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Freudian slip?
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Emil Jeřábek
Emil Jeřábek
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Yes. The field of formal PuisexPuiseux 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 PuisexPuiseux 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 PuisexPuiseux 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.

Yes. The field of formal Puisex 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 Puisex 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 Puisex 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.

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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Will Sawin
Will Sawin
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Yes. The field of formal Puisex 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 Puisex 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 Puisex 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.