What do we call the algebraic closure of $\mathbb{R}(x)$, and how can we describe it? $\mathbb{R}(x)$ is not algebraically closed, as can easily be seen by considering the equation $xy^2=1$; there is no rational function $r(x)$ with the property that $xr(x)^2=1$.  However, everyone knows how to solve this equation for $y$, obtaining
$$y=\frac{1}{\sqrt{x}}, y=-\frac{1}{\sqrt{x}}$$
as the two solutions. So the solutions exist, but in order to find them we have to go beyond $\mathbb{R}(x)$ to a larger collection of functions.
The algebraic closure of $\mathbb{R}(x)$ would contain not only all rational functions, but also roots of all orders, plus some things we don't have names or notation for (solutions of polynomials with coefficients in $\mathbb{R}(x)$ of degree 5 or higher that are not solvable by radicals , for example).  (The word "functions", here, is used somewhat casually, as most of these things are partial functions at best.)  I guess it would have to contain $\mathbb{C}$ as well, come to think of it.  
What is the algebraic closure called?  Does it have a conventional name?  I think of it as the "field of algebraic functions" but I don't know if that's standard terminology or if there's another, more commonly-used name.  Is there a notation for it?  Does it contain any other exotic things, besides what I have already mentioned?
 A: Let's change to $\mathbb{C}(x)$ (we might as well, since certainly the algebraic closure of $\mathbb{R}[x]$ will contain a square root of negative one).
Finite extensions of $\mathbb{C}(x)$ are in bijection with pairs of a smooth proper algebraic curves $X$ together with a meromorphic function $\phi: X \to \mathbb{P}^1$. The easy direction of the bijection is that a curve together with a $\phi$ gives a map of fields $\mathbb{C}(x) \to \mathbb{C}(X)$ (the field of meromorphic functions on $X$) given by pre-composing a function with $\phi$. The interesting direction is that the data of a field extension actually determines an algebraic curve (together with a meromorphic function on the curve), so that every field extension of $\mathbb{C}(x)$ arises in this way.
This correspondence is such that if $C_1 \to C_2$ is a map of curves, then $\mathbb{C}(C_2) \hookrightarrow \mathbb{C}(C_1)$.
Since the algebraic closure is the union of finite extensions, you can therefore think of it as the space of meromorphic functions on the inverse limit of all the algebraic curves; concretely, an element of it is a choice of curve (together with map to $\mathbb{P}^1$) and a meromorphic function on that curve. So for instance, $\sqrt{x}$ is "really" a function on the pullback of $\mathbb{P}^1$ under $x \to x^2$.
A: For $L/\mathbb{C}(z)$ a Galois extension, pick a simply connected open set $U \subset \mathbb{C}$ such that $L$ is a field of meromorphic functions $U \to \mathbb{C}$.
Let $\gamma(t),t \in [0,1]$ be a closed curve in the Riemann sphere with $\gamma(0)= \gamma(1) \in U$. 
For $f \in L$, continue $f$ analytically along $\gamma$, assume it is well-defined and that the resulting function is meromorphic on $U$, call it $f^\gamma$.

The map $f \mapsto f^\gamma$ is an element of $G=Gal(L/\mathbb{C}(z))$.

Proof : analytic continuation is clearly compatible with products, addition, quotient, so it is a field morphism, and if $\sum_{j=0}^J \frac{P_j(z)}{Q_j(z)} f(z)^j = 0$ then $\sum_{j=0}^J \frac{P_j(z)}{Q_j(z)} f^\gamma(z)^j = 0$ so the image lies in $L$, and since the map is inversible, the image is exactly $L$, so the map is $\in Aut(L)$, and it leaves $\mathbb{C}(z)$ fixed so it is $\in G$.

If $f \in L$ and $V$ is a simply connected open set then there exists at most $|G|$ different meromorphic functions $V \to \mathbb{C}$ that are analytic continuations of $f$. 

Write $L = \mathbb{C}(z,h(z))$ to obtain that for some set of curves $\prod_{\gamma \in \Gamma} h^\gamma \in \mathbb{C}(z) \implies \forall f \in L,\prod_{\gamma \in \Gamma} f^\gamma \in \mathbb{C}(z)$ and hence $G = \Gamma$.

If $f \in L$ then any analytic continuation of $f$ is meromorphic everywhere (on the Riemann sphere) except at finitely many points where $f(a+(z-a)^{|G|})$ is meromorphic (if $a = \infty$ check that $f(1/z^{|G|})$ is meromorphic at $0$)

For a branch point $a$ of $f$, continue it analytically by rotating $|G|$ times around $a$ to obtain that $a$ isn't a branch point of $f(a+(z-a)^{|G|})$. And $a$ isn't an essential singularity of $f(a+(z-a)^{|G|})$ because $\sum_{j=0}^J \frac{P_j(z)}{Q_j(z)} f(z)^j = 0$, thus $f(a+(z-a)^{|G|})$ is meromorphic at $a$.

Those criterion are a characterization of the algebraic functions : a meromorphic function $f : U \to \mathbb{C}$ is in $\overline{\mathbb{C}(z)}$ iff $f$ has finitely many analytic continuation and those are meromorphic everywhere except at finitely many branch points where $f(a+(z-a)^n)$ is meromorphic.

Proof : note that $\prod_{\gamma \in \Gamma} (y-f^\gamma(z)) \in R[y]$ where $R$ is the ring of meromorphic functions on the Riemann sphere, that is $R = \mathbb{C}(z)$ and hence $f \in \overline{\mathbb{C}(z)}$.
