# How to describe Algebraic Closure of $\mathbb{C}(x)$?

Let $\mathbb{C}$ be the set of complex numbers, and $x$ be an indeterminate. Let $\overline{\mathbb{C}(x)}$ be an algebraic closure of $\mathbb{C}(x)$. Then what are the elements of $\overline{\mathbb{C}(x)}$?

Obviously, elements like $\sqrt[n]{x}, \sqrt[n]{f(x)}$ (where $f(x)\in\mathbb{C}(x)$), etc. will be in the set. But is it easy to describe the full set?

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I'm quite sure there's no "easy" description of the elements of this field, for any interpretation of "easy" I can imagine. –  Alon Amit Aug 26 '11 at 7:30
"There are more things in heaven and earth, Horatio, Than are dreamt of in your philosophy." (Shakespeare, Hamlet, I,5) –  Georges Elencwajg Aug 26 '11 at 7:58
I doubt it's any easier than "the Puiseux series that are roots of some polynomial with coefficients in the range of the obvious embedding from $\mathbb{C}$". –  Ricky Demer Aug 26 '11 at 9:01
@Ricky Demer: yes, it is true that the requested algebraic closure is the set of Puiseux series algebraic over $\mathbb C(t)$. But that doesn't say anything against "easiness".Actually there is a huge literature on explicit algorithmic procedures in that field (!), the key words being "Newton polygon". The book by Gerd Fischer Plane Algebraic Curves is a nice introduction to the rich ties of Puiseux series with Geometry: Newton polygons, desingularization of curves, ramified coverings, ... –  Georges Elencwajg Aug 26 '11 at 11:43
It might be noted that the 'inverse galois' problem holds over $\mathbb{C}(t)$ i.e. that for any finite group $G$ there exists a finite extension $L/\mathbb{C}(t)$ such that $Gal(L/\mathbb{C}(t)) \cong G$ –  jspecter Aug 27 '11 at 1:49

Yes, it possible to describe the elements of $\overline{\mathbb{C}(x)}$ very explicitly.
They are called Puiseux series and consist of infinite series of the form $$ax^{-1/12}+bx^{-7/3}+cx^{4/11}+\ldots$$ with $a,b, c,\ldots \in \mathbb{C}$. The infinitely many fractions which appear as exponents must have a common denominator and only finitely many of them may be negative.The rules for addition and multiplication are the evident ones, analogous to those for ordinary power series with integral exponents.
Of course this is an informal definition: officially you decree that a Puiseux series is a map $\mathbb Q \to \mathbb C$ whose support (=set of rationals where the map is $\neq 0$) is both bounded below and in $\frac{1}{n} \mathbb Z$ for some integer $n\gt 0$ (depending on the series).
The key result is that the Puiseux series form an algebraically closed field.
Hence the set of Puiseux series algebraic over $\mathbb{C}(x)$ gives an algebraic closure $\overline{\mathbb{C}(x)}$ of $\mathbb C(x)$.
For example, the equations $y^3=x$ or $z^2=1+x$ have solutions $y=x^{1/3}$ and $z=\Sigma_{n\geq 0}\binom {1/2}{n}x^n$.
The exact same procedure describes an algebraic closure of $F(x)$ for $F$ an arbitrary algebraically closed field of characteristic zero.

Here is a link to the Wikipedia article on the subject.

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The Puiseux series do form an algebraically closed field. IIRC they form the algebraic closure of the subfield $K$ consisiting of the field of fractions of $\mathbf{C}[[x]]$. But surely not the algebraic closure of the subfield $\mathbf{C}(X)$? For example $e^x$ is a Puiseux series, but it is transcendental over $\mathbf{C}(X)$, hence doesn't belong to the algebraic closure of $\mathbf{C}(X)$ inside the field of Puiseux series. –  Jyrki Lahtonen Aug 26 '11 at 10:17
Yes, Jyrki, yours is a nice example showing that we really do have to take the algebraic Puiseux series to get an algebraic closure of $\mathbb C(t)$ . (The difficult result being that the Puiseux series form an algebraically closed field) –  Georges Elencwajg Aug 26 '11 at 11:17

It really depends on what counts as a description. e.g. for some purposes, "the algebraic closure of $\mathbb{C}(x)$" is already a rather good and easy description!

For explicit computation, it's often good enough to just construct extension fields of $\mathbb{C}(x)$ as you need them -- e.g. to start, you might decree that $\alpha$ satisfies $f(x, \alpha) = 0$, and then continue by working in the field $\mathbb{C}(x, \alpha)$ (which is, of course, isomorphic to $\mathbb{C}(x)[y] / f(x, y)$.

As a twist on this, there's probably some way to use Riemann surfaces (together with a local choice of branch) embedded in $\mathbb{C}^2$ to name elements of the algebraic closure, but it's not immediately obvious to me if it would work.

The field of complex Puiseaux series is not an algebraic closure -- it is the algebraic closure of $\mathbb{C}((x))$ and thus "too big" -- but it does contain an algebraic closure of $\mathbb{C}(x)$, so you can name things this way.

As an aside, writing things like $\sqrt[n]{x}$ is tricky, since the algebraic closure actually contains $n$ elements that can rightfully claim that name, and without choosing a specific representation of the set, it is impossible to specify which of those $n$ elements you mean. (Of course, you can get around this by simply decreeing that we choose one before-hand, and use $\sqrt[n]{x}$ to refer to it)

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Dear Hurkyl, writing $\sqrt[n]{x}$ is indeed tricky and should not be done. On the other hand, $x^{1/n}$ is a perfectly unambiguously defined Puiseux series, with the property $(\sqrt[n]{x})^n=x$ : this comes from the definition of Puiseux series and their ring structure. The set of $n$-th roots of the Puiseux series $x$ is equal to the $n$-element set of Puiseux series of the form $\mathbb \zeta x^{1/n}$ where $\zeta\in \mu_n(\mathbb C)=$ group of $n$-th roots of unity in $\mathbb C$. (to be continued) –  Georges Elencwajg Aug 28 '11 at 9:22
(continued) This is in complete analogy with the fact that $i$ is a perfectly well defined complex number, but if you want to describe the square roots of $-1$, you shouldn't write $\sqrt {-1}$, but say that they form the set $\{i,-i\}$. –  Georges Elencwajg Aug 28 '11 at 9:24