0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\sigma|_L \in Aut(L)$. $\endgroup$ – reuns Oct 17 '17 at 0:48
0
$\begingroup$

Reuns addresses your second point concisely and accurately so let me focus on the inverse limit part.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.