The Stacks project

Proof. More precisely, set $X_{A'} = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A')$. Let $\mathcal{F}_{A'} = p^*\mathcal{F}$ where $p : X_{A'} \to X$ is the projection. By Cohomology of Schemes, Lemma 30.5.2 we have $H^0(X_{k'}, \mathcal{F}_{A'}) = H^0(X, \mathcal{F}) \otimes _ A A'$. Thus if $s_ i$, $i \in I$ are generators for $H^0(X, \mathcal{F})$ as an $A$-module, then their images in $H^0(X_{A'}, \mathcal{F}_{A'})$ are generators for $H^0(X_{A'}, \mathcal{F}_{A'})$ as an $A'$-module. Thus we have to show that the map $\alpha : \bigoplus _{i \in I} \mathcal{O}_ X \to \mathcal{F}$, $(f_ i) \mapsto \sum f_ i s_ i$ is surjective if and only if $p^*\alpha $ is surjective. This we may check over an affine open $U = \mathop{\mathrm{Spec}}(B)$ of $X$. Then $\mathcal{F}|_ U$ corresponds to a $B$-module $M$ and $s_ i|_ U$ to elements $x_ i \in M$. Thus we have to show that $\bigoplus _{i \in I} B \to M$ is surjective if and only if the base change $\bigoplus _{i \in I} B \otimes _ A A' \to M \otimes _ A A'$ is surjective. This is true because $A \to A'$ is faithfully flat. $\square$

Proof. If $n = 0$ the result is trivial, hence we assume $n > 0$. By definition of a very ample invertible sheaf, the lemma immediately reduces to the case where $X = \mathbf{P}^ r_ k$ for some $r > 0$ and $\mathcal{L} = \mathcal{O}_ X(1)$. Write $\mathbf{P}^ r_ k = \text{Proj}(k[T_0, \ldots , T_ r])$. Set $V = H^0(X, \mathcal{L}) = kT_0 \oplus \ldots \oplus kT_ r$. Since $x$ is a $k$-rational point, we see that the set $s \in V$ which vanish at $x$ is a codimension $1$ subspace $W \subset V$ and that $W$ generates the homogeneous prime ideal corresponding to $x$. Since $x_ i \not= x$ the corresponding homogeneous prime $\mathfrak p_ i \subset k[T_0, \ldots , T_ r]$ does not contain $W$. Since $k$ is infinite, we then see that $W \not= \bigcup W \cap \mathfrak q_ i$ and the proof is complete. $\square$

Proof. Throughout the proof we will use that for every $x \in X$ the linear map

is nonzero. The proof is by induction on $\dim (X)$.

The base case is $\dim (X) = 0$. In this case $X$ has finitely many points $X = \{ x_1, \ldots , x_ n\} $ (see for example Lemma 33.20.2). Since $k$ is infinite there exists a vector $v \in V$ such that $\psi _{x_ i}(v) \not= 0$ for all $i$. Then $W = k\cdot v$ does the job.

Assume $\dim (X) > 0$. Let $X_ i \subset X$ be the irreducible components of dimension equal to $\dim (X)$. Since $X$ is Noetherian there are only finitely many of these. For each $i$ pick a point $x_ i \in X_ i$. As above choose $v \in V$ such that $\psi _{x_ i}(v) \not= 0$ for all $i$. Let $Z \subset X$ be the zero scheme of the image of $v$ in $\Gamma (X, \mathcal{L})$, see Divisors, Definition 31.14.8. By construction $\dim (Z) < \dim (X)$. By induction we can find $W \subset V$ with $\dim (W) \leq \dim (X)$ such that $W$ generates $\mathcal{L}|_ Z$. Then $W + k\cdot v$ generates $\mathcal{L}$. $\square$