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Let $A \subset \mathbb{C}$ be a subfield of $\mathbb{C}$.

According to a result of P.B. Yale, Theorem 7, any automorphism of $A$ can be extended to an automorphism of $\mathbb{C}$; see also this question (and its good comments and answers).

My question: If we know, in addition, that a given automorphism of $A$ is of finite order (for example, of order $2$, namely an involution), is it possible to know that its extension to $\mathbb{C}$ is of the same order (or at least also of finite order)? Or is it hopeless, and the best we can obtain is 'just' that it has an extension to $\mathbb{C}$, probably of infinite order?

In other words, can we adjust Zorn's Lemma to guarantee a finite order extension? (I am afraid that the answer is no.. I hope I am wrong).

Thanks for any comments.

Community
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user237522
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1 Answers1

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It is usually not possible to extend a finite order automorphism of $A$ to an automorphism of $\mathbb{C}$ of finite order. More precisely, it follows from the Artin-Schreier theorem that any nontrivial finite order automorphism of $\mathbb{C}$ has order $2$, maps $i$ to $-i$, and has a fixed field which is real closed. So if an automorphism of $A$ has order greater than $2$, it cannot have any finite order extension to $\mathbb{C}$. Even if an automorphism of $A$ has order $2$, it may not extend to a finite order automorphism of $\mathbb{C}$ (for instance, if $i\in A$ and the automorphism fixes $i$).

Eric Wofsey
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  • Please, can one tell when there is a chance to extend an involution $g$ of $A$ to an automorphism of $\mathbb{C}$ of finite order $m$?. If $i \in A$, is $g(i)^2=-1$ the only necessary condition? I guess there may exist additional necessary conditions, is it possible to find them? Perhaps this depends on more information we should have about $A$? (And what if $i \notin A$?). – user237522 Apr 01 '17 at 23:03
  • I don't know a simple answer to that off the top of my head, but sending $i$ to $-i$ is not sufficient. I suggest asking that as a new question. – Eric Wofsey Apr 02 '17 at 00:56
  • Thank you very much for your answer and comment. (I will probably ask a new question, as you suggested). – user237522 Apr 02 '17 at 01:47