# Automorphism group of separable closure

Let $k$ be a field and $k_s$ its separable closure. I would like to understand why $\mathrm{Aut}_k(k_s)$ is an inverse limit of the groups $\mathrm{Gal}(L/k)$, where $L$ is a finite Galois extension of $k$.

The proof says that every automorphism of $k_s$ fix each finite Galois extension of $k$. Why is this true? How do I conclude the result then?

Thanks!

• This is really the definitions. If $\sigma \in Gal(k_s/k)$ then $\sigma(k(\alpha)) = k(\sigma(\alpha))$ where $\alpha,\sigma(\alpha)$ have the same minimal polynomial $f \in k[x]$. If $\alpha \in L$ and $L/k$ is Galois then $f$ splits completely in $L$, and since $f$ has at most $\deg(f)$ roots in the field $k_s$, those are all in $L$ and $\sigma(\alpha) \in L$ ie. $\sigma|_L \in Aut(L)$. – reuns Oct 17 '17 at 0:48

The point is that an element $\sigma \in \mathrm{Aut}_k(k_s)$ is determined by what it does to elements $\alpha\in k_s$. Now you know that each such $\alpha$ is separable and algebraic so that the extension $k(\alpha)$ is a finite separable extension. So the element $\sigma$ is really determined by it's restriction to all the finite separable extensions $k(\alpha)/k$ and these have to be compatible, i.e. if $k\subseteq K\subseteq L$, then $(\sigma|_{L})|_{K}$ needs to be the same as $\sigma|_{K}$. Conversely any such compatible system of $\{\sigma_K\}$ with $\sigma_K \in \mathrm{Aut}_k(K)$ for $K$ ranging over the finite separable extensions of $k$, will together determine an element in $\mathrm{Aut}_k(k_s)$ (This is just an application of Zorn's lemma)
If you now see the definition of inverse limit, you will see that we have proven that $$\mathrm{Aut}_k(k_s) = \varprojlim_{K/k \text{ fin } \text{sep}} \mathrm{Aut}_k(K)$$ where the limit is taken over the directed system of finite separable extensions of $k$ ordered by inclusion.