I am studying etale fundamental groups, and I am confused with the basic example: $\pi_1(\text{Spec}\ K)$ for a field $K$. I tried to read several explanations of this example, but none of them provided a very detailed answer to this question.

Choose a geometric point $\overline{x}: \text{Spec }\overline{K}\rightarrow \text{Spec }K$, or equivalently an embedding $K\hookrightarrow \overline{K}$. The "Galois coverings" of the scheme $\text{Spec }K$ together with the geometric point $\overline{x}$ are just the morphisms $\text{Spec }L\rightarrow \text{Spec }K$, where $L$ is a finite Galois extension of $K$. The fiber of this covering over $\overline{x}$ is the set of geometric points $\overline{y}: \text{Spec }\overline{K}\rightarrow \text{Spec }L$ which lies over $\overline{x}$, or equivalently, the field embeddings $L\hookrightarrow \overline{K}$ which extends the embedding $K\hookrightarrow \overline{K}$ corresponding to $\overline{x}$. Once we fix an embedding $K\hookrightarrow \overline{K}$, the image in $\overline{K}$ of the finite Galois extension $L$ is also fixed regardless of the choice of the embedding. Hence, the fiber $Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$ is in one-to-one correspondence with the Galois group $Gal(L/K)$. The etale fundamental group $\pi_1(\text{Spec }K, \overline{x})$ is, by definition, the automorphism group of the fiber functor $Fib_\overline{x}$ from the category of (finite) etale coverings of $\text{Spec }K$ to the category of sets.

I know that, by definition, an automorphism of $Fib_\overline{x}$ is a collection of automorphisms (in the category of sets; so just bijections) of $Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$ which is compatible with the morphisms $Fib_\overline{x}(\text{Spec }L_1 \rightarrow \text{Spec }L_2)$ coming from each pair of finite Galois extensions $(L_1, L_2)$ of $K$ and a field homomorphism $L_2\rightarrow L_1$ between them. This morphism $Fib_\overline{x}(\text{Spec }L_1 \rightarrow \text{Spec }L_2)$ maps $\overline{y}:\text{Spec }\overline{K}\rightarrow \text{Spec }L_1$ to $\overline{y}': \text{Spec }\overline{K}\rightarrow \text{Spec }L_1\rightarrow \text{Spec }L_2$. Hence, an automorphism $\alpha$ maps each embedding $\iota: L\hookrightarrow \overline{K}$ to another embedding $\alpha_L(\iota):L\hookrightarrow \overline{K}$, and if $\varphi:L'\rightarrow L$ is an embedding of another finite Galois extension $L'$ into $L$, then $\alpha_{L'}(\iota\circ \varphi) = \alpha_{L}(\iota)\circ \varphi$.

Now I am stuck here: obviously the elements of the absolute Galois group $Gal(K^{sep}/K)$ give such automorphisms, but conversely how do I show that this automorphism $\alpha$ is actually coming from an element of $Gal(K^{sep}/K)$? In other words, why is the map $\alpha_L:Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)\rightarrow Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$ not just a bijection of sets (i.e. an element of the symmetric group $S_{[L:K]}$) but actually a composition with an element of $Gal(L/K)$?

Actually, while writing this question, I could think a possible strategy: focus on the actual subfields $L$ of $\overline{K}$ (viewing $K$ as a subfield of $\overline{K}$), and use the identification of $Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$ with $Gal(L/K)$ as above, to assign to each $\alpha\in Aut(Fib_\overline{x})$ an element of $Gal(K^{sep}/K)$. So we can identify the inclusion $\iota_0:L\hookrightarrow \overline{K}$ with the identity element of $Gal(L/K)$. Then $\alpha_L(\iota_0)$ is another element of $Gal(L/K)$. If $\varphi_0: L'\hookrightarrow L$ is the inclusion, then $\alpha_{L'}(\iota_0\circ\varphi_0) = \alpha_L(\iota_0)\circ\varphi_0$ is the restriction of $\alpha_L(\iota_0)\in Gal(L/K)$ to $L'$ (as $L'$ is Galois over $K$ so $\alpha_L(\iota_0)(L')=L'$). So $\alpha$ at each inclusion of the subfields of $\overline{K}$ which are finite Galois over $K$ defines a collection of elements of the Galois groups $Gal(L/K)$ for such extensions $L$, and this elements are compatible with restrictions. Therefore they can be glued to define an element of $Aut(\overline{K}/K) =Gal(K^{sep}/K)$.

Does this argument actually work? Also, is everything I wrote correct?



You must log in to answer this question.