# Reference for the proof of this lemma. "Double group trick"

I am reading from "Topics in Galois Theory" by Serre.

I want to study the proof of this lemma. He has not provided the proof. Kindly mention some reference.

This follows from the Riemann-Hurwitz formula. Namely, if you have a finite degree $N$ map of curves $Y\rightarrow X$ over a number field of genus $g_Y,g_X$ respectively, then we have: $$2g_Y-2 = N(2g_X-2) + \sum_{P\in Y}(e_P-1)$$ where $e_P$ is the ramification index at $P$.

You can find this formula all over the place. In your case, $X = \mathbb{P}^1_k$ with function field $k(T)$. If $H$ is index 2 in $G$, then the fixed field of $H$ is the function field of a curve $Y$ of degree 2 over $X$, ramified at most above 3 points. Since the map is degree 2, the ramification indices are all either 1 or 2, and so you get: $$2g_Y - 2 = 2\cdot(-2) + (0\text{ or }1) + (0\text{ or }1) + (0\text{ or }1)$$ since $g_Y$ is a nonnegative integer, this shows that in fact $Y$ is actually only ramified over 2 points, and $g_Y = 0$.

Note that if you allow ramification over 4 points, then you can get a genus 1 degree 2 cover - namely, the quotient of an elliptic curve by its involution $[-1]$.

EDIT: If $K$ is a field, then a place of $K$ is by definition an equivalence class of absolute values on $K$, where two absolute values are equivalent if they define the same (metric) topology on $K$. An easy starting point is:

https://en.wikipedia.org/wiki/Algebraic_number_field#Places

If $X$ is a curve over a field $k$, then locally it is the Spec of a $k$-algebra $A$, e.g. $A = k[x]$, or $A = k[x,y]/(f)$...etc. A $k$-rational point is just a maximal ideal $m$ with residue field $k$. Given such a maximal ideal, the localization $A_m$ is a discrete valuation ring (wikipedia this), which naturally comes with a "valuation map" $v_m : A_m\rightarrow\mathbb{N}$, which extends to a valuation $$K := \text{Frac }A_m\stackrel{v}{\longrightarrow}\mathbb{Z}$$ The properties of this valuation ensure that if we define: $$|x| := 2^{-v_m(x)}\qquad\forall x\in K$$ then $|\cdot|$ is a (nonarchimedean) "absolute value" on $K$, where "2" is an arbitrary constant, and can be chosen to be any real number $> 1$ without affecting the equivalence class of the corresponding absolute value. Here, $K$ is precisely the function field of $X$, and hence every $k$-point $m$ defines a "$k$-rational place" of $K$, and different maximal ideals/points yield inequivalent places.

In a sentence, there is a bijection between $k$-rational points of a projective algebraic curve $X$ and $k$-rational places of its function field. Here, projective basically means "complete", in the sense that there are no points missing. If $X$ is an affine curve, you can always consider its "completion" by adding in the missing points. Note that however the function field of a projective curve is the same as the function field of any open subset, and hence there may be places of $\text{Frac}(A)$ which do not come from a maximal ideal of $A$. This is the connection between places and points.

• I think I don't have enough background to understand this. I am reading this book on my own. If you can clarify some points, it will be helpful for me. What does he mean by the following "places which are rational over $k$" and "..fixed field is rational". I have never see this terminology before. can you please explain . Nov 14, 2016 at 20:29
• I am 100% sure that your proof is correct, But I am not able to understand it completely. Can you mention some reference, from where I can read about this ramification thing. I also want to understand How we concluded that fixed field of $H$ is the function field of a curve $Y$ of degree $2$ over $X$. Nov 14, 2016 at 20:36
• @Sri Also, locally (in the complex topology) near any point $x$ in the domain, the map of curves in a small neighborhood of this point looks like the map $\mathbb{C}\rightarrow\mathbb{C}$ given by $z\mapsto z^n$ for some integer $n$. In this case we can define the ramification index to be $n$. Nov 14, 2016 at 21:04