TITLE: Are there countable index subrings of the reals?
QUESTION [18 upvotes]: Does ${\mathbb R}$ have proper, countable index subrings?  By countable I mean finite or countably infinite.  By subring I mean any additive subgroup which is closed under multiplication (I don't care if it contains $1$.)  By index, I mean index as an additive subgroup.
Given some real number $x$, when is it possible to find a countable index subring of ${\mathbb R}$ which does not contain $x$?

REPLY [12 votes]: Simon Thomas's approach answers Question 2 too.  The answer is that for any nonzero real number $x$ there exists such a subring (possibly without 1) not containing $x$.
Proof: Let $K$ be the Puiseux series field $\overline{\mathbf{Q}}((t^{\mathbf{Q}}))$, let $A$ be its valuation ring, and let $\mathfrak{m}$ be its maximal ideal.
If $x \in \mathbf{R}$ is not algebraic over $\mathbf{Q}$, then choose an identification $\mathbf{C} \simeq K$ sending $x$ to the transcendental element $1/t$, and use the subring $\mathbf{R} \cap A$.
If $x \in \mathbf{R}^\times$ is algebraic over $\mathbf{Q}$, then choose an identification $\mathbf{C} \simeq K$ again, and use $\mathbf{R} \cap \mathfrak{m}$ (a subring of $\mathbf{R}$ without $1$). $\square$
Remark: If one insists on using subrings with $1$, then the answer is that such a subring not containing $x$ exists if and only if $x \notin \mathbf{Z}$.  
Proof: Repeat the argument above, but in the case where $x$ is algebraic (and outside $\mathbf{Z}$), use $\mathbf{R} \cap (\mathbf{Z} + \mathfrak{m})$.