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$\mathbb{R}(x)$ 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?

mweiss
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  • Possibly related: https://math.stackexchange.com/questions/2609504, https://math.stackexchange.com/questions/318690 – Watson Nov 20 '18 at 08:45

2 Answers2

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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 $x \to x^2$.

hunter
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    Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $\mathbb{C}(x)$ exists without the axiom of choice. – Eric Wofsey Nov 20 '18 at 04:30
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    Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $\mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though. – Eric Wofsey Nov 20 '18 at 04:45
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    I guess geometrically this corresponds to considering only curves over $\mathbb{P}^1$ which are obtained by first taking the $n$th power map $\mathbb{P}^1\to\mathbb{P}^1$ and then taking a branched cover of $\mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$. – Eric Wofsey Nov 20 '18 at 05:20
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    @EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $\mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it. – Tobias Kildetoft Nov 20 '18 at 07:07
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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 $f \in \overline{\mathbb{C}(z)}$.

reuns
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  • There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $\mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once. – Eric Wofsey Nov 20 '18 at 07:35
  • As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$. – Eric Wofsey Nov 20 '18 at 07:51
  • As an example of the fallacy of just blindly writing things like $f(z) \in \overline{\mathbb{C}(z)}$ without first defining $\overline{\mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other. – Eric Wofsey Nov 20 '18 at 07:56
  • @EricWofsey Hi, I tried to repair the sloppy parts – reuns Nov 21 '18 at 12:38
  • What is $\overline{\mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $\in \mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $\overline{\mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions. – reuns Nov 21 '18 at 14:35
  • Um, OK, you can make that definition of $\overline{\mathbb{C}(z)}$ if you want, but the question was asking about an actual ordinary algebraic closure. It's not clear how to extract an actual algebraic closure from your definition (nothing in what you wrote requires a root of any polynomial in $\mathbb{C}(z)[y]$ to exist). And if you add some sort of "algebraically closed" condition to your definition, then it is not at all obvious that any such gadget actually exists. – Eric Wofsey Nov 21 '18 at 15:03
  • @EricWofsey Otherwise an alternative to the idea of Puiseux series would be to identify each finite extension $L/\mathbb{C}(z)$ with a field of meromorphic functions on $\Re(z) \in (0,T_L)$. – reuns Nov 21 '18 at 16:08
  • @EricWofsey For the existence, with $U$ a simply connected open where the $u_j$ have no poles and $u_0$ has no zeros I find $ \sum_{j=0}^J u_j(z) f(z)^j=0$ $ \implies \sum_{j=0}^J u_j'(z) f(z)^j + f'(z)\sum_{j=1}^J u_j(z) f(z)^{j-1} =0$ $ \implies f'(z) =\frac{-\sum_{j=0}^J u_j'(z) f(z)^j}{\frac{-u_0(z)}{f(z)}}$ so $f$ is holomorphic on $U$. – reuns Nov 21 '18 at 17:22