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How can we show that any automorphism of $\mathbb{Q}$ under addition is of the form
$x \to qx$ ,for some q in $\mathbb{Q}$.

edited- I found the same question was asked by @tattwamasi-amrutam tagged automorphism from $\mathbb{Q}$ to $\mathbb{Q}$.there is some useful answer.

spectraa
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  • @ Alex G. any automorphism must preserve the binary operation.how to show that x being sent to qx is the only map preserving the binary operation of Q under addition? – spectraa Aug 24 '14 at 02:45
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    Well, suppose that you somehow know this is the case. If I give you an automorphism $\varphi: \mathbb{Q} \to \mathbb{Q}$, how can you find out what $q$ is, corresponding to $\varphi$? – Alex G. Aug 24 '14 at 02:49
  • Traditionally, one shows that this is true for simply the integers then by using a rational form $p/q$, one can extend the results. Extending such results from the integers to the rationals is typically not too difficult. However it very often fails to extend to the reals in general. If you do not require the automorphism on $\Bbb R$ to be continuous, it does not need to be of this form. – Cameron Williams Aug 24 '14 at 03:11

1 Answers1

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Here is a way that you can follow: $f:\Bbb Q\to \Bbb Q$ an automorphim.

  • Show that $f(nx)=nf(x)$ for all $n\in \Bbb N$ and $x\in \Bbb Q$,
  • Show that $f(-x)=-f(x)$ for all $x\in \Bbb Q$ ,
  • Show that $f(nx)=nf(x)$ for all $n\in \Bbb Z$ and $x\in \Bbb Q$,
  • Let $r=\dfrac{p}{q}\in \Bbb Q$, with $(p,q)\in \Bbb N\times \Bbb Z^*$, show that $qf(r)=pf(1)$.
    Now your turn.
AnotherPerson
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Hamou
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