Let $K$ be a field and $L$ an algebraic field extension of $K$ contained in an algebraic closure $\overline{K}$ of $K$. Let $\sigma$ be an automorphism of $L$ fixing $K$. I read in numerous places that there is a natural extension of $\sigma$ to the an automorphism $\tilde{\sigma}$ of $\overline{K}$. But I do not have a proof of this statement. Can somebody explain what is this natural extension or suggest a reference.

I am interested in whether the following commutativity property is true: given two automorphism $\sigma, \tau$ of $L$, fixing $K$, does the extensions $\tilde{\sigma}$ and $\tilde{\tau}$ satisfy the property that $\tilde{\sigma}\tilde{\tau}=\widetilde{\sigma\tau}$?

  • $\begingroup$ Can you cite some of the places where you read this? Context helps a lot in questions like this. $\endgroup$ – Nefertiti Aug 24 '17 at 8:31
  • $\begingroup$ @Nefertiti Do you think this is false? If I understand correctly, it is implicit in the answer math.stackexchange.com/questions/219869/… There are other references. $\endgroup$ – user43198 Aug 24 '17 at 9:12
  • $\begingroup$ No, I am not saying it is false --- just looking for context, as I said. $\endgroup$ – Nefertiti Aug 24 '17 at 9:38
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    $\begingroup$ @user43198 Read carefully the linked answer. There is no mention of uniqueness when it comes to extend $\tau$ from $F(S)$ to $A$. In general, for $L/K$ Galois the map $\operatorname{Gal}(\overline{K}/K)\rightarrow\operatorname{Gal}(L/K), \sigma\mapsto\sigma_{|L}$ is onto, but there is no natural choice of extension. It is also possible that there is no section at all, so that the equality $\tilde{\sigma}\tilde{\tau}=\tilde{\sigma\tau}$ will always fail for some $\sigma$ and $\tau$. $\endgroup$ – Roland Aug 24 '17 at 9:58
  • $\begingroup$ The argument for the existence of an extension is a typical application of Zorn's lemma. You can find it in many textbooks. I learned it from Jacobson's Basic Algebra I-II. A building block is that if you have already extended it to a field $F$, and $a\in \overline{K}\setminus F$, then the theory of finite extensions lets you extend it to $F(a)$ (or its normal closure over $F$ inside $\overline{K}$). So there is nothing that can stop you from extending it all the way! Except possibly if you work with a set theory that doesn't have the axiom of choice (when Zorn's lemma is not available). $\endgroup$ – Jyrki Lahtonen Aug 25 '17 at 6:41

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