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The following question came up in my arithmetic geometry course yesterday. Suppose $\alpha$ is an irrational real algebraic integer, and suppose $\epsilon >0$ is given. Then by Roth's theorem there are at most finitely many rational numbers $\frac{h}{q}$ with $\gcd(h,q)=1$, $q>1$, such that $$ \left| \alpha - \frac{h}{q}\right| < \frac{1}{q^{2+\epsilon}}. $$ Are there any results on how large such $q$ can be? Thanks.

edited Feb 9 2012 at 0:22

Charles
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asked Mar 15 2011 at 20:17

Ramin
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I presume not, for two reasons. 1) The proof of Roth's theorem involves many solutions h/q, H/Q, with Q much larger than q and concludes that they don't exist. Knowing effectively one h/q would lead to an effective Roth's theorem, which is still unknown. 2) In the related context of Mordell's conjecture, Szpiro has put forward a "small points conjecture" asserting that a curve has a point whose height is not too large. Again, it implies the yet unknown effective Mordell. – ACL Mar 15 2011 at 20:21

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Ramin, you probably wanted to ask "how large the largest such $q$ can be?" ($q=1$ works for any $\alpha$ as you can choose $h$ to make $|\alpha-h|\leq 1/2$.) As far as I know there is no known bound, this is a major open problem in number theory. If $\alpha$ is special (e.g. a cubic irrational) then there are bounds for $\epsilon$ not too small. – GH Mar 15 2011 at 21:07

@GH: You are right. Thank you. – Ramin Mar 16 2011 at 17:01

Felipe Voloch · Accepted Answer · 2011-03-17 05:41:12Z UTC

My other answer was for the first version of this question. The question has now been changed completely. As Antoine mentioned in his comment, an effective Roth's theorem is not known in general. Finding such a result is the probably the main open problem in Diophantine approximation. There some instances in which a non-trivial effective result can be proved. Most notable is the Baker-Feldman theorem which provides such a result with an exponent of the form $\deg \alpha - \epsilon$ (instead of $2+\epsilon$) for a suitable small positive $\epsilon$.

Felipe Voloch · Answer 2 · 2011-03-15 20:25:42Z UTC

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Of course, there may be none or there may be a very small one (if $|\alpha|<1$ then $q=1,h=0$ works). If there is a small very good approximation, the others must be very large. There are lots of papers on effective diophantine approximation dealing with this. I recall there was a paper of Davenport and Roth which first established such a bound.

answered Mar 15 2011 at 20:25

Felipe Voloch
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Davenport and Roth's paper deals with bounding the number of $h/q$ such that the inequality is satisfied. They do say though that they do not know how to answer the question I'm asking. I was wondering if there has been any progress in this direction. I added the restriction that $\alpha$ is an algebraic integer. – Ramin Mar 15 2011 at 20:48

I'm not sure I understand how making $\alpha$ an algebraic integer will change anything, you can still have $|\alpha|<1$ and $q=1,h=0$. You might want to think more carefully about what it is exactly that you want to ask. – Felipe Voloch Mar 15 2011 at 23:31

I added $q>1$ to the statement of the problem. As Antoine points out this came out of the discussion to figure out if it was possible to make the proof of Roth's theorem effective (or improve the Davenport-Roth bound). – Ramin Mar 16 2011 at 2:39

You still are not making sense. I can still have $|\alpha| < 1/8$ and let $q=2,h=0$ which works for any $\epsilon < 1$. If you want to make Roth's theorem effective, you need to ask how big $q$ can be in terms of $\alpha$, not how small. There is also something called the gap principle which shows that good approximations to a fixed algebraic number cannot be close together. – Felipe Voloch Mar 16 2011 at 2:48

Correction, of course $2,0$ are not coprime. Take $q=2,h=1$ and $|\alpha - 1/2| < 1/8$. – Felipe Voloch Mar 16 2011 at 2:51

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