[proofplan] We use the Riemann--Roch theorem for compact Riemann surfaces to construct a divisor $D$ of sufficiently large degree whose associated complete linear system has enough sections. For $\deg D \geq 2g+1$, Riemann--Roch shows that the sections of $\mathcal{O}_X(D)$ separate points and tangent directions. These sections define a holomorphic map $X \to \mathbb{P}^N_{\mathbb{C}}$ that is injective and immersive. Since $X$ is compact and projective space is Hausdorff, the map is a topological embedding; combined with immersion, it is a holomorphic embedding. [/proofplan]

[step:Choose a divisor of degree large enough for Riemann--Roch to control all relevant section spaces]Let $g$ denote the genus of the compact Riemann surface $X$. Choose distinct points $p_1,\dots,p_d \in X$ with $d \geq 2g+1$, and define the effective divisor \begin{align*} D &:= p_1+\cdots+p_d. \end{align*} Let $\mathcal{O}_X(D)$ denote the holomorphic line bundle associated to $D$, and set \begin{align*} V &:= H^0(X,\mathcal{O}_X(D)). \end{align*} Let $\operatorname{Div}(X)$ denote the abelian group of divisors on $X$. Define the dimension function \begin{align*} \ell: \operatorname{Div}(X) &\to \mathbb{Z}_{\geq 0} \\ E &\mapsto \dim_{\mathbb{C}} H^0(X,\mathcal{O}_X(E)). \end{align*} We use the Riemann--Roch theorem for compact Riemann surfaces, recorded here as an external prerequisite because it is not yet available as an Androma theorem page, which states that \begin{align*} \ell(E)-\ell(K_X-E)&=\deg E+1-g, \end{align*} where $K_X$ is a canonical divisor on $X$. Since $\deg K_X=2g-2$, whenever $\deg E>2g-2$ we have $\deg(K_X-E)<0$, hence $H^0(X,\mathcal{O}_X(K_X-E))=0$. Therefore \begin{align*} \ell(E)&=\deg E+1-g \end{align*} for every divisor $E$ with $\deg E>2g-2$. In particular, for every point $p \in X$ and every pair of points $p,q \in X$, possibly equal, the divisors $D-p$, $D-p-q$, and $D-2p$ have degree at least $2g-1$ whenever they occur below. Thus Riemann--Roch gives \begin{align*} \ell(D)-\ell(D-p)&=1 \end{align*} and \begin{align*} \ell(D)-\ell(D-p-q)&=2. \end{align*}[/step]

[guided]The role of the divisor $D$ is to provide many meromorphic functions, or equivalently many holomorphic sections of a line bundle, with controlled poles. We choose $D$ effective of degree $d \geq 2g+1$ so that subtracting one point or two points still leaves a divisor of degree strictly larger than $2g-2$, the degree of a canonical divisor. Let $K_X$ be a canonical divisor on $X$. The Riemann--Roch theorem for compact Riemann surfaces, used here as an external prerequisite because it is not yet available as an Androma theorem page, says \begin{align*} \ell(E)-\ell(K_X-E)&=\deg E+1-g. \end{align*} If $\deg E>2g-2$, then \begin{align*} \deg(K_X-E)&=2g-2-\deg E<0. \end{align*} A line bundle attached to a divisor of negative degree has no nonzero holomorphic sections, because a nonzero meromorphic section has an effective divisor of zeros and poles whose total degree is forced to be nonnegative. Hence \begin{align*} \ell(K_X-E)&=0, \end{align*} and Riemann--Roch reduces to \begin{align*} \ell(E)&=\deg E+1-g. \end{align*} Now take $E=D$ and $E=D-p$. Since $\deg D=d$ and $\deg(D-p)=d-1$, both degrees exceed $2g-2$. Therefore \begin{align*} \ell(D)&=d+1-g \end{align*} and \begin{align*} \ell(D-p)&=d-g, \end{align*} so \begin{align*} \ell(D)-\ell(D-p)&=1. \end{align*} Similarly, for any two points $p,q \in X$, with $p=q$ allowed when we write $D-2p$, we have \begin{align*} \deg(D-p-q)&=d-2 \geq 2g-1>2g-2. \end{align*} Thus \begin{align*} \ell(D-p-q)&=d-1-g, \end{align*} and hence \begin{align*} \ell(D)-\ell(D-p-q)&=2. \end{align*} These two dimension drops are the numerical heart of the proof: losing one point imposes exactly one independent condition, and losing two first-order conditions imposes exactly two independent conditions.[/guided]

[step:Use the dimension drops to prove that the complete linear system has no base points] For $p \in X$, the inclusion of sheaves \begin{align*} \mathcal{O}_X(D-p)&\subset \mathcal{O}_X(D) \end{align*} induces the subspace \begin{align*} H^0(X,\mathcal{O}_X(D-p)) &\subset H^0(X,\mathcal{O}_X(D)). \end{align*} The quotient has dimension \begin{align*} \dim_{\mathbb{C}} V/H^0(X,\mathcal{O}_X(D-p))&=1. \end{align*} This quotient is naturally the value space of sections at $p$, namely the one-dimensional fiber $\mathcal{O}_X(D)_p$. Therefore the evaluation map \begin{align*} \operatorname{ev}_p: V &\to \mathcal{O}_X(D)_p \end{align*} is surjective. Hence for every $p \in X$ there exists $s \in V$ with $s(p)\neq 0$. Thus the complete linear system $V$ has no base points. [/step]

[step:Show that the complete linear system separates distinct points]Let $p,q \in X$ with $p\neq q$. Consider the restriction map \begin{align*} \rho_{p,q}: V &\to \mathcal{O}_X(D)_p \oplus \mathcal{O}_X(D)_q. \end{align*} Its kernel consists precisely of the sections vanishing at both $p$ and $q$, so \begin{align*} \ker \rho_{p,q}&=H^0(X,\mathcal{O}_X(D-p-q)). \end{align*} Since both the source quotient and the target have complex dimension $2$, \begin{align*} \dim_{\mathbb{C}} V/\ker \rho_{p,q} &= \ell(D)-\ell(D-p-q) \\ &= 2 \\ &= \dim_{\mathbb{C}}\bigl(\mathcal{O}_X(D)_p \oplus \mathcal{O}_X(D)_q\bigr), \end{align*} the map $\rho_{p,q}$ is surjective. Therefore there exists a section $s \in V$ such that \begin{align*} s(p)&=0, & s(q)&\neq 0. \end{align*} Thus the complete linear system separates the two distinct points $p$ and $q$.[/step]

[guided]To separate two points, we need a section whose values at those two points are independently controllable. Define \begin{align*} \rho_{p,q}: V &\to \mathcal{O}_X(D)_p \oplus \mathcal{O}_X(D)_q \end{align*} by sending a section to its pair of values at $p$ and $q$. The kernel consists exactly of those sections that vanish at both points, which is \begin{align*} H^0(X,\mathcal{O}_X(D-p-q)). \end{align*} The previous Riemann--Roch computation gives \begin{align*} \ell(D)-\ell(D-p-q)&=2. \end{align*} Thus the image of $\rho_{p,q}$ has dimension $2$. Since the target is the direct sum of two one-dimensional complex vector spaces, it also has dimension $2$. Hence $\rho_{p,q}$ is surjective. Surjectivity means we may prescribe the value at $p$ and the value at $q$ independently. In particular, we choose the prescribed values to be $0$ in $\mathcal{O}_X(D)_p$ and a nonzero element of $\mathcal{O}_X(D)_q$. The resulting section $s \in V$ satisfies \begin{align*} s(p)&=0, & s(q)&\neq 0. \end{align*} This is exactly point separation.[/guided]

[step:Show that the complete linear system separates tangent directions] Fix $p \in X$. Let $\mathfrak{m}_p \subset \mathcal{O}_{X,p}$ be the maximal ideal of [holomorphic function](/page/Holomorphic%20Function) germs vanishing at $p$, and let $\mathcal{O}_X(D)_{p,\mathrm{stalk}}$ denote the stalk of the line bundle $\mathcal{O}_X(D)$ at $p$. The first jet space of $\mathcal{O}_X(D)$ at $p$ is the two-dimensional complex [vector space](/page/Vector%20Space) \begin{align*} J_p^1(\mathcal{O}_X(D)) &:= \mathcal{O}_X(D)_{p,\mathrm{stalk}}/\mathfrak{m}_p^2\mathcal{O}_X(D)_{p,\mathrm{stalk}}. \end{align*} Indeed, after choosing any local holomorphic frame of $\mathcal{O}_X(D)$ near $p$, this quotient identifies with $\mathcal{O}_{X,p}/\mathfrak{m}_p^2$, which has basis represented by $1$ and by a local coordinate vanishing at $p$. The first-jet evaluation map is \begin{align*} j_p^1: V &\to J_p^1(\mathcal{O}_X(D)), \end{align*} and its kernel is the space of sections vanishing to order at least $2$ at $p$, namely \begin{align*} \ker j_p^1&=H^0(X,\mathcal{O}_X(D-2p)). \end{align*} Since \begin{align*} \dim_{\mathbb{C}} V/\ker j_p^1 &= \ell(D)-\ell(D-2p) \\ &= 2 \\ &= \dim_{\mathbb{C}} J_p^1(\mathcal{O}_X(D)), \end{align*} the map $j_p^1$ is surjective. Choose $a \in V$ with first jet equal to a nonzero constant jet at $p$, and choose $b \in V$ with first jet equal to a nonzero linear jet and zero constant term at $p$. Then $a(p)\neq 0$ and $b$ vanishes to order exactly $1$ at $p$. Let $U \subset X$ be a coordinate neighbourhood of $p$ with holomorphic coordinate \begin{align*} z: U &\to z(U)\subset \mathbb{C} \end{align*} satisfying $z(p)=0$, and choose a holomorphic frame $e$ of $\mathcal{O}_X(D)$ over $U$. Write \begin{align*} a|_U&=a_U e, & b|_U&=b_U e \end{align*} for holomorphic maps $a_U,b_U:U\to\mathbb{C}$. Since $a_U(p)\neq 0$ and $b_U$ has a simple zero at $p$, the holomorphic map \begin{align*} h: U &\to \mathbb{C} \\ x &\mapsto \frac{b_U(x)}{a_U(x)} \end{align*} has a simple zero at $p$. Hence its derivative at $p$ is nonzero. Thus the linear system separates tangent directions at $p$. [/step]

[step:Define the projective map from a basis of sections] The Riemann--Roch computation above gives \begin{align*} \dim_{\mathbb{C}}V&=\ell(D)=d+1-g<\infty. \end{align*} Choose a basis \begin{align*} s_0,\dots,s_N \end{align*} of the finite-dimensional complex vector space $V=H^0(X,\mathcal{O}_X(D))$, where $N+1=\dim_{\mathbb{C}}V$. Define \begin{align*} \Phi: X &\to \mathbb{P}^N_{\mathbb{C}} \end{align*} by \begin{align*} \Phi(x)&:=[s_0(x):\cdots:s_N(x)]. \end{align*} This expression is interpreted after choosing any local holomorphic frame of $\mathcal{O}_X(D)$ near $x$; changing the frame multiplies all local representatives by the same nonzero holomorphic function, so the point of projective space is unchanged. Since the linear system has no base points, not all $s_i(x)$ vanish at any $x \in X$, and therefore $\Phi$ is defined everywhere. In local frames, the homogeneous coordinates are holomorphic functions, so $\Phi$ is holomorphic. [/step]

[step:Prove that the projective map is injective and immersive] Let $p,q \in X$ with $p\neq q$. Since the linear system separates points, there exists $s \in V$ with \begin{align*} s(p)&=0, & s(q)&\neq 0. \end{align*} Choose local holomorphic frames $e_p$ of $\mathcal{O}_X(D)$ near $p$ and $e_q$ of $\mathcal{O}_X(D)$ near $q$. For each basis section $s_i$, write $s_i=u_{i,p}e_p$ near $p$ and $s_i=u_{i,q}e_q$ near $q$, where $u_{i,p}$ and $u_{i,q}$ are holomorphic complex-valued functions on the corresponding neighbourhoods. If $\Phi(p)=\Phi(q)$, then the two nonzero coordinate vectors \begin{align*} (u_{0,p}(p),\dots,u_{N,p}(p)) \end{align*} and \begin{align*} (u_{0,q}(q),\dots,u_{N,q}(q)) \end{align*} represent the same point of $\mathbb{P}^N_{\mathbb{C}}$, so they are proportional by some scalar $\lambda \in \mathbb{C}^{\times}$. Express $s=\sum_{i=0}^N c_i s_i$ with $c_i \in \mathbb{C}$. Then \begin{align*} 0=s(p)&=\left(\sum_{i=0}^N c_i u_{i,p}(p)\right)e_p(p) \end{align*} implies $\sum_{i=0}^N c_i u_{i,p}(p)=0$, and proportionality gives \begin{align*} \sum_{i=0}^N c_i u_{i,q}(q)&=\lambda \sum_{i=0}^N c_i u_{i,p}(p)=0. \end{align*} Hence $s(q)=0$, contradicting $s(q)\neq 0$. Therefore $\Phi(p)\neq \Phi(q)$, so $\Phi$ is injective. Now fix $p \in X$. From tangent separation, choose $a,b \in V$ such that $a(p)\neq 0$ and $b$ vanishes to order exactly $1$ at $p$. In the affine chart of $\mathbb{P}^N_{\mathbb{C}}$ where the coordinate corresponding to $a$ is nonzero, the map $\Phi$ has a holomorphic coordinate function equal to $b/a$. The derivative of $b/a$ at $p$ is nonzero, so \begin{align*} d\Phi_p:T_pX&\to T_{\Phi(p)}\mathbb{P}^N_{\mathbb{C}} \end{align*} is injective. Since $p$ was arbitrary and $\dim_{\mathbb{C}}T_pX=1$, $\Phi$ is a holomorphic immersion. [/step]

[step:Use compactness to upgrade the injective immersion to an embedding] The space $X$ is compact by hypothesis, and $\mathbb{P}^N_{\mathbb{C}}$ is Hausdorff. Therefore the continuous injective map \begin{align*} \Phi:X&\to \mathbb{P}^N_{\mathbb{C}} \end{align*} is a homeomorphism from $X$ onto its image $\Phi(X)$ by the compact-to-Hausdorff continuous bijection theorem, recorded here as an external prerequisite because it is not yet available as an Androma theorem page. It remains to justify the holomorphic submanifold structure on the image and the holomorphicity of the inverse. Fix $p \in X$. Since $d\Phi_p$ is injective and $\dim_{\mathbb{C}}X=1$, the holomorphic constant-rank local embedding theorem, also used here as an external prerequisite, gives an open neighbourhood $U_p \subset X$ of $p$ such that $\Phi(U_p)$ is a complex submanifold of $\mathbb{P}^N_{\mathbb{C}}$ and \begin{align*} \Phi|_{U_p}:U_p&\to \Phi(U_p) \end{align*} is a biholomorphism. Since $\Phi:X\to\Phi(X)$ is already a homeomorphism, these local inverse maps agree on overlaps and assemble to a holomorphic inverse \begin{align*} \Phi(X)&\to X. \end{align*} Thus $\Phi(X)$ is a complex submanifold of $\mathbb{P}^N_{\mathbb{C}}$, and $\Phi$ is a holomorphic embedding. Consequently $X$ admits a holomorphic embedding into complex projective space, so $X$ is projective. This proves that every compact connected complex manifold of dimension $1$ is projective. [/step]