Main Question
I was searching for a calculable working example of algebraically closed field of characteristic $0$, other than $\mathbb{C}$ or $\overline{\mathbb{Q}}$. Then I came across the following question: What is the algebraic closure of $\mathbb{C}(x)$?
Basics: Field of Fractions, Rational Functions
Given an integral domain $R$, let $R^* = R - \{0\}$ be the set of nonzero elements. Define an equivalence relation $\sim$ on $R \times R^*$ by letting $(n,d) \sim (m,c)$ to denote $nc = md$. Define $\dfrac{n}{d}$ denote $(n,d)$. Then $\mathrm{Frac}(R) = (R\times R^*)/\sim$ has the structure of field and called the field of fraction of $R$.
Given a field $\mathbb{K}$, let $\mathbb{K}[x]$ denote the polynomial ring in one indeterminate. Then $\mathbb{K}[x]$ is an integral domain, so we can consider the field of fraction. Let $\mathbb{K}(x)$ denote $\mathrm{Frac}(\mathbb{K}[x])$ and is called the field of rational functions.
Not Algebraically Closed
Let $\mathbb{F}$ be the field of rational functions over complex numbers: $\mathbb{F} = \mathbb{C}(x)$. Is it algebraically closed? No, it isn't. Let $p(x) \in \mathbb{Q}[x] \subset \mathbb{F}$ be a polynomial over rational number. For example, let $T^2 - p(x) \in \mathbb{F}[T]$ be a polynomial over $\mathbb{F}$. Consider the solution for $T$ of $T^2 - p(x) = 0$. Then we obtain $T = \pm\sqrt{p(x)}$, but $\pm\sqrt{p(x)}\notin\mathbb{F}$. So $\mathbb{F}$ is not algebraically closed. Then my question is: What is the algebraic closure of $\mathbb{F}$? Let $\overline{\mathbb{F}}$ be the algebraic closure of $\mathbb{F}$.
Examples of Fields of Characteristic Zero
The algebraic closure of $\mathbb{R}$ is $\mathbb{C}$, both uncountable cardinality.
The algebraic closure of $\mathbb{Q}$ is the field of algebraic numbers $\overline{\mathbb{Q}}$, both countable cardinality.
There are many countable algebraically closed field between $\overline{\mathbb{Q}}$ and $\mathbb{C}$, for example, $\overline{\mathbb{Q}(\pi)}$
$\overline{\mathbb{F}} = \overline{\mathbb{C}(x)}$ is equal to $\overline{\mathbb{R}(x)}$.
Characterize Elements of $\mathbb{F}$
The question what is $\mathbb{F}$ can be translated to how to characterize each element of $\mathbb{F}$. In the case of $\overline{\mathbb{Q}}=\mathbb{C}$, we can characterize an element of $r\in\mathbb{R}$ by a convergent series of elements of $\mathbb{Q}$, and then we can characterize an element of $c\in\mathbb{C}$ by a pair of elements of $\mathbb{R}$. How about $\mathbb{F}$? In order to use the same convergent series strategy, we need to define distance between rational functions, but I don't know it well.
Algebraic Function [PEND: ToDo]
Although a rational function $r(x)\in\mathbb{F}$ can be considered a complex function, the definition does not restrict what the domain indeterminate $x$ resides. Moreover, $f\in\overline{\mathbb{F}}$ may have branches, for example, $x^2 + T^2 = 1$ imply $T$ is a two valued function.
At last, the Question is:
What is the set $\overline{\mathbb{C}(x)}$ the algebraic closure of the field $\mathbb{C}(x)$?
Any ideas?