Exercise 1.6.21 ($\varphi_k: q \mapsto kq$ is an automorphism of $\mathbb{Q}$)

Question

Prove that for each fixed nonzero $k \in \mathbb{Q}$ the map from $\mathbb{Q}$ to itself defined by $a \mapsto kq$ is an automorphism of $\mathbb{Q}$ (cf. Exercise 20).

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

\begin{proof} 
Consider for every $ k \in \mathbb{Q}^*$ the map
$$
\varphi_k
\left\{
\begin{array}{ccl}
\Q & 	o &\Q\
q & \mapsto & kq.
\end{array}
ight.
$$
Then for all $q \in \Q$,
$$(\varphi_{k^{-1}} \circ \varphi_k)(q) = k^{-1}(k q) = q,$$
thus $\varphi_{k^{-1}} \circ \varphi_k = \mathrm{Id}_{\Q}$, and similarly $ \varphi_k  \circ  \varphi_{k^{-1} }= \mathrm{Id}_{\Q}$. This shows that $\varphi_k$ is bijective, and $\varphi_k ^{-1} = \varphi_{k^{-1}}$.

\bigskip
Moreover, for all $q,\, l \in \Q$,
$$\varphi_k(q + l) = k (q+l) = kq + kl = \varphi_k(q )+\varphi_k(l),$$
therefore $\varphi_k \in \mathrm{Aut}(\Q)$.
\end{proof}

Exercise 1.6.21 ($\varphi_k: q \mapsto kq$ is an automorphism of $\mathbb{Q}$)

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Exercise 1.6.21 ($\varphi_k: q \mapsto kq$ is an automorphism of $\mathbb{Q}$)

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