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I am reading from Topics in Galois Theory by Serre. I have the following Question:

Let us say that $G$ has property $Gal_{T}$ if there is a regular $G$-covering $C\longrightarrow P^{1}$ defined over $\mathbb{Q}$.

Now there is proposition, which says that.

Proposition :Let $A$ be a finite abelian group. There exists a torus $S$ over $\mathbb{Q}$, and an embedding of $A$ in $S(\mathbb{Q})$, such that the quotient $S^{'} = S/A$ is a permutation torus.(In particular $S^{'}$ is a $\mathbb{Q}$-rational variety.)

According to the author the above proposition implies that $A$ has property $Gal_{T}$.

I did not had any previous knowledge about algebraic groups, torus, isogeny,etc. I read the definitions first and then tried to understand how the above proposition implies that abelian groups has property $Gal_{T}$ But still I am not able to understand.

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  • $\begingroup$ books.google.co.in/…. Here is the link. Section 4.2, Proposition 4.2.1 $\endgroup$ – Tensor_Product Oct 18 '16 at 4:19
  • $\begingroup$ You need to be way more specific about exactly what you don't understand. $\endgroup$ – oxeimon Oct 21 '16 at 4:34
  • $\begingroup$ @nesos I don't understand how the proposition stated above implies that Abelian groups have property $Gal_{T}$. $\endgroup$ – Tensor_Product Oct 21 '16 at 6:02
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Let $p : S\rightarrow S/A = S'$ be the quotient map. The fact that $A$ is embedded in $S(\mathbb{Q})$ means that $A$ acts freely on $S$, the action is defined over $\mathbb{Q}$, and hence $A = Gal(p)$.

The fact that $S'$ is a $\mathbb{Q}$-rational variety means that it's birational to $\mathbb{P}^n_{\mathbb{Q}}$ for some $n$, so there is an open subscheme $U\subset S'$ which is isomorphic to an open subscheme of $\mathbb{P}^n$. By restricting onto some 1-dimensional subscheme, you obtain a subscheme $V\subset S'$ such that $V$ is isomorphic to some open subscheme of $\mathbb{P}^1$. At this point you can already apply Hilbert Irreducibility to deduce that there is a $\mathbb{Q}$-rational point of $V$ with connected fiber (ie, the fiber is a $A$-Galois cover of $\mathbb{Q}$).

But, if you really want to get a $A$-Galois cover of $\mathbb{P}^1$, then let $S_V$ be the preimage of $S$ over $V$, then $S_V\rightarrow V$ is finite etale and Galois with Galois group $A$.

To get a $A$-Galois cover of $\mathbb{P}^1$, simply embed $V$ in $\mathbb{P}^1$, and take the normalization of $\mathbb{P}^1$ inside $S_V$.

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