0
$\begingroup$

Let $H\leq G=\operatorname{Gal}(K/F)$ ($K/F$ is a finite galois extension), why is the following map well defined:

$\varphi:G/H\to\Gamma_F(K^H,K)$ defined by $\sigma H\mapsto\sigma|_{K^H}$ ,where $\Gamma_{F}(K^H,K)$ denotes all homomorphisms from $K^H$ to $K$ that fixes $F$.

My lecture wrote : Let $\sigma\in G$ ,if $\tau\in H$ then $\tau|_{K^H}=\operatorname{Id}_{K^H}$hence $\sigma\tau|_{K^H}=\sigma|_{K^H}$. Why does this imply (what I understand that need to be shown): $\sigma_1|_{K^H}=\sigma_2|_{K^H}\implies\sigma_2^{-1}\sigma_1\in H$ ?

Please note I was not told $H$ is a normal subgroup of $G$.

  • 0
    Sometimes, in some books. In others, it is reserved for the quotient group.2012-06-08

1 Answers 1

3

Note that $G/H$ represents the left cosets of $H$ in $G$.

Let $\sigma\in G$. Then for every $\tau\in H$ we have that the value of $\sigma$ and of $\sigma\tau$ on $K^H$, the field fixed by $H$, is the same. Thus, every element of $\sigma H$ determines the same map $K^H\to K$. This means that the map $G\to\mathrm{Hom}_F(K^H,K)$ given by restriction actually factors through the cosets.

Moreover, if $\sigma_1,\sigma_2\in G$ are such that $\sigma_1|_{K^H} = \sigma_2|_{K^H}$, then for every $a\in K^H$ we have $\sigma_1(a)=\sigma_2(a)$, hence $\sigma_2^{-1}\sigma_1(a) = a$. Thus, $\sigma_2^{-1}\sigma_1$ fixes every $a\in K^H$, hence lies in $H$ (since the extension is Galois, the stabilizer of $K^H$ is exactly $H$). Thus, $\sigma_1H = \sigma_2H$. Hence, the restriction map factors through the cosets but not through any larger subgroup.

  • 0
    @Belgi: Indeed I should. Fixed.2012-06-08