On the definition of algebraic dimension In the book of complex geometry by D. Huybrechts, the algebraic dimension of a compact connected complex manifold $X$ is $\mathrm{a}(X):=\mathrm{trdeg}\mathcal{K}(X)$  over $\mathbb{C}$, where $\mathbb{C}$ is the complex number field and $\mathcal{K}(X)$ the function field of $X$. By Siegel's theorem, one knows that $\mathrm{a}(X)\leq \mathrm{dim}(X)$, so what if we omit the condition of compactness? In the case, how to define algebraic dimension of a connected complex manifold? In other words, my question is that whether the compactness condition of above definition is necessary or not.
Thanks in advance!
 A: WE can define the algebraic dimension of $X$ as follows
For any compact connected
complex manifold $X$ the algebraic dimension $a(X)$ is described as
$$a(X) = max\{\kappa(L);L ∈ PicX\}$$
where $\kappa(L)$ is the Iitaka dimension of the line bundle $L$ and $ PicX $ is the Picard
variety of $X$ and the Iitaka dimension can be defined as follows
$$\kappa(L)=\limsup_{m\to \infty}\frac{\log dim H^0(X,L^{\otimes m})}{\log m}
$$
Moreover, $a(X)=dim X$ if $X$ is projective algebraic.
In fact $a(X)$  measures how many meromorphic
functions exist on $ X$
For any $ X$, there exists a projective algebraic manifold $ Y $ and a surjective
meromorphic map $ f : X → Y $ which induces an isomorphism for the meromorphic function fields of $ Y $ and
$X, $ so that $dim Y = a(X)$; the meromorphic map $f : X → Y $ is called the algebraic reduction of $X$, and is known
to be unique under a bimeromorphic equivalence. Moreover, fibers of $f $ are necessarily connected.
See Daniel Barlet paper 
https://arxiv.org/abs/1610.02238
