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It is an amusing coincidence (at least it appears to be a coincidence to me) that any completion of the field $\mathbb{Q}$ has trivial automorphism group as an abstract field, i.e. when ignoring the topology.

For $\mathbb{R}$, this is obtained as follows: the condition $x\ge 0$ is equivalent to $\exists y:y^2=x$, hence the ordering on $\mathbb{R}$ is preserved by any field automorphism. Since such an automorphism is the identity on $\mathbb{Q}$, it must be the identity anywhere.

For $\mathbb{Q}_p$, one can look at this very short paper. The trick is as follows: $x\in\mathbb{Q}_p^{\times }$ is a unit in $\mathbb{Z}_p$ if and only if it has an $m$-th root in $\mathbb{Q}_p$ for all $m$ prime to $p(p-1)$. Hence $\mathbb{Z}_p^{\times }$ is set-wise preserved by any field automorphism of $\mathbb{Q}_p$, and it is easy to deduce that such an automorphism must be the identity.

Now if we go on to finite extensions, something dramatic happens in the archimedean case: since $\mathbb{C}$ is algebraically closed, its automorphism group is huge.

So, two questions:

(a) Is there any conceptual explanation why $\operatorname{Aut}_{fields }(K)$ is trivial for any completion $K$ of $\mathbb{Q}$?

(b) What happens for finite extensions of $\mathbb{Q}_p$?

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It seems to me that the same argument as you gives that all automorphisms for finite extensions of $\mathbb Q_p$ are continuous in the $p$-adic topology (and hence are the identity on $\mathbb Q_p$. –  Torsten Ekedahl Nov 12 '11 at 14:43
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I guess the analogue of your "something dramatic" is that $\mathbb{C}_p$ -- the completion of an algebraic closure of $\mathbb{Q}_p$ -- has many automorphisms that are not continuous. –  David Loeffler Nov 12 '11 at 15:43
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Torsten is right about (b). On the other hand, for some infinite algebraic extensions there are automorphisms that are not continuous (i.e. that do not preserve $\mathbb Q_p$ setwise). I don't know how general this is. Concerning (a), note that any such explanation might reasonably be expected to apply also to all totally real finite extensions of $\mathbb Q$, but had better not apply to the others. –  Tom Goodwillie Nov 12 '11 at 16:04
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The automorphisms are not just continuous, but isometries. In particular, the argument goes through with the automorphism assumed only to be a homomorphism: surjectivity is never used and is a consequence. That is true for the argument with the real numbers too. –  KConrad Nov 12 '11 at 17:09
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See also mathoverflow.net/questions/22897/… –  François Brunault Nov 12 '11 at 21:16
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