Why is : $ \pi_1 ( X , \overline{x} ) = \mathrm{Gal} ( \overline{k} / k ) $?

Question

Let $ k $ be a field, $ X = \mathrm{Spec} (k) $, and $ \overline{x} = \mathrm{Spec} ( \overline{k} ) $ such that : $ \overline{k} $ is a separable closure of $ k $.

How to show that : $$ \pi_1 ( X , \overline{x} ) = \mathrm{Gal} ( \overline{k} / k ) $$ ?.

Thanks in advance for your help.

Answer 1

You need to recall the definition of $\pi_1$. We have

$$\pi_1(X,\overline{x}) := \varprojlim_{(Y,\overline{y})}\operatorname{Gal}(Y/X)$$

where $Y$ is a finite connected Galois cover of the scheme $X$, and $\overline{y}$ is a geometric point of $Y$ over $\overline{x}$. However because $Y$ is just the spectrum of a Galois extension, the group $\pi_1(X,\overline{x})$ is isomorphic to the inverse limit of all the Galois groups of the finite Galois extensions of $k$. You should be able to convince yourself that this is indeed $\operatorname{Gal}(\overline{k}/k)$.

Answer 2

Hint: the etale coverings of $Spec(k)$ are of the form $Spec(L)$ where is a finite products of separable extensions of $k$ and the Galois group of a separable extension is the group of covering morphisms, so $\pi_1(Spec(K))$ is the inverse limit of the Galois goups of the separable extensions.