Welcome to MathOverflow

Question

3

1

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.

edited Oct 9 2011 at 16:26

GH
14.9k●1●31●53

asked Oct 9 2011 at 14:05

Taicho
85●4

see mathoverflow.net/questions/71885/… – a-fortiori Oct 9 2011 at 14:14

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 $K=\mathbf Q(\zeta_m)$. – a-fortiori Oct 9 2011 at 15:06

Got it! That's nicer than what I'd come up with. – David Speyer Oct 9 2011 at 15:17

That is pretty nice indeed. If you post it as an answer I could accept it. – Taicho Oct 9 2011 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 $\phi(m)$. – a-fortiori Oct 9 2011 at 16:06

show 3 more comments

a-fortiori · Accepted Answer · 2011-10-09 15:40:48Z UTC

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.

Felipe Voloch · Answer 2 · 2011-10-09 20:44:27Z UTC

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?

Richard Stanley · Answer 3 · 2011-10-09 16:17:26Z UTC

7

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.

answered Oct 9 2011 at 16:17

Richard Stanley
13.3k●2●60●94

	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 2011 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 2011 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 2011 at 15:51

For any $n$, does there exist a number field with at least $n$ solutions to the unit equation

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.