The Stacks project

Proof. Choose a closed immersion $i : X \to \mathbf{P}^ n_ k$ (Lemma 33.43.4). Setting $\mathcal{L} = i^*\mathcal{O}_{\mathbf{P}^ n}(d)$ for $d \gg 0$ we see that there exists an invertible sheaf $\mathcal{L}$ with $H^0(X, \mathcal{L}) \not= 0$ and $H^1(X, \mathcal{L}) = 0$ (see Cohomology of Schemes, Lemma 30.17.1 for vanishing and the references therein for nonvanishing). We will finish the proof of (1) by descending induction on $t = \dim _ k H^0(X, \mathcal{L})$. The base case $t = 1$ is trivial. Assume $t > 1$.

Let $U \subset X$ be the nonempty open subset of nonsingular points studied in Lemma 33.25.8. Let $s \in H^0(X, \mathcal{L})$ be nonzero. There exists a closed point $x \in U$ such that $s$ does not vanish in $x$. Let $\mathcal{I}$ be the ideal sheaf of $i : x \to X$ as in Lemma 33.43.8. Look at the short exact sequence

Observe that $H^0(X, i_*i^*\mathcal{L}) = H^0(x, i^*\mathcal{L})$ has dimension $1$ as $x$ is a $k$-rational point ($k$ is algebraically closed). Since $s$ does not vanish at $x$ we conclude that

is surjective. Hence $\dim _ k H^0(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}) = t - 1$. Finally, the long exact sequence of cohomology also shows that $H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}) = 0$ thereby finishing the proof of the induction step.

To get an invertible sheaf as in (2) take an invertible sheaf $\mathcal{L}$ as in (1) and do the argument in the previous paragraph one more time. $\square$