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.

  • $\begingroup$ 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

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 $S_V$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.