Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
  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 $x$?

share|improve this question
3  
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
1  
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
add comment

3 Answers

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.

share|improve this answer
    
+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
add comment

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

share|improve this answer
    
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
2  
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
add comment

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 $b = u - v$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.