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I am stuck on the following question:

Let $\mathcal{A}=\{c_0,c_1,F_+,F_\times,P_<\}$. Let $\mathcal{R}=(\mathbb{R},0,1,+,\cdot,<)$ be the structure of $\mathcal{L}_\mathcal{A}$ with universe $\mathbb{R}$ equipped with the standard addition, multiplication and ordering.

Show that if $\sigma : \mathcal{R} \to \mathcal{R}$ is an automorphism, then $\sigma$ is the identity function of $\mathbb{R}$.

Sincere thanks for any help!

Once again, I apologize for asking 3 Logic questions in a row. This is the last one.

yoyostein
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1 Answers1

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Hopefully you already know how to show that any automorphism of a field fixes everything in the prime subfield (which for $\Bbb{R}$ is the rational numbers). Then use the fact that your automorphism also respects the order and the fact that every real number is determined by which rationals are above and below it.

The harder version of this question, which you should attempt, is to show that the result still holds if we reduce the language to only have symbols for $+$ and $\cdot$.

Chris Eagle
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  • Is the harder question related to proving Aut($\mathbb{R}/\mathbb{Q}$)=1? – yoyostein Apr 25 '13 at 17:58
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    Both statements could be expressed in that way. The question is whether we're treating $\Bbb{R}$ and $\Bbb{Q}$ as ordered fields (the easy version) or as fields (the hard version). In general less structure means more possible automorphisms, so it takes more work to show that in fact there aren't any. – Chris Eagle Apr 25 '13 at 18:45