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We know that $(\mathbb{Q},+,\times)$ is a subfield of $(\mathbb{R},+,\times)$. It is easy to see that the automorphism of $\mathbb{Q}$ is only the identity. For a quick proof, lets go through the main steps:

  1. $f(1)=1 \Longrightarrow f(n)=n$ for all $n \in \mathbb{N}$.
  2. $f(-1)=-1$ which says that $f(x)=x$ for all $x \in \mathbb{Z}$.

  3. $\displaystyle f \Bigl(\frac{p}{q}\Bigr) = \frac{p}{q}$, where $q \neq 0$.

One, then uses the continuity of $f$ and the denseness of $\mathbb{Q}$ to prove that the Automorphism of $\mathbb{R}$ is also trivial.

My Question Given a subfield $K$ of $\mathbb{C}$ is it and an automorphism of $K$ can it be extended to the whole of $\mathbb{C}$.

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3 Answers 3

up vote 8 down vote accepted

Firstly, you need a little more to show that the only automorphism of $\mathbb R$ is the identity. You have to prove that any such automorphism is continuous. For a proof, let $x$ be a positive number; so it is the square of some number; so it is taken to another positive number under the automorphism. Since positive numbers go to positive numbers, the automorphism is order-preserving, and hence the identity.

Secondly, this result is not true if you replace $\mathbb R$ by $\mathbb C$. The automorphism group of $\mathbb C$ over $\mathbb Q$ is uncountable. This is because $\mathbb C$ can be constructed by attaching uncountably many algebraicaly-independent transcendentals to $\mathbb Q$ and then taking the algebraic closure. Any automorphism of $\mathbb Q$ can be extended to an automorphism of $\mathbb C$, by first extending to the algebraic closure, and then to $\mathbb C$. The same argument would work for any algebraic number field.

More general subfields are obtained as follows: You attach some number of transcendentals to $\mathbb Q$, and then take an algebraic extension of it. An automorphism of this field can be similarly extended to the whole of $\mathbb C$ as above.

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@George S: I know it. Its just i didnt want to type it out. – anonymous Aug 12 '10 at 13:38
@George S: $\mathbb{R}$ is subfield of $\mathbb{C}$ and take the trivial automorphism. We can extend it so that it remains a Automorphism on $\mathbb{Q}$ as well. – anonymous Aug 12 '10 at 13:42

This and related questions about automorphism groups of algebraically closed fields (a topic I find interesting and have spent some time thinking about) are discussed in Section 9.1 of

Specifically, Theorem 77 answers the OP's question affirmatively.

(These notes are still very rough. In particular there is not yet a bibliography. When this gets remedied, a citation to Paul Yale's paper will be in order: it was definitely something I read when writing these notes.)

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L.Clark: Thank you sir! – anonymous Aug 12 '10 at 19:58

First of all thanks a lot to George S. Secondly I recently found a paper entitled, which discusses something on Automorphisms of Complex Numbers.

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