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Whilst reading P.M. Cohn's "Skew Field Constructions" (LMS LNS 27; Cambridge 1977), I found myself unable to follow the author's argument on p. 69 (proof of Prop. 3.5.4, specifically lines 15-16). Most of my troubles would disappear if I could prove the following: given an automorphism $\sigma$ of a (commutative) field $K$ of order $n$ and a primitive $n$-th root of unity $\omega\in K$ (i.e. an element of order $n$ in the multiplicative group of $K$), where $1<n\in\mathbb N$, one necessarily has $\sigma(\omega)=\omega$. Now I'm not even sure whether this statement is true, though a look at Ex. § 11, 4) c), Chapter V of Bourbaki's Algebra II makes me hopeful. Any help, one way or the other, would be greatly appreciated !

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    $\begingroup$ Unless I've misunderstood something, I have an easy counter-example. For $K = \mathbb{C}$, let $\sigma(z) = \overline{z}$. Certainly $\sigma^4 = \mathrm{id}$, and $i = \sqrt{-1}$ is a primitive $4^{th}$ root of unity. However, $\sigma(i) = -i \neq i$. $\endgroup$ – Shaun Ault Sep 5 '12 at 11:19
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    $\begingroup$ If you could provide a scan of p.69, perhaps the problem can be spotted. Is $\sigma$ supposed to be order $n$ (which is stronger than saying $\sigma^n = \mathrm{id}$)? $\endgroup$ – Shaun Ault Sep 5 '12 at 11:23
  • $\begingroup$ Sorry about that Shaun, I've corrected the question; naturally $\sigma$ should have order $n$. (A scan of the page in question wouldn't really be of any help, because my question doesn't appear there as such - an answer, however, would help in making the remaining arguments comprehensible. The reference was given simply to provide context.) $\endgroup$ – bonnbaki Sep 5 '12 at 12:30
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I think this is a counterexample. Let $K={\bf Q}(\zeta)$, where $\zeta$ is a primitive 20th root of 1. Define $\sigma$ by $\sigma(\zeta)=\zeta^3$. Then $\sigma$ is of order 4, and $\sigma(i)=-i$.

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  • $\begingroup$ Thanks very much Gerry - looks good to me ! (And unfortunately means that I still don't understand Cohn's argument in his book - but that is another question ...) - Kind regards ! $\endgroup$ – bonnbaki Sep 5 '12 at 16:35

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