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

Let $n$ be a positive integer.

Does there exist a number field $K$ such that the number of solutions of the unit equation $$a+b =1, \quad a,b\in O_{K}^\ast$$ is at least $n$? Can we write down such a number field explicitly?

I know that the number of solutions is always finite in a fixed number field.

share|improve this question
    
5  
For arbitrary $n$, you can find $m$ odd with $\phi(m)\geq n$. Then, $a=\zeta_m+1$, $b=-\zeta_m$ gives $\phi(m)\geq n$ solutions for $K=\mathbf Q(\zeta_m)$. –  user2035 Oct 9 '11 at 15:06
    
Got it! That's nicer than what I'd come up with. –  David Speyer Oct 9 '11 at 15:17
    
That is pretty nice indeed. If you post it as an answer I could accept it. –  Taicho Oct 9 '11 at 15:21
1  
I do not understand this last point: if $n$ is even, $n+1$ is odd...? Also $m=3^k$ are odd numbers having very explicitly unbounded $\phi(m)$. –  user2035 Oct 9 '11 at 16:06
show 3 more comments

3 Answers

up vote 9 down vote accepted

A slightly more general form of the above mentioned lemma states: whenever $m$ has at least two distinct prime factors and $\zeta_m$ is a primitive $m$-th root of unity, $1-\zeta_m$ is a unit in $\mathbf Z[\zeta_m]$.

Choosing $a=1-\zeta_m$ and $b=\zeta_m$ for the various primitive roots of unity, we get $\varphi(m)$ solutions for $K=\mathbf Q(\zeta_m)$. So any such $m$ satisfying $\varphi(m)\geq n$ will do.

share|improve this answer
add comment

Another answer: let $f(x)$ be any monic polynomial with integer coefficients satisfying $f(0)=\pm 1$ and $f(1)=\pm 1$. Then all zeros $u$ of $f(x)$ are units (in the splitting field of $f(x)$), and each $1-u$ is also a unit.

share|improve this answer
    
This construction has a-fortiori's answer as a special case: when $f(x)$ is the $m$-th cyclotomic polynomial and $m$ has at least two distinct prime factors, $f(0) = 1$ and $f(1) = 1$. Of course there is then also the task of directly constructing such polynomials beyond cyclotomic examples... –  KConrad Oct 9 '11 at 18:29
    
Keith, what do you mean by constructing such polynomials? It is trivial to construct polynomials satisfying $f(0)=\pm 1$ and $f(1)=\pm 1$. –  Richard Stanley Oct 11 '11 at 1:19
    
It is easy to see that the polynomials f in K[X] such that f(0) = 1 and f(1) = 1 is precisely the set of polynomials 1+ x(x-1) g, where g is an element of K[X]. (It's similar for $\pm 1$.) –  Shaye Oct 13 '11 at 15:51
add comment

Someone (Elkies?) pointed out recently here on MO that, if $u$ is a unit, then the roots $a,b$ of $x(1-x)=u$ are units satisfying $a+b=1$. Start with your favorite $u$ and iterate.

Bonus question: It's known that the number of solutions of the unit equation is bounded in terms of the rank of the group of units, hence the degree. What's the smallest degree of a number field where the unit equation has $n$ solutions?

share|improve this answer
2  
Elkies's comment is here: mathoverflow.net/questions/76206/… –  user2035 Oct 9 '11 at 20:53
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.