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

Question

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 $H^0(X,\mathscr{L}^{-1})=0$.

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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.

Answer 1

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.

Answer 2

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.