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  1. 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.

  2. Given some real number $x$, when is it possible to find a countable index subring of ${\mathbb R}$ which does not contain $x$?

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Not measurable ones. Every measurable proper subgroup of ${\mathbf R}$ has measure zero, and thus so does a union of countably many of its cosets. – Jonas Meyer Mar 20 '10 at 21:26
For Borel (or analytic) subrings you can say even more ... a proper one has Hausdorff dimension zero. – Gerald Edgar Mar 20 '10 at 22:01
up vote 27 down vote accepted

Perhaps surprisingly, it turns out that such subrings do exist. This was proved in Section 2 of my paper:

Simon Thomas, Infinite products of finite simple groups II, J. Group Theory 2 (1999), 401--434.

The basic idea of the proof is quite simple. Clearly the ring of $p$-adic integers has countable index in the field of $p$-adic numbers. Now the $p$-adic integers are the valuation ring of the obvious valuation on the field of $p$-adic numbers ... and it turns out to be enough to show that $\mathbb{C}$ has an analogous valuation. This is true because $\mathbb{C}$ is isomorphic to the field of Puiseux series over the algebraic closure of $\mathbb{Q}$, which has an appropriate valuation.

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+1: Very nice, Simon! – François G. Dorais Mar 21 '10 at 3:06
+1 This is great! – Harry Gindi Mar 21 '10 at 3:23
Nice answer! This is really interesting (I'm reading your paper now.) – Fabrizio Polo Mar 21 '10 at 9:06

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 $\mathbf{R} \cap (\mathbf{Z} + \mathfrak{m})$.

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By the way, rings really should have a 1, so that one can take products of any finite sequence of elements, including the empty sequence! – Bjorn Poonen Mar 21 '10 at 5:24
Thanks for the help. I have to award the answer to Simon, but your argument definitely handles question 2 so I gave you a vote. I knew a while ago that if I could miss a given transcendental then I could miss any other, but I had no idea if I could miss even one. I also didn't know if the algebraic case needed to be treated separately. I could easily get addicted to this website. It's dangerous. – Fabrizio Polo Mar 21 '10 at 9:12
You did the right thing by accepting Simon's answer; my answer is really just a small addendum to his, so I would have felt guilty if you had accepted mine! And yes, this site is very dangerous... – Bjorn Poonen Mar 21 '10 at 18:10

I think the answer is no, but I didn't get anywhere. [Edit: I used to think the answer was no, but Simon Thomas convinced me otherwise.] Here is the condensed version of what I posted earlier, which seems to put serious constraints on $S$.

Let $R$ be any field and let $S$ be an additive subgroup of $R$ which is closed under multiplication. If $S$ has index less than $|S|$ as an additive subgroup of $R$, then every element of $R$ is of the form $a/b$ for $a, b \in S$. To see this, pick $r \in R$ and consider the multiples $ur$ for $u \in S$. Since $S$ has index less than $|S|$, there must be $u \neq v$ such that $a = ur - vr \in S$ then $r = a/b$ where $b = u - v$.

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Note that there are plenty of proper additive subgroups $S$ of $R$ closed under multiplication such that every element of $R$ is of the form $a/b$ for some $a,b \in S$. So my observation is not that helpful. – François G. Dorais Mar 21 '10 at 2:40

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