[proofplan] Choose finitely many meromorphic functions on $X$ that generate the field $\mathbb{C}(X)$ over $\mathbb{C}$. These functions define a meromorphic map from $X$ to projective space; taking the Zariski closure of its image gives an irreducible projective variety whose rational function field pulls back to $\mathbb{C}(X)$. Finally, normalize this image to obtain a normal projective variety without changing the function field. [/proofplan]

[step:Choose meromorphic generators for the function field]Let \begin{align*} K := \mathbb{C}(X) \end{align*} be the field of meromorphic functions on $X$. Since $X$ is compact and connected, the analytic version of Siegel's finite generation theorem gives that $K$ is a finitely generated [field extension](/page/Field%20Extension) of $\mathbb{C}$ with finite transcendence degree \begin{align*} a(X) = \operatorname{trdeg}_{\mathbb{C}} K. \end{align*} Here we are citing a result not yet in the wiki: finite generation of the field of meromorphic functions on a compact complex manifold. Choose meromorphic functions \begin{align*} f_1,\dots,f_m \in K \end{align*} such that \begin{align*} K = \mathbb{C}(f_1,\dots,f_m). \end{align*} If $K = \mathbb{C}$, we take $m=0$.[/step]

[guided]The main input is that compactness forces the [meromorphic function](/page/Meromorphic%20Function) field to behave like the function field of an algebraic variety: it is finitely generated over $\mathbb{C}$. We denote this field by \begin{align*} K := \mathbb{C}(X). \end{align*} By the analytic form of Siegel's finite generation theorem, $K$ is a finitely generated extension field of $\mathbb{C}$. This theorem is not yet available in the wiki, so the citation is recorded explicitly here as a missing prerequisite: finite generation of the field of meromorphic functions on a compact complex manifold. Because $K$ is finitely generated, we may choose meromorphic functions \begin{align*} f_1,\dots,f_m \in \mathbb{C}(X) \end{align*} whose field of rational expressions is all of $K$: \begin{align*} K = \mathbb{C}(f_1,\dots,f_m). \end{align*} The number $m$ is not required to be minimal. If $X$ has no nonconstant meromorphic functions, then $K=\mathbb{C}$ and we take the empty generating family, so $m=0$.[/guided]

[step:Construct the projective meromorphic map from the generators]Define a meromorphic map \begin{align*} F: X \dashrightarrow \mathbb{P}^m_{\mathbb{C}} \end{align*} by \begin{align*} x \longmapsto [1:f_1(x):\cdots:f_m(x)] \end{align*} on the common domain where all $f_i$ are holomorphic and finite. Let \begin{align*} Z := \overline{F(X)}^{\,\mathrm{Zar}} \subset \mathbb{P}^m_{\mathbb{C}} \end{align*} be the Zariski closure of the image. Since $X$ is connected, the image of $F$ is analytically irreducible on its domain of definition, and therefore $Z$ is an irreducible projective variety. The map $F$ is dominant as a meromorphic map \begin{align*} F: X \dashrightarrow Z \end{align*} by the definition of $Z$ as the Zariski closure of its image.[/step]

[guided]The chosen generators give a map to projective space by recording their values as homogeneous coordinates. More precisely, define \begin{align*} F: X \dashrightarrow \mathbb{P}^m_{\mathbb{C}} \end{align*} by \begin{align*} x \longmapsto [1:f_1(x):\cdots:f_m(x)]. \end{align*} This formula is understood on the open subset of $X$ where all the meromorphic functions $f_i$ are holomorphic and finite. Since the first homogeneous coordinate is $1$, the map lands in the affine chart $\{Z_0 \neq 0\}$ wherever it is defined. Now define \begin{align*} Z := \overline{F(X)}^{\,\mathrm{Zar}} \subset \mathbb{P}^m_{\mathbb{C}}. \end{align*} Because $\mathbb{P}^m_{\mathbb{C}}$ is projective, every closed subvariety is projective, so $Z$ is projective. Because $X$ is connected, the analytic image of the domain of definition of $F$ has a single irreducible analytic closure, and its Zariski closure is therefore irreducible. Thus $Z$ is an irreducible projective variety. Finally, if we regard the same map as \begin{align*} F: X \dashrightarrow Z, \end{align*} then it is dominant by construction: the Zariski closure of its image is exactly $Z$.[/guided]

[step:Identify the function field of the image with the generated meromorphic field]Let $T_0,\dots,T_m$ denote the homogeneous coordinate functions on $\mathbb{P}^m_{\mathbb{C}}$. Let $\mathbb{A}^1_{\mathbb{C}}$ denote the affine line over $\mathbb{C}$. For each $i \in \{1,\dots,m\}$, define the rational function \begin{align*} u_i: Z \dashrightarrow \mathbb{A}^1_{\mathbb{C}} \end{align*} by restricting the affine coordinate $T_i/T_0$ to $Z \cap \{T_0 \neq 0\}$. By the definition of $F$, \begin{align*} F^*(u_i) = f_i \end{align*} for every $i \in \{1,\dots,m\}$. Since $F: X \dashrightarrow Z$ is dominant, pullback gives an injective field homomorphism \begin{align*} F^*: \mathbb{C}(Z) \longrightarrow \mathbb{C}(X). \end{align*} Its image contains $\mathbb{C}(f_1,\dots,f_m)$ because it contains every $f_i$. Conversely, every rational function on $Z \subset \mathbb{P}^m_{\mathbb{C}}$ pulls back to a rational expression in the coordinate functions $T_i/T_0$, hence its pullback belongs to $\mathbb{C}(f_1,\dots,f_m)$. Therefore \begin{align*} F^*\mathbb{C}(Z) = \mathbb{C}(f_1,\dots,f_m) = \mathbb{C}(X). \end{align*} Thus $F^*: \mathbb{C}(Z) \to \mathbb{C}(X)$ is an isomorphism onto $\mathbb{C}(X)$.[/step]

[guided]The point of using the map \begin{align*} x \longmapsto [1:f_1(x):\cdots:f_m(x)] \end{align*} is that the original generators can be recovered from projective coordinates. Let $T_0,\dots,T_m$ be the homogeneous coordinates on $\mathbb{P}^m_{\mathbb{C}}$. Let $\mathbb{A}^1_{\mathbb{C}}$ denote the affine line over $\mathbb{C}$. On the affine chart $\{T_0 \neq 0\}$, define rational coordinate functions by \begin{align*} u_i := \frac{T_i}{T_0}\bigg|_Z \end{align*} for $i \in \{1,\dots,m\}$. These are rational functions on $Z$, meaning \begin{align*} u_i \in \mathbb{C}(Z). \end{align*} By the defining formula for $F$, their pullbacks are exactly the chosen meromorphic functions: \begin{align*} F^*(u_i)=f_i. \end{align*} Because $F: X \dashrightarrow Z$ is dominant, a rational function on $Z$ that pulls back to zero on $X$ must vanish on a Zariski [dense subset](/page/Dense%20Subset) of $Z$, hence must be the zero rational function. Therefore the pullback map \begin{align*} F^*: \mathbb{C}(Z) \longrightarrow \mathbb{C}(X) \end{align*} is injective. We now identify its image. Since the image contains every $f_i$, it contains the field generated by them: \begin{align*} \mathbb{C}(f_1,\dots,f_m) \subset F^*\mathbb{C}(Z). \end{align*} For the reverse inclusion, every rational function on the projective subvariety $Z \subset \mathbb{P}^m_{\mathbb{C}}$ is locally represented by a quotient of homogeneous polynomials in the coordinates $T_0,\dots,T_m$ of the same degree. On the affine chart $T_0 \neq 0$, this quotient is a rational expression in \begin{align*} \frac{T_1}{T_0},\dots,\frac{T_m}{T_0}. \end{align*} Pulling back by $F$ replaces these coordinate functions by \begin{align*} f_1,\dots,f_m. \end{align*} Hence every element of $F^*\mathbb{C}(Z)$ lies in $\mathbb{C}(f_1,\dots,f_m)$. Therefore \begin{align*} F^*\mathbb{C}(Z)=\mathbb{C}(f_1,\dots,f_m)=\mathbb{C}(X). \end{align*} So the pullback identifies $\mathbb{C}(Z)$ with $\mathbb{C}(X)$.[/guided]

[step:Compute the dimension of the image]For an irreducible projective variety, the dimension equals the transcendence degree of its rational function field over $\mathbb{C}$. Applying this to $Z$ gives \begin{align*} \dim Z = \operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(Z). \end{align*} Since $F^*: \mathbb{C}(Z) \to \mathbb{C}(X)$ is an isomorphism, \begin{align*} \dim Z = \operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(X) = a(X). \end{align*}[/step]

[guided]The dimension of an irreducible algebraic variety is measured by the size of its rational function field. Specifically, for an irreducible projective variety $Z$ over $\mathbb{C}$, \begin{align*} \dim Z = \operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(Z). \end{align*} This is a standard result from algebraic geometry, not yet resolved to a theorem entry in the wiki. From the previous step, the pullback map gives an isomorphism of fields \begin{align*} F^*: \mathbb{C}(Z) \longrightarrow \mathbb{C}(X). \end{align*} Isomorphic field extensions have the same transcendence degree over the base field, so \begin{align*} \operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(Z) = \operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(X). \end{align*} By the definition of algebraic dimension, \begin{align*} a(X)=\operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(X). \end{align*} Combining these equalities gives \begin{align*} \dim Z = a(X). \end{align*}[/guided]

[step:Normalize the projective image without changing its function field]Let \begin{align*} \nu: Y \longrightarrow Z \end{align*} be the normalization of the irreducible projective variety $Z$. Since normalization is finite and birational, $Y$ is an irreducible normal projective variety and the induced pullback \begin{align*} \nu^*: \mathbb{C}(Z) \longrightarrow \mathbb{C}(Y) \end{align*} is an isomorphism. Moreover, \begin{align*} \dim Y = \dim Z = a(X). \end{align*} Define the meromorphic map \begin{align*} r: X \dashrightarrow Y \end{align*} as the meromorphic lift of $F: X \dashrightarrow Z$ through the birational morphism $\nu$. On function fields, its pullback is \begin{align*} r^* = F^* \circ (\nu^*)^{-1}: \mathbb{C}(Y) \longrightarrow \mathbb{C}(X). \end{align*} Since both $F^*$ and $\nu^*$ are isomorphisms onto their respective targets, $r^*$ is an isomorphism. Thus $Y$ and $r$ have all required properties.[/step]

[guided]The variety $Z$ already has the correct function field, but it may not be normal. To impose normality, take the normalization \begin{align*} \nu: Y \longrightarrow Z. \end{align*} Normalization is a finite birational morphism from a normal irreducible variety $Y$ to $Z$. Because $Z$ is projective and $\nu$ is finite, $Y$ is also projective. Because $\nu$ is birational, it does not change the rational function field: \begin{align*} \nu^*: \mathbb{C}(Z) \longrightarrow \mathbb{C}(Y) \end{align*} is an isomorphism. A finite birational morphism preserves dimension, so \begin{align*} \dim Y = \dim Z. \end{align*} The previous step proved $\dim Z=a(X)$, hence \begin{align*} \dim Y=a(X). \end{align*} It remains to define the desired meromorphic map into the normal variety. Since $\nu$ is birational, it has a rational inverse defined on dense open subsets. Compose $F$ with this rational inverse to obtain a meromorphic map \begin{align*} r: X \dashrightarrow Y. \end{align*} Equivalently, and more invariantly, define $r$ by its pullback on function fields: \begin{align*} r^* = F^* \circ (\nu^*)^{-1}: \mathbb{C}(Y) \longrightarrow \mathbb{C}(X). \end{align*} Both maps on the right are isomorphisms, so $r^*$ is an isomorphism. Therefore $Y$ is an irreducible normal projective variety of dimension $a(X)$, and the meromorphic map \begin{align*} r: X \dashrightarrow Y \end{align*} identifies $\mathbb{C}(Y)$ with $\mathbb{C}(X)$ by pullback.[/guided]