[proofplan] We use positivity of $L$ to invoke the analytic jet-separation form of Kodaira's theorem: for all sufficiently large tensor powers, global sections prescribe arbitrary first jets at one point and arbitrary values at two distinct points. These two separation properties imply that the complete linear system of $L^m$ is base-point-free, separates points, and separates tangent directions. The associated evaluation map into the projective space of hyperplanes is therefore a holomorphic injective immersion, and compactness of $X$ upgrades it to a holomorphic embedding. [/proofplan]

[step:Use positivity to obtain uniform separation of values and first jets]Choose a smooth Hermitian metric $h$ on $L$ whose Chern curvature form is positive; this is exactly the positivity hypothesis for the holomorphic line bundle $L \to X$. For each integer $m \ge 1$, define the finite-dimensional complex [vector space](/page/Vector%20Space) \begin{align*} V_m := H^0(X,L^m), \end{align*} where $L^m$ denotes the $m$-fold tensor power of $L$ and $H^0(X,L^m)$ denotes the vector space of global holomorphic sections $s: X \to L^m$. We use the [Kodaira jet separation theorem](/page/Kodaira%20Jet%20Separation%20Theorem) in the following form. Since $X$ is compact complex and $L$ is positive, there exists an integer $m_0 \ge 1$ such that for every integer $m \ge m_0$ the following two assertions hold: \begin{align*} &\text{for every } x \in X, \text{ the first-jet map } J_{m,x}: V_m \to (L^m)_x \oplus ((T_x^*X)^{1,0} \otimes (L^m)_x) \text{ is surjective,} \\ &\text{for every pair } x,y \in X \text{ with } x \ne y, \text{ the value map } E_{m,x,y}: V_m \to (L^m)_x \oplus (L^m)_y \text{ is surjective.} \end{align*} Here $(L^m)_x$ is the fiber of $L^m$ at $x$, $(T_x^*X)^{1,0}$ is the holomorphic cotangent space at $x$, and the maps are defined by \begin{align*} J_{m,x}(s) &= (s(x), \partial s(x)), \\ E_{m,x,y}(s) &= (s(x),s(y)), \end{align*} where $\partial s(x)$ denotes the holomorphic first derivative of $s$ at $x$ in any local holomorphic frame for $L^m$. The definition of $\partial s(x)$ in the first-jet quotient is independent of the chosen frame.[/step]

[guided]The positivity hypothesis is the analytic input. A positive holomorphic line bundle $L \to X$ admits a smooth Hermitian metric $h$ whose Chern curvature form is positive. The Kodaira analytic package, proved from the Bochner-Kodaira-Nakano identity and Hörmander $L^2$ estimates for the operator $\bar\partial: \Omega^{0,q}(X,L^m) \to \Omega^{0,q+1}(X,L^m)$, turns this curvature positivity into global holomorphic sections with prescribed finite jets. We record exactly what is needed. For each integer $m \ge 1$, define \begin{align*} V_m := H^0(X,L^m). \end{align*} This is the complex vector space of global holomorphic sections $s: X \to L^m$. The [Kodaira jet separation theorem](/page/Kodaira%20Jet%20Separation%20Theorem) applies because $X$ is compact complex and $L$ is positive. It gives an integer $m_0 \ge 1$ such that, for every integer $m \ge m_0$, global sections of $L^m$ separate both first jets at a single point and values at two distinct points. For a point $x \in X$, define the first-jet map \begin{align*} J_{m,x}: V_m &\to (L^m)_x \oplus ((T_x^*X)^{1,0} \otimes (L^m)_x), \\ s &\mapsto (s(x),\partial s(x)). \end{align*} Here $(L^m)_x$ is the fiber at $x$, $(T_x^*X)^{1,0}$ is the holomorphic cotangent space at $x$, and $\partial s(x)$ is the holomorphic first derivative of $s$ at $x$, computed in a local holomorphic frame and interpreted as a first jet. The theorem says that $J_{m,x}$ is surjective. For distinct points $x,y \in X$, define the two-point evaluation map \begin{align*} E_{m,x,y}: V_m &\to (L^m)_x \oplus (L^m)_y, \\ s &\mapsto (s(x),s(y)). \end{align*} The same theorem says that $E_{m,x,y}$ is surjective. These two surjectivity statements are the precise bridge from curvature positivity to projective geometry.[/guided]

[step:Construct the projective map from global sections] Fix an integer $m \ge m_0$. Since $J_{m,x}$ is surjective for every $x \in X$, the projection of $J_{m,x}$ onto $(L^m)_x$ is surjective. Therefore, for every $x \in X$, there exists $s \in V_m$ with $s(x) \ne 0$. Thus the evaluation functional at $x$ is not the zero functional. Define \begin{align*} \Phi_m: X &\to \mathbb P(V_m^*) \end{align*} by assigning to $x \in X$ the hyperplane of sections vanishing at $x$: \begin{align*} \Phi_m(x) := \{s \in V_m : s(x)=0\} \in \mathbb P(V_m^*). \end{align*} Equivalently, after choosing any local holomorphic frame $e: U \to L^m$ over an [open set](/page/Open%20Set) $U \subset X$ and a basis $s_0,\dots,s_N$ of $V_m$, writing $s_j|_U=f_j e$ with holomorphic functions $f_j: U \to \mathbb C$, the map is locally \begin{align*} \Phi_m|_U: U &\to \mathbb P^N, \\ x &\mapsto [f_0(x):\cdots:f_N(x)]. \end{align*} Because the sections do not vanish simultaneously at any point, this local formula is defined everywhere. Since changes of holomorphic frame multiply all coordinates by the same nowhere-zero [holomorphic function](/page/Holomorphic%20Function), the local formulas agree on overlaps. Hence $\Phi_m$ is a well-defined holomorphic map. [/step]

[step:Separate distinct points by two-point evaluation] Let $x,y \in X$ with $x \ne y$. The surjectivity of \begin{align*} E_{m,x,y}: V_m \to (L^m)_x \oplus (L^m)_y \end{align*} allows us to choose a section $s \in V_m$ such that \begin{align*} s(x)=0, \qquad s(y) \ne 0. \end{align*} Thus $s \in \Phi_m(x)$ but $s \notin \Phi_m(y)$, so the hyperplanes $\Phi_m(x)$ and $\Phi_m(y)$ are distinct. Therefore $\Phi_m$ is injective. [/step]

[step:Separate tangent directions by first-jet evaluation]Let $x \in X$ and let $v \in T_xX$ be a nonzero tangent vector. Since $J_{m,x}$ is surjective, choose $s \in V_m$ such that \begin{align*} s(x)=0, \qquad \partial s(x)(v) \ne 0. \end{align*} Choose $t \in V_m$ with $t(x) \ne 0$, which exists by base-point-freeness. In a local holomorphic frame $e: U \to L^m$ near $x$, write \begin{align*} s|_U=f e, \qquad t|_U=g e, \end{align*} where $f,g: U \to \mathbb C$ are holomorphic and $g(x) \ne 0$. On the affine chart of $\mathbb P(V_m^*)$ where the coordinate corresponding to $t$ is nonzero, the ratio \begin{align*} \rho_{s,t}: U &\to \mathbb C, \\ z &\mapsto \frac{f(z)}{g(z)} \end{align*} is a holomorphic coordinate function of $\Phi_m$. Since $f(x)=0$ and $g(x) \ne 0$, the quotient rule gives \begin{align*} d(\rho_{s,t})_x(v)=\frac{\partial f(x)(v)g(x)-f(x)\partial g(x)(v)}{g(x)^2}=\frac{\partial f(x)(v)}{g(x)} \ne 0. \end{align*} Therefore $d(\Phi_m)_x(v) \ne 0$. Since $v$ was arbitrary and nonzero, $d(\Phi_m)_x$ is injective for every $x \in X$, so $\Phi_m$ is a holomorphic immersion.[/step]

[guided]To prove that $\Phi_m$ is an immersion, we must show that its differential kills no nonzero tangent vector. Fix $x \in X$ and a nonzero vector $v \in T_xX$. First-jet separation gives a section $s \in V_m$ with prescribed value and first derivative: \begin{align*} s(x)=0, \qquad \partial s(x)(v) \ne 0. \end{align*} Base-point-freeness gives another section $t \in V_m$ with $t(x) \ne 0$. Choose a local holomorphic frame \begin{align*} e: U \to L^m \end{align*} on an open neighbourhood $U \subset X$ of $x$. Then there are holomorphic functions $f,g: U \to \mathbb C$ such that \begin{align*} s|_U=f e, \qquad t|_U=g e. \end{align*} The conditions above become $f(x)=0$, $g(x) \ne 0$, and $\partial f(x)(v) \ne 0$. On the affine projective chart where the coordinate defined by $t$ is nonzero, one coordinate function of the projective map is \begin{align*} \rho_{s,t}: U &\to \mathbb C, \\ z &\mapsto \frac{f(z)}{g(z)}. \end{align*} This function is holomorphic near $x$ because $g(x) \ne 0$ and hence, after shrinking $U$ if necessary, $g$ is nowhere zero on $U$. Now compute the derivative in the direction $v$. The quotient rule gives \begin{align*} d(\rho_{s,t})_x(v)=\frac{\partial f(x)(v)g(x)-f(x)\partial g(x)(v)}{g(x)^2}. \end{align*} Since $f(x)=0$, this reduces to \begin{align*} d(\rho_{s,t})_x(v)=\frac{\partial f(x)(v)}{g(x)}. \end{align*} The numerator is nonzero by the choice of $s$, and the denominator is nonzero by the choice of $t$. Hence $d(\rho_{s,t})_x(v) \ne 0$. A coordinate function of $\Phi_m$ has nonzero derivative on $v$, so $d(\Phi_m)_x(v) \ne 0$. This proves that $\Phi_m$ is an immersion.[/guided]

[step:Upgrade the injective immersion to an embedding] We have shown that $\Phi_m: X \to \mathbb P(V_m^*)$ is a holomorphic injective immersion. Since $X$ is compact and $\mathbb P(V_m^*)$ is Hausdorff, the continuous map $\Phi_m$ is a homeomorphism from $X$ onto its image. Combining this topological fact with the injectivity of $d(\Phi_m)_x$ for every $x \in X$, the holomorphic [inverse function theorem](/page/Inverse%20Function%20Theorem) gives that $\Phi_m$ is a holomorphic embedding onto the complex submanifold $\Phi_m(X) \subset \mathbb P(V_m^*)$. Thus, for every integer $m \ge m_0$, the global holomorphic sections of $L^m$ define a holomorphic embedding \begin{align*} X \hookrightarrow \mathbb P(H^0(X,L^m)^*). \end{align*} This is the desired conclusion. [/step]