Can $ℂ$ be viewed as a (nontrivial) field of fractions?

Question

Is there an interesting ring $S ⊂ ℂ$ such that $ℂ = Q(S)$? I’m thinking no, but how can I prove it?

Answer 1

There is an isomorphism $\mathbf{C} \cong \mathbf{C}_p$ -- i.e. you can extend the p-adic absolute value to the complexes.

$\mathbf{C}_p$ is the field of fractions of its ring of integers: the subring of numbers whose $p$-adic absolute value is less than or equal to $1$.

Answer 2

Depends what you mean by "interesting". The smallest examples of rings with $\mathbb{C}$ as their field of fractions would be $$R[\{t_\alpha\}_{\alpha\in A}]$$ where $R$ denotes the ring of algebraic integers, and $\{t_\alpha\}_{\alpha\in A}$ is a (edit: pure) transcendence basis for the extension $\mathbb{C}/F$, where $F$ is the field of algberaic numbers i.e. the field of fractions of $R$.

(Pure transcendence basis means that the $t_\alpha$ are transcendental numbers, algebraically independent, and $\mathbb{C}=F(\{t_\alpha\}_{\alpha\in A})$.)

Of course they are just isomorphic to $R$ with continuum-many variables adjoined. I don't think these rings are terribly useful, but they could arguably be "interesting".