[proofplan] The positivity of $L$ implies that $L$ is ample, so high tensor powers of $L$ satisfy uniform Serre vanishing for the ideal sheaves of all zero-dimensional subschemes of length at most two. Applying this vanishing to the exact restriction sequence for such a subscheme makes the restriction map on global sections surjective. Length-one subschemes give base point freeness, reduced length-two subschemes give separation of distinct points, and nonreduced length-two subschemes in a tangent direction give separation of tangent vectors. [/proofplan] [step:Convert positivity into uniform vanishing on length-two subschemes] We use two standard external results: the [Kodaira Ampleness Criterion for Positive Line Bundles](/theorems/???) and [Uniform Serre Vanishing for Bounded Families of Coherent Sheaves](/theorems/???). Since $L \to X$ is positive, the Kodaira ampleness criterion applies to the compact complex manifold $X$ and implies that $L$ is ample and that $X$ is projective. Let $H := \operatorname{Hilb}_{\leq 2}(X)$ denote the Hilbert scheme parametrising zero-dimensional closed subschemes $Z \subset X$ of length at most $2$. Because $X$ is projective, $H$ is projective, hence compact. Let \begin{align*} \pi: X \times H &\to H \end{align*} be the projection, let \begin{align*} \mathcal{Z} \subset X \times H \end{align*} be the universal closed subscheme, and let \begin{align*} \mathcal{I}_{\mathcal{Z}} \subset \mathcal{O}_{X \times H} \end{align*} be its coherent ideal sheaf. The morphism $\pi$ is projective because $X$ is projective, and the line bundle $\operatorname{pr}_1^*L$ is $\pi$-ample because $L$ is ample on $X$. Thus the family of coherent sheaves \begin{align*} \mathcal{I}_Z := \mathcal{I}_{\mathcal{Z}}\big|_{X \times \{Z\}}, \qquad Z \in H, \end{align*} is bounded by the single coherent sheaf $\mathcal{I}_{\mathcal{Z}}$ on the projective family $\pi: X \times H \to H$. Uniform Serre vanishing for bounded families therefore gives an integer $m_0 \in \mathbb{N}$ such that, for every $m \geq m_0$ and every zero-dimensional closed subscheme $Z \subset X$ of length at most $2$, \begin{align*} H^1(X,\mathcal{I}_Z \otimes L^m)=0. \end{align*} [guided] The point of this step is to obtain one integer $m_0$ that works simultaneously for all points, pairs of points, and first-order tangent directions. Positivity of a holomorphic line bundle is an analytic curvature condition, while the separation statements are algebraic conditions on global sections. We apply the [Kodaira Ampleness Criterion for Positive Line Bundles](/theorems/???). Its hypothesis is exactly that $L \to X$ is a positive holomorphic line bundle over a compact complex manifold. Hence $L$ is ample, and $X$ is projective. We need vanishing not just for one ideal sheaf, but uniformly for every ideal sheaf cutting out either one point, two distinct points, or a first-order infinitesimal tangent direction. These are all zero-dimensional closed subschemes of length at most $2$. Let \begin{align*} H := \operatorname{Hilb}_{\leq 2}(X) \end{align*} denote the Hilbert scheme parametrising them. Since $X$ is projective, this Hilbert scheme is projective, hence compact. Now define the projective family to which uniform Serre vanishing will be applied. Let \begin{align*} \pi: X \times H &\to H \end{align*} be the projection, let \begin{align*} \mathcal{Z} \subset X \times H \end{align*} be the universal closed subscheme, and let \begin{align*} \mathcal{I}_{\mathcal{Z}} \subset \mathcal{O}_{X \times H} \end{align*} be its coherent ideal sheaf. For each point $Z \in H$, the fibre of the universal family is the corresponding finite subscheme $Z \subset X$, and its ideal sheaf is \begin{align*} \mathcal{I}_Z := \mathcal{I}_{\mathcal{Z}}\big|_{X \times \{Z\}}. \end{align*} We verify the hypotheses of [Uniform Serre Vanishing for Bounded Families of Coherent Sheaves](/theorems/???). The morphism $\pi$ is projective because $X$ is projective. The line bundle $\operatorname{pr}_1^*L$ is $\pi$-ample because $L$ is ample on $X$. The sheaves $\mathcal{I}_Z$ form a bounded family because they are all fibres of the single coherent sheaf $\mathcal{I}_{\mathcal{Z}}$ on $X \times H$. Therefore uniform Serre vanishing gives a single integer $m_0 \in \mathbb{N}$ such that, for every $m \geq m_0$ and every $Z \in H$, \begin{align*} H^1(X,\mathcal{I}_Z \otimes L^m)=0. \end{align*} This uniformity is the essential point: ordinary Serre vanishing for a fixed $Z$ would give an integer depending on $Z$, which would not prove the theorem. [/guided] [/step] [step:Surject global sections onto every subscheme of length at most two] Fix $m \geq m_0$ and a zero-dimensional closed subscheme $Z \subset X$ of length at most $2$. The defining exact sequence of $Z$ is \begin{align*} 0 \longrightarrow \mathcal{I}_Z \longrightarrow \mathcal{O}_X \longrightarrow \mathcal{O}_Z \longrightarrow 0. \end{align*} Tensoring by the locally free sheaf $L^m$ preserves exactness, giving \begin{align*} 0 \longrightarrow \mathcal{I}_Z \otimes L^m \longrightarrow L^m \longrightarrow L^m|_Z \longrightarrow 0. \end{align*} Taking sheaf cohomology yields the exact segment \begin{align*} H^0(X,L^m) \longrightarrow H^0(Z,L^m|_Z) \longrightarrow H^1(X,\mathcal{I}_Z \otimes L^m). \end{align*} By the vanishing from the previous step, the final term is zero. Hence the restriction map \begin{align*} \rho_Z: H^0(X,L^m) &\to H^0(Z,L^m|_Z) \end{align*} is surjective. [guided] We now translate cohomology vanishing into the concrete ability to prescribe values on small subschemes. Fix $m \geq m_0$ and a zero-dimensional closed subscheme $Z \subset X$ of length at most $2$. Its ideal sheaf $\mathcal{I}_Z$ fits into \begin{align*} 0 \longrightarrow \mathcal{I}_Z \longrightarrow \mathcal{O}_X \longrightarrow \mathcal{O}_Z \longrightarrow 0. \end{align*} Because $L^m$ is locally free of rank one, tensoring by $L^m$ is exact. Thus \begin{align*} 0 \longrightarrow \mathcal{I}_Z \otimes L^m \longrightarrow L^m \longrightarrow L^m|_Z \longrightarrow 0. \end{align*} Taking cohomology gives an exact sequence whose relevant part is \begin{align*} H^0(X,L^m) \longrightarrow H^0(Z,L^m|_Z) \longrightarrow H^1(X,\mathcal{I}_Z \otimes L^m). \end{align*} The last group vanishes by the uniform choice of $m_0$. Exactness therefore says precisely that every section of $L^m|_Z$ is the restriction of a global holomorphic section of $L^m$. In other words, the restriction map \begin{align*} \rho_Z: H^0(X,L^m) &\to H^0(Z,L^m|_Z) \end{align*} is surjective. [/guided] [/step] [step:Use length-one subschemes to prove base point freeness] Let $x \in X$, and take $Z=\{x\}$ with its reduced scheme structure. Since $L^m|_Z$ is a one-dimensional complex [vector space](/page/Vector%20Space), choose a nonzero element \begin{align*} \ell_x \in H^0(Z,L^m|_Z). \end{align*} By surjectivity of $\rho_Z$, there exists $s \in H^0(X,L^m)$ such that $\rho_Z(s)=\ell_x$. Therefore $s(x)=\ell_x\neq 0$. Hence $H^0(X,L^m)$ is base point free. [guided] We must prove that no point $x \in X$ is a base point of the linear system $H^0(X,L^m)$. Fix $x \in X$ and let $Z=\{x\}$ with its reduced scheme structure. Then $Z$ is a zero-dimensional closed subscheme of length $1$, so the restriction map \begin{align*} \rho_Z: H^0(X,L^m) &\to H^0(Z,L^m|_Z) \end{align*} is surjective by the previous step. The vector space $H^0(Z,L^m|_Z)$ is naturally the fibre $L_x^m$, hence is one-dimensional over $\mathbb{C}$. Choose \begin{align*} \ell_x \in H^0(Z,L^m|_Z) \end{align*} with $\ell_x \neq 0$. Surjectivity of $\rho_Z$ gives a global section $s \in H^0(X,L^m)$ such that $\rho_Z(s)=\ell_x$. Since restriction to the reduced point records the value at $x$, this means \begin{align*} s(x)=\ell_x \neq 0. \end{align*} Thus for the arbitrary point $x$, there is a global section of $L^m$ not vanishing at $x$. Therefore $H^0(X,L^m)$ is base point free. [/guided] [/step] [step:Use reduced length-two subschemes to separate distinct points] Let $x,y \in X$ with $x \neq y$, and let \begin{align*} Z=\{x,y\} \end{align*} with its reduced scheme structure. Then \begin{align*} H^0(Z,L^m|_Z) \cong L_x^m \oplus L_y^m. \end{align*} Choose an element \begin{align*} \ell_{x,y} \in H^0(Z,L^m|_Z) \end{align*} whose component in $L_x^m$ is $0$ and whose component in $L_y^m$ is nonzero. By surjectivity of $\rho_Z$, there exists $s \in H^0(X,L^m)$ such that $\rho_Z(s)=\ell_{x,y}$. Hence $s(x)=0$ and $s(y)\neq 0$. Thus $H^0(X,L^m)$ separates distinct points. [guided] We now prove point separation. Let $x,y \in X$ be distinct points, and let \begin{align*} Z=\{x,y\} \end{align*} with its reduced scheme structure. This is a zero-dimensional closed subscheme of length $2$, so the restriction map \begin{align*} \rho_Z: H^0(X,L^m) &\to H^0(Z,L^m|_Z) \end{align*} is surjective. Because $Z$ is the disjoint union of the two reduced points $x$ and $y$, a section over $Z$ is exactly a pair of fibre values. Thus \begin{align*} H^0(Z,L^m|_Z) \cong L_x^m \oplus L_y^m. \end{align*} Choose \begin{align*} \ell_{x,y} \in H^0(Z,L^m|_Z) \end{align*} whose component in $L_x^m$ is $0$ and whose component in $L_y^m$ is nonzero. Surjectivity of $\rho_Z$ gives a global section $s \in H^0(X,L^m)$ such that $\rho_Z(s)=\ell_{x,y}$. Reading the two components of this equality gives \begin{align*} s(x)=0, \qquad s(y)\neq 0. \end{align*} Therefore the global sections of $L^m$ separate the distinct points $x$ and $y$. Since the pair was arbitrary, $H^0(X,L^m)$ separates distinct points. [/guided] [/step] [step:Use first infinitesimal neighbourhoods to separate tangent vectors] Let $x \in X$ and let $v \in T_xX$ be nonzero. Let $\mathfrak{m}_x \subset \mathcal{O}_{X,x}$ be the maximal ideal of germs vanishing at $x$. Define the length-two local Artinian algebra \begin{align*} A_v := \mathcal{O}_{X,x}/\mathfrak{a}_v, \end{align*} where \begin{align*} \mathfrak{a}_v := \{ f \in \mathfrak{m}_x : df_x(v)=0 \}. \end{align*} Let $Z_v \subset X$ be the closed subscheme supported at $x$ with local ring $A_v$. Since $v \neq 0$, there exists a germ $g \in \mathfrak{m}_x$ with $dg_x(v)\neq 0$, and therefore \begin{align*} \dim_{\mathbb{C}} A_v = 2. \end{align*} Choose a local holomorphic frame $e$ of $L^m$ near $x$. Over $Z_v$, this frame identifies $L^m|_{Z_v}$ with $\mathcal{O}_{Z_v}$. Let \begin{align*} \tau_v \in H^0(Z_v,L^m|_{Z_v}) \end{align*} be the section corresponding under this trivialisation to the class of $g$ in $A_v$. By construction, $\tau_v$ has value $0$ at $x$ and first derivative along $v$ equal to $dg_x(v)\neq 0$. By surjectivity of \begin{align*} \rho_{Z_v}: H^0(X,L^m) \to H^0(Z_v,L^m|_{Z_v}), \end{align*} there exists $s \in H^0(X,L^m)$ with $\rho_{Z_v}(s)=\tau_v$. Writing $s=f e$ near $x$, the equality of restrictions to $Z_v$ gives $f(x)=0$ and $df_x(v)=dg_x(v)\neq 0$. This nonvanishing derivative is independent of the local frame. Indeed, if $e'$ is another local holomorphic frame near $x$, then $e'=u e$ for a nowhere-vanishing [holomorphic function](/page/Holomorphic%20Function) $u$ near $x$. Writing $s=f' e'$, we have $f'=f/u$. Since $f(x)=0$ and $u(x)\neq 0$, the product and quotient rules give \begin{align*} df'_x(v)=\frac{df_x(v)u(x)-f(x)du_x(v)}{u(x)^2}=\frac{df_x(v)}{u(x)}\neq 0. \end{align*} Hence the section $s$ separates the tangent vector $v$ in the intrinsic sense, and $H^0(X,L^m)$ separates tangent vectors. [guided] To separate tangent vectors, it is not enough to prescribe values at points; we must prescribe a first derivative in a chosen tangent direction. This is encoded by a nonreduced length-two subscheme supported at the point. Let $x \in X$ and let $v \in T_xX$ be nonzero. Denote by $\mathfrak{m}_x \subset \mathcal{O}_{X,x}$ the maximal ideal of holomorphic germs vanishing at $x$. Define \begin{align*} \mathfrak{a}_v := \{ f \in \mathfrak{m}_x : df_x(v)=0 \}. \end{align*} Then \begin{align*} A_v := \mathcal{O}_{X,x}/\mathfrak{a}_v \end{align*} records the value of a germ and its first derivative in the direction $v$. Because $v \neq 0$, there is a germ $g \in \mathfrak{m}_x$ with $dg_x(v)\neq 0$; for example, in a holomorphic coordinate chart at $x$, some coordinate differential evaluates nontrivially on $v$. Therefore $A_v$ has one dimension from the constant term and one dimension from the directional first-order term, so \begin{align*} \dim_{\mathbb{C}} A_v = 2. \end{align*} Let $Z_v \subset X$ be the closed subscheme supported at $x$ with local ring $A_v$. Choose a local holomorphic frame $e$ of $L^m$ near $x$. This identifies sections of $L^m|_{Z_v}$ with elements of $A_v$. Let \begin{align*} \tau_v \in H^0(Z_v,L^m|_{Z_v}) \end{align*} be the section corresponding to the class of the germ $g$. Since $g \in \mathfrak{m}_x$, the value of $\tau_v$ at $x$ is zero. Since $dg_x(v)\neq 0$, its first derivative in the direction $v$ is nonzero. The restriction map \begin{align*} \rho_{Z_v}: H^0(X,L^m) \to H^0(Z_v,L^m|_{Z_v}) \end{align*} is surjective by the length-two vanishing already proved. Hence there is a global section $s \in H^0(X,L^m)$ restricting to $\tau_v$. Writing $s=f e$ near $x$, equality on $Z_v$ means that $f$ and $g$ have the same value and the same first derivative in the direction $v$. Thus \begin{align*} f(x)=0, \qquad df_x(v)=dg_x(v)\neq 0. \end{align*} We also check that this statement is not an artefact of the chosen frame $e$. Let $e'$ be any other local holomorphic frame near $x$. Then there is a nowhere-vanishing holomorphic function $u$ near $x$ such that $e'=u e$. If $s=f' e'$, then $f'=f/u$. Since $f(x)=0$ and $u(x)\neq 0$, differentiating the quotient in the direction $v$ gives \begin{align*} df'_x(v)=\frac{df_x(v)u(x)-f(x)du_x(v)}{u(x)^2}=\frac{df_x(v)}{u(x)}\neq 0. \end{align*} Thus every local frame records a nonzero directional derivative for $s$ along $v$. This is exactly separation of the tangent vector $v$ at $x$. [/guided] [/step] [step:Combine the three separation properties] The integer $m_0$ was chosen independently of the point, pair of points, tangent vector, and integer $m \geq m_0$. The previous three steps show that for every $m \geq m_0$, the space $H^0(X,L^m)$ is base point free, separates distinct points, and separates tangent vectors. This proves the theorem. [/step]