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I am tryingx to check whether the groups $\operatorname{Aut}_{\mathbb{Q}}(\overline{\mathbb{Q}})$ and $\operatorname{Aut}_{\mathbb{Q}}(\mathbb{R})$ are abelian or not. Can anyone help? Thanks!

Grigory M
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user120553
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  • Often people call $Aut_{\mathbb{Q}}\bar{\mathbb{Q}}$ "the Galois group". It is also often also written as $Gal(\bar{Q}/\mathbb{Q})$. Here is an MO question on the Galois group,http://mathoverflow.net/questions/2791/understanding-gal-bar-q-q . The Galois group is indeed a monster that is hard is slay! – Baby Dragon Jan 11 '14 at 07:24

2 Answers2

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The answer given by another user appears incorrect to me.

It is correct with respect to $\overline{\mathbf{Q}}$, but the only field automorphism of $\mathbf{R}$ is the identity. The difficulty is that $\mathbf{R}$ is not a normal extension of $\mathbf{Q}$, so automorphisms cannot necessarily be extended.

Any automorphism $\phi$ of $\mathbf{R}$ must be the identity on the prime field $\mathbf{Q}$, so it is immaterial whether we consider automorphisms over $\mathbf{Q}$ or not. But $\phi$ must also be increasing, for if $x \leq y$, we have $$\phi(x) \leq \phi(x) + \phi(\sqrt{y-x})^2 = \phi(y).$$

It follows easily that $\phi$ is the identity on $\mathbf{R}$.

user120557
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  • The only continuous field automorphism of $\mathbb{R}$ is the identity. The $\phi$ that you mention does not have to be the identity. For instance, pick a transcendental number (say $\pi$). Then take all of the real numbers that are real solutions (in mathbb{R}) to all polynomials in the field $\mathbb{Q}(\pi)$. Call this new field $\bar{\mathbb{Q}(\pi)}$. Now take a transcendental not in $\bar{\mathbb{Q}(\pi)}$, call it $\eta$. Now consider the field automorphism of $\mathbb{R}$ that swaps $\pi$ and $\eta$ and leaves everything else fixed. such automorphisms are wildly discontinuous. – Baby Dragon Jan 11 '14 at 07:21
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    @Baby Dragon I don't agree that such an automorphism exists. There is no continuity condition, as my proof shows. If you don't agree with the proof I gave, maybe you will trust others. The determination of all field automorphisms of $\mathbf{R}$ is a question that has been answered before: http://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map – user120557 Jan 11 '14 at 07:26
  • Ok I stand down ;) My general intuition on such things (which is wrong) is that $\mathbb{R}$ is large and $\mathbb{Q}$ is small, so then their must be many autmorphisms, all constructed with the axiom of choice. Indeed $Gal(\mathbb{C}/\bar{\mathbb{Q}})$ is similarly huge, so even though your statement is true, then if you take the elements of $Gal(\mathbb{C}/\bar{\mathbb{Q}})$ that fix a very small subset of elements, you get a trivial group. It looks like the quotient of some $\mathbb{Z}/2$ action away from triviality. This makes this simple theorem look quite strange to me! – Baby Dragon Jan 11 '14 at 07:38
  • Thanks. I noticed that the same argument did not apply for $\mathbb{R}$ so I looked at other posts. Any automorphism in this case must indeed be the identity. – user120553 Jan 11 '14 at 08:13
  • Thank you for your correction. Consider adding comments to the flawed answer, this will ensure the writer will get informed of his mistake. – Ewan Delanoy Jan 11 '14 at 08:14
  • I didn't have enough points to make comments. – user120557 Jan 11 '14 at 08:17
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Hint : consider $K$ such that $G=Gal(K/{\mathbb Q})$ is nonabelian (for example, isomorphic to $S_3$ ; you can take the decomposition field of $X^3-2$). Any $\mathbb Q$-automorphism in $G$ can be extended to a full automorphism of $\bar{\mathbb Q}$.

See the other answer for the case of $\mathbb R$.

Ewan Delanoy
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