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Let $\phi: \mathbb{R}\rightarrow \mathbb{R}$ be an automorphism. Suppose $p=\frac{m}{n}$ is a rational number. Then is it true that $\phi(p)=\frac{\phi(m)}{\phi(n)}$?

I got this problem while doing an algebra problem. In the original problem it has been asked to prove that $\phi(\mathbb{Q})=\mathbb{Q}$.

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    $\begingroup$ Note that $\phi(m)=\phi(np)=\phi(n)\phi(p)$. $\endgroup$
    – Pedro ♦
    Dec 25, 2014 at 22:25
  • $\begingroup$ As an aside, it turns out that the only ring endomorphism of $\mathbb{R}$ is the identity automorphism! $\endgroup$
    – user14972
    Dec 25, 2014 at 22:26
  • $\begingroup$ But there are fields containing $\mathbb Q$ and contained in $\mathbb R$ which have lots and lots of automorphisms :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Dec 25, 2014 at 22:27
  • $\begingroup$ If $\phi$ is the identity map on $\mathbb{Q}$ and $\phi$ maps positive to positive, does it imply that $\phi$ is the identity map? Then how? $\endgroup$
    – Extremal
    Dec 25, 2014 at 23:14

2 Answers 2

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Hint:

Write down $\phi (p)=\phi(\frac m n)=\phi (m\cdot n^{-1})$

Now, since $\phi$ is an automorphism, in particular it is a homomorphism. What homomorphism properties can you use?

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Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ be an automorphism. So $\phi (1) = 1.$ Thus, for each $n \in \mathbb{N}, \phi (n) = \phi(1 + 1 + \cdots + 1) = \phi (1) + \phi (1) + \cdots + \phi(1) = n \phi(1) = n.$ Also $\phi(-n) = - \phi(n) = -n.$ Hence $\phi(n) = n, \forall n \in \mathbb{Z}.$

Now let $\dfrac{m}{n} \in \mathbb{Q}.$ So $\phi (m) = m\phi (1) = m \Rightarrow \phi(n. \dfrac{m}{n}) = m \Rightarrow n\phi(\dfrac{m}{n}) = m \Rightarrow \phi(\dfrac{m}{n}) = \dfrac{m}{n}.$

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