Showing $\operatorname{Aut}(\mathbb{R})$ is abelian

Question

I have an exercise (not for a class) that asks whether $\operatorname{Aut}(\mathbb{R})$ (field automorphisms) is abelian. I know that $\operatorname{Aut}(\mathbb{R})$ is just the trivial group, but is there a nice way to see that it is abelian without knowing the group?

This is Part (c) of a question and Parts (a) and (b) were that $\operatorname{Aut}(\overline{\mathbb{Q}})$ is infinite and nonabelian.

You mean Aut(R) is trivial ? in the second line you wrote R is the trivial group.. — Belgi, Oct 16 '14 at 15:04
possible duplicate of Is an automorphism of the field of real numbers the identity map? — Dietrich Burde, Oct 16 '14 at 15:04
@DietrichBurde This is not a duplicate. I specifically want to answer this question without demonstrating the group is trivial (as stated in the question). — Jacob Bond, Oct 16 '14 at 15:09
I see. But I think the best way to see that the group is abelian is given in the answers there. — Dietrich Burde, Oct 16 '14 at 15:14

Mic.R · Answer 1 · 2014-10-16T16:47:45.570

-1

$\mathbb R$ is a $\mathbb Q$-vector space, so every $\mathbb Q$-linear function is in $Aut(\mathbb R)$ and $End_{\mathbb Q}(\mathbb R)$ isn't abelian, so $Aut(\mathbb R)$ isn't abelian.

If you mean $Aut(\mathbb R)$ like the group of ring automorphism of $\mathbb R$, the only one is the identity.

edited Oct 16 '14 at 16:47

answered Oct 16 '14 at 16:41

Mic.R

195

2

The OP wants field automorphisms. – lhf Oct 16 '14 at 16:42

Showing $\operatorname{Aut}(\mathbb{R})$ is abelian

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