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.
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 $S_V$.