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While working in a question about the affine group $\text{Aff}(\mathbb{R})$, I have come up with the following strange question about the real numbers:

Question: Do there exist a non-trivial decomposition $\mathbb{R} = T \oplus T'$ as an additive $\mathbb{Q}$-vector space and another decomposition $\mathbb{R}^{+} = SS'$ as a multiplicative vector space (i.e. $\log(S) \oplus \log(S') = \mathbb{R}$), satisfying the following properties:

1) $S'$ and $T'$ are countable. (Therefore $S$, $T$ are uncountable)

2) $T$ is $S$-invariant, i.e. $ST = T$.

Comment: One observation worth making is that if such a decomposition exist, then $V = \mathbb{R}/T$ is a countable $\mathbb{Q}$-vector space and there is a linear action of $S$ on $V$ (because $T$ is $S$-invariant). Now, if $\text{dim}_{\mathbb{Q}}{V} < \infty$, being the set $S$ uncountable and the group of linear transformation $GL(V)$ countable, there is $s\neq1$ in $S$ such that $sv = v$ for all $v \in V$, which implies that $(s-1)V = 0$ and therefore $(s-1)\mathbb{R} \subset T$, which implies that $T = \mathbb{R}$. In conclusion, such non-trivial decomposition can exist only if $\text{dim}_{\mathbb{Q}}(V) = \infty$.

Any information related (or vaguely related) to this question is greatly appreciated. Also, I don't know if I have the right tags for this question. Thanks.

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    $\begingroup$ Does there even exist a proper subring $A\subset \mathbb{R}$ such that $\mathbb{R}^\times/A^\times$ is countable? If your $S$ and $T$ exist, then the subring generated by $S$ has this property. $\endgroup$ May 29 '14 at 7:54
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Yes. The field of formal Puiseux series over $\overline{\mathbb Q}$ is an algebraically closed field of characteristic $0$ with the cardinality of the continuum, hence is isomorphic to $\mathbb C$. It is easy to check that the ring of Puiseux series with nonnegative valuation does not contain $\mathbb R$, hence its intersection with $\mathbb R$ is a proper subring of $\mathbb R$. Let that ring be $T$ and its group of units be $S$. Then $\mathbb R^+/T^+$ is contained in the field of Puiseux series modulo the power series with nonnegative valuation, which is a vector space of countable dimension over $\overline{\mathbb Q}$, hence of countable dimension over $\mathbb Q$. $\mathbb R^\times/S^\times$ is contained in the value group of the field of Puisex series, which is $\mathbb Q$. Choose an arbitrarily lifting of those to get $S'$ and $T'$

Proof that the ring does not contain $\mathbb R$: Let $z$ be an element of positive valuation, then if $\bar{z}$ has negative valuation then $z+ \bar{z}$ is a real of negative valuation, and if $\bar{z}$ has positive valuation $1/z\bar{z}$ is a real of negative valuation.

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    $\begingroup$ Thanks a lot for your very nice construction, I did not you can think about the complex numbers in that way. $\endgroup$
    – shurtados
    May 30 '14 at 19:39

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