Permutations of Hamel bases of $\mathbb{R}$ over $\mathbb{Q}$ Consider $\mathbb{R}$ with the usual addition and multiplication operations as a vector space over $\mathbb{Q}$. Do there exist a Hamel basis $\{ v_i \}_{i \in I}$ of $\mathbb{R}$ and a permutation $\phi:I \rightarrow I$ different from the identity map, such that $v_{\phi(i)}=cv_i$ for some constant $c \in \mathbb{R}$ and all $ i \in I$?
Thank you very much in advance for your help.
 A: Suppose you have such a basis $\{v_i\}$, permutation $\phi$, and $c\in\mathbb{R}$.  It is clear that every orbit of $\phi$ must be infinite, so $c$ is transcendental (since otherwise $c$ would have to be $-1$, but then $v_i$ is not linearly independent from $v_{\phi(i)}$).  Choose a set of representatives $B\subset\{v_i\}$ for the orbits of $\phi$.  Then the set $B$ is a basis for $\mathbb{R}$ as a module over the ring $\mathbb{Q}[c,c^{-1}]$.
However, I claim that $\mathbb{R}$ cannot be free over $\mathbb{Q}[c,c^{-1}]$.  For instance, every element of $\mathbb{R}$ is infinitely divisible by $c+1$.  But since $c$ is transcendental, no nonzero element of the ring $\mathbb{Q}[c,c^{-1}]$ is infinitely divisible by $c+1$, and hence no nonzero element of any free module can be infinitely divisible by $c+1$.  Thus no such basis exists.
More generally, this argument applies with $\mathbb{Q}\subset\mathbb{R}$ replaced by any field extension $K\subset L$ (if $L$ contains roots of unity not in $K$, you need to use the fact that the powers of a root of unity cannot be linearly independent over $K$ since their sum is $0$).
