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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 7 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

http://math.uga.edu/~pete/FieldTheory.pdf

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.

Here is the link for that paper: mathdl.maa.org/images/upload_library/22/Ford/PaulBYale.pdf

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