Lemma 33.22.3. Let $k$ be an infinite field. Let $X$ be an algebraic $k$-scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $V \to \Gamma (X, \mathcal{L})$ be a linear map of $k$-vector spaces whose image generates $\mathcal{L}$. Then there exists a subspace $W \subset V$ with $\dim _ k(W) \leq \dim (X) + 1$ which generates $\mathcal{L}$.

**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 $\mathcal{L}$. $\square$

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