# Correspondence between coverings and field extensions

I am self reading from Groups as Galois Group by Helmut Volklein There is a result on page 94(section 5.4)

Let $G$ be a finite group. Let $P\subset P^{1}$ finite and $q\in P^{1}\P$. There is a natural $1-1$ correspondence between

1. The $\mathbb{C}(x)$-isomorphism classes of Galois extensions $L/\mathbb{C}(x)$ with Galois group isomorphic to $G$ and with branch points contained in $P$.
2. The equivalence class of Galois coverings $f:R\rightarrow P^{1}\P$ with deck transformation group isomorphic to $G$.
3. The normal subgroups of the fundamental group $\pi_{1}(P^{1}\p,q)$ with quotient isomorphic to $G$.

I am interested in Inverse Galois problem over $\mathbb{Q}(T)$ as then by Hilbert's Irreducibility theorem I can study IGP over $\mathbb{Q}$

Is there similar theorem(result) that states the correspondence between field extensions of $\mathbb{Q}(T)$ and coverings of $P^{1}_{Q}$.

Edits

Why I think there is such result

See the attached image [ Topics in Galois Theory, by Jean-Pierre Serre]

He also mentions the same idea, He has written the statement but I want to convince myself (mathematically by a proof) that it is actually true.

## 1 Answer

Chapter 2 Remark 2.5 of Silvermans' The Arithmetic of Elliptic Curves book gives you a bijection between covers of $\mathbb{P}^1_{\mathbb{Q}}$ by curves over $\mathbb{Q}$ and regular field extensions $K$ over $\mathbb{Q}(T)$ - that is, field extensions such that $\mathbb{Q}\cap K = \mathbb{Q}$.

The problem is that producing such covers is generally a difficult problem. Over $\mathbb{C}$ (or more generally $\overline{\mathbb{Q}}$), one can easily produce such covers topologically using quotients $\pi_1(\mathbb{P}^1_{\mathbb{C}}-S)$ for some varying finite sets of points $S$, then use Riemann's existence theorem to make it algebraic (but still over $\mathbb{C}$). However, to do this over $\mathbb{Q}$, one would need to analyze the finite quotients of $\pi_1(\mathbb{P}^1_{\mathbb{Q}}-D)$ where $D$ is now a $\mathbb{Q}$-rational effective divisor on $\mathbb{P}^1_{\mathbb{Q}}$, and this fundamental group is much more difficult to understand, at least when $D$ contains at least three geometric points. To see this, it's well known that there is a 'homotopy exact sequence' $$1\rightarrow\pi_1(\mathbb{P}^1_{\overline{\mathbb{Q}}} - D)\rightarrow\pi_1(\mathbb{P}^1_{\mathbb{Q}}-D)\rightarrow\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow 1$$ If $D$ contains $n$ geometric points, then the first term is just the profinite completion of the free group $F_{n-1}$ of rank $n-1$, but to produce covers of $\mathbb{P}^1_{\mathbb{Q}}$, one needs to understand the finite quotients of the middle term, which as you can see contains all the complexity of the absolute Galois group of $\mathbb{Q}$. This isn't so bad when $n = 1,2$, but once $n\ge 3$, producing (nonabelian) covers of $\mathbb{P}^1_{\mathbb{Q}}$ through this method becomes very difficult in general. Most of the progress on the IGP uses more ingenious but indirect methods.

• +1 Thanks for the clear and detailed answer. Feb 28, 2017 at 5:46