}$$ is ample, we may assume $$\mathscr{L}^n$$ generated by global sections and very ample which induce $$i:X\to\mathbb{P}^N$$ such that $$\mathscr{L}^n=i^*\mathscr{O}(1)$$. But I don't know which can be used.

First, note that $$\Gamma(X, \mathscr{O}_X)$$ is always a finite dimensional $$k$$-vector space, as pointed out in the Stacks Project. You already proved that $$\Gamma(X, \mathscr{L}^k) \subseteq \Gamma(X, \mathscr{O}_X)$$ for all $$k \geq 0$$. This way, it suffices to show that $$\dim_k\Gamma(X, \mathscr{L}^k)$$ goes to infinity.
One way to think about this is refering to the Hilbert polynomial of $$X$$. Let us assume that $$\mathscr{L}$$ is very ample for simplicity, so that $$\mathscr{L}$$ induces an embedding $$X \subseteq \mathbb{P}^N$$. We have that $$\chi(\mathscr{L}^n)=p(n)$$ is $$dn^s/s!+\text{(lower order terms)}$$, where $$d$$ is the degree (of the embedding) and $$s$$ the dimension (which is $$\geq 1$$!). And by Serre vanishing, we get $$\chi(\mathscr{L}^n)=h^0(X, \mathscr{L}^n)$$. This way we get that $$h^0(X, \mathscr{L}^n)$$ goes to $$+\infty$$, as claimed.
Alternatively, one may prove that $$\mathcal{O}_X(X)$$ is a field. Therefore for any nonzero section $$u\in\mathcal{L}^n$$, the nonvanishing scheme $$X_u$$ is precisely $$X$$. Combining this with the proof of theorem II.7.6 which shows that for any $$x\in X$$ there is an $$n\geq 0$$ and an $$s\in\mathcal{L}^n$$ with $$X_s$$ an affine open neighborhood of $$x$$, we observe that $$X$$ must be finite over $$k$$ since it is both affine and projective. But this contradicts the fact that $$X$$ is of positive dimension.