[proofplan] The lower bound follows because the constant meromorphic functions embed $\mathbb{C}$ into $\mathbb{C}(X)$. For the upper bound, we take an arbitrary finite family of meromorphic functions on $X$ and use it to build a meromorphic map from $X$ to a product of projective lines. The analytic graph of this map has compact image in $(\mathbb{P}^1)^m$; Remmert's theorem and [Chow's theorem](/theorems/3886) convert that image into a projective algebraic variety of dimension at most $n$. Since the chosen functions factor through the function field of this variety, their transcendence degree is bounded by its dimension, hence by $n$. [/proofplan] [step:Embed the constant functions to obtain the lower bound] Since $X$ is connected, every constant function $c \in \mathbb{C}$ defines a [meromorphic function](/page/Meromorphic%20Function) \begin{align*} \underline{c}: X &\dashrightarrow \mathbb{P}^1 \\ x &\mapsto c . \end{align*} Thus the assignment $c \mapsto \underline{c}$ embeds $\mathbb{C}$ as a subfield of $\mathbb{C}(X)$. Therefore \begin{align*} a(X)=\operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(X) \geq 0. \end{align*} [/step] [step:Associate a meromorphic map to a finite family of meromorphic functions] Let $m \in \mathbb{N}$ and let $f_1,\dots,f_m \in \mathbb{C}(X)$ be meromorphic functions. Each $f_j$ is equivalently a meromorphic map \begin{align*} f_j: X &\dashrightarrow \mathbb{P}^1 . \end{align*} Define the meromorphic map \begin{align*} F: X &\dashrightarrow (\mathbb{P}^1)^m \\ x &\mapsto (f_1(x),\dots,f_m(x)) \end{align*} on the common domain where all $f_j$ are holomorphic. Let $\Gamma \subset X \times (\mathbb{P}^1)^m$ denote the analytic closure of the graph of $F$ over this common domain. By the graph theorem for meromorphic maps between compact complex spaces (citing a result not yet in the wiki: analytic graph theorem for meromorphic maps), $\Gamma$ is a compact analytic subset of $X \times (\mathbb{P}^1)^m$, and the projection \begin{align*} \pi_X: \Gamma &\to X \end{align*} is a proper modification over the domain of definition of $F$. [guided] We start with an arbitrary finite list $f_1,\dots,f_m \in \mathbb{C}(X)$ because the algebraic dimension is the largest possible size of an algebraically independent family of meromorphic functions. A meromorphic function on a complex manifold is the same kind of object as a meromorphic map to $\mathbb{P}^1$, so the list gives one meromorphic map \begin{align*} F: X &\dashrightarrow (\mathbb{P}^1)^m \\ x &\mapsto (f_1(x),\dots,f_m(x)). \end{align*} This map may fail to be holomorphic on an analytic indeterminacy set, so we replace it by its graph. Let $\Gamma$ be the analytic closure in $X \times (\mathbb{P}^1)^m$ of the ordinary graph over the common holomorphic domain of the $f_j$. Since $X$ is compact and $(\mathbb{P}^1)^m$ is compact, the ambient product $X \times (\mathbb{P}^1)^m$ is compact. The analytic graph theorem for meromorphic maps between compact complex spaces says that this closure is a compact analytic subset and that the projection \begin{align*} \pi_X: \Gamma &\to X \end{align*} is proper and agrees with the original graph over the domain where $F$ is defined. This replaces the meromorphic map by an honest analytic space on which projection to $(\mathbb{P}^1)^m$ is holomorphic. [/guided] [/step] [step:Apply Remmert and Chow to make the image algebraic] Let \begin{align*} \pi_P: \Gamma &\to (\mathbb{P}^1)^m \end{align*} be the restriction of the second projection. Since $\Gamma$ is compact, $\pi_P$ is proper. By Remmert's proper mapping theorem (citing a result not yet in the wiki: Remmert proper mapping theorem), the image \begin{align*} Y := \pi_P(\Gamma) \subset (\mathbb{P}^1)^m \end{align*} is a compact analytic subset. Moreover \begin{align*} \dim_{\mathbb{C}} Y \leq \dim_{\mathbb{C}} \Gamma \leq \dim_{\mathbb{C}} X = n. \end{align*} By Chow's theorem for analytic subsets of projective space, applied after the Segre embedding of $(\mathbb{P}^1)^m$ into projective space (citing a result not yet in the wiki: Chow's theorem), $Y$ is a projective algebraic subvariety of $(\mathbb{P}^1)^m$. [guided] The projection \begin{align*} \pi_P: \Gamma &\to (\mathbb{P}^1)^m \end{align*} is holomorphic because it is the restriction of the holomorphic projection from the product. It is proper because $\Gamma$ is compact and the target is Hausdorff. Therefore Remmert's proper mapping theorem applies: the image of a proper holomorphic map from a complex analytic space is an analytic subset of the target. Hence \begin{align*} Y := \pi_P(\Gamma) \end{align*} is a compact analytic subset of $(\mathbb{P}^1)^m$. The dimension estimate comes from the fact that holomorphic images cannot have larger complex dimension than their source: \begin{align*} \dim_{\mathbb{C}} Y \leq \dim_{\mathbb{C}} \Gamma. \end{align*} The graph $\Gamma$ is a modification of $X$ over a dense [open set](/page/Open%20Set), so every irreducible component relevant to the meromorphic map has dimension at most $\dim_{\mathbb{C}} X=n$. Thus \begin{align*} \dim_{\mathbb{C}} Y \leq n. \end{align*} Now we use algebraicity. The space $(\mathbb{P}^1)^m$ is projective via the Segre embedding, and Chow's theorem says that every compact analytic subset of projective space is algebraic. Applying Chow's theorem to the Segre image of $Y$, and then pulling back through the Segre embedding, shows that $Y$ is a projective algebraic subvariety of $(\mathbb{P}^1)^m$. [/guided] [/step] [step:Factor the chosen meromorphic functions through the function field of the image] For each index $j \in \{1,\dots,m\}$, let \begin{align*} p_j: (\mathbb{P}^1)^m &\to \mathbb{P}^1 \end{align*} be the $j$-th projection. Its restriction to $Y$ defines a meromorphic function \begin{align*} g_j: Y &\dashrightarrow \mathbb{P}^1 . \end{align*} On the common domain where $F$ is holomorphic, we have \begin{align*} f_j = g_j \circ F. \end{align*} Therefore the field generated by $f_1,\dots,f_m$ over $\mathbb{C}$ is a quotient of, equivalently is algebraically controlled by, the subfield of $\mathbb{C}(Y)$ generated by $g_1,\dots,g_m$. Hence \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(f_1,\dots,f_m) \leq \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(Y). \end{align*} Since $Y$ is an algebraic variety, its function field has transcendence degree equal to its algebraic dimension: \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(Y) = \dim_{\mathbb{C}} Y \leq n, \end{align*} where the equality is the standard [dimension theorem](/theorems/915) for irreducible algebraic varieties, applied componentwise to the image carrying the functions (citing a result not yet in the wiki: transcendence degree equals dimension for algebraic varieties). [guided] The point of introducing $Y$ is that the functions $f_j$ become pullbacks of the simplest possible functions on $Y$. For each $j$, define \begin{align*} p_j: (\mathbb{P}^1)^m &\to \mathbb{P}^1 \end{align*} to be the projection onto the $j$-th factor. Restricting $p_j$ to the algebraic subvariety $Y$ gives a meromorphic function \begin{align*} g_j: Y &\dashrightarrow \mathbb{P}^1 . \end{align*} By construction of $F$, wherever the original meromorphic map is defined, \begin{align*} f_j = g_j \circ F. \end{align*} Thus any polynomial relation among the $g_j$ over $\mathbb{C}$ pulls back to the same polynomial relation among the $f_j$. Equivalently, the field generated by $f_1,\dots,f_m$ cannot have larger transcendence degree than the function field of $Y$: \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(f_1,\dots,f_m) \leq \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(Y). \end{align*} Because $Y$ is algebraic, we may use the algebraic dimension theorem: the transcendence degree of the function field of an irreducible algebraic variety over $\mathbb{C}$ equals its dimension. If $Y$ has several irreducible components, the functions $g_j$ restrict to the component through which the image of the graph is dense, and the same estimate applies on that component. Therefore \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(Y) = \dim_{\mathbb{C}} Y \leq n. \end{align*} This is the step where analytic compactness has been converted into the algebraic bound on the number of independent meromorphic functions. [/guided] [/step] [step:Conclude that no algebraically independent family has more than $n$ elements] The preceding argument applies to every finite family $f_1,\dots,f_m \in \mathbb{C}(X)$ and gives \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(f_1,\dots,f_m) \leq n. \end{align*} If $f_1,\dots,f_m$ are algebraically independent over $\mathbb{C}$, then \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathbb{C}(f_1,\dots,f_m)=m. \end{align*} Hence $m \leq n$. Taking the supremum over all finite algebraically independent families in $\mathbb{C}(X)$ yields \begin{align*} a(X)=\operatorname{trdeg}_{\mathbb{C}}\mathbb{C}(X)\leq n. \end{align*} Together with the lower bound, this proves \begin{align*} 0 \leq a(X) \leq n. \end{align*} [/step]