Here, in page $1$, the absolute Galois group is defined by $$G_{\mathbb{Q}}:=\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})=\{\sigma: \bar{\mathbb{Q} }\to \bar{\mathbb{Q}}, \ \text{field automorphism} \}$$ is a profinite group. Then the article defines for any Galois extension $K$ of $\mathbb{Q}$, the Galois group by $$\text{Gal}(K/\mathbb{Q}) \cong G_{\mathbb{Q}}/{\{\sigma \in G_{\mathbb{Q}}: \ \sigma|_K=id_K \}}$$ to be the quotient group.
My question- Why is $\text{Gal}(K/\mathbb{Q}) \cong G_{\mathbb{Q}}/{\{\sigma \in G_{\mathbb{Q}}: \ \sigma|_K=id_K \}}$ ?
Because we know by definition of Galois extension $\text{Gal}(K/\mathbb{Q}) = \{\sigma \in \text{Aut}(K): \ \sigma(a)=a, \ \forall a \in \mathbb{Q} \}$.
So the question-
How to see the relation $ \{\sigma \in \text{Aut}(K): \ \sigma(a)=a, \ \forall a \in \mathbb{Q} \} \cong G_{\mathbb{Q}}/{\{\sigma \in G_{\mathbb{Q}}: \ \sigma|_K=id_K \}}$ ?
How to see the isomorphism ?
Best Answer
There is a map $$ G_\mathbb{Q} \to \text{Gal}(K/\mathbb{Q}) $$ given by restriction. It is plain that the kernel of this map is $$ \{ \sigma \in G_\mathbb{Q} : \sigma|_K = \text{id}_K\}, $$ and so you need to know that this map is surjective, i.e. that every automorphism of $K$ lifts to an automorphism of the algebraic closure. This is one of the more difficult lemmas of Galois theory; see a proof here for instance: Extension of field automorphism to automorphism of algebraic closure