MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K$ be a number field with ring of integers $O_K$. Is there a section of $\mathbf{P}^1_{O_K}$ over $O_K$ whose image is disjoint from $0$, $1$ and $\infty$? If $K=\mathbf{Q}$ this is not possible because any integer $n>1$ is divisible by a prime number. What if $K \neq \mathbf{Q}$?

share|cite|improve this question
This looks like a fancy way of asking whether the unit equation in $K$ has solutions: a "section" other than $\infty$ is some field element $u$; it's disjoint from $0,\infty$ iff it's a unit; and it's then disjoint from $1$ iff $u-1$ is a unit too. This can certainly happen, e.g. in a field where $u^2 + u = \pm 1$ has a solution, though it's known that the number of solutions is finite for each $K$. – Noam D. Elkies Sep 23 '11 at 13:50
To expand on Noam's comment: Let $\omega$ be a primitive cube root of unity. Then $-\omega$ and $-\omega-1=\omega^2$ are both units, so $-\omega$ works. – David Speyer Sep 23 '11 at 17:23
To expand on David's expansion: I should have written $u^2 - u = \pm 1$, which is equivalent (replace $u$ by $-u$) but more transparent (if $u(u-1) = \pm 1$ then both $u$ and $u-1$ are units). David described the $-1$ case; $+1$ makes $u$ the golden ratio $(1+\sqrt{5})/2$. Somewhat more generally, if $x^{m+n}-x^m = \pm 1$ for positive integers $m,n$ then both $x$ and $x^n-1$ are units, so we may take $u=x^n$.$$ $$P.S. I edited the question only to add the "number-fields" and "diophantine-equations" tags, which are at least as relevant as "intersection-theory" and "ag.algebraic-geometry". – Noam D. Elkies Sep 24 '11 at 4:00
Great! So I'm looking for number fields such that the finite set of solutions to the unit equation is non-empty. Such number fields are $\mathbf{Q}(\sqrt{5}$, $\mathbf{Q}(\zeta_3)$, etc. If you post your comment as an answer I can accept it. – Tamed Sep 26 '11 at 10:22
@Tamed: OK; see below. – Noam D. Elkies Oct 2 '11 at 3:30
up vote 12 down vote accepted

Such sections are tantamount to solutions of the unit equation $u + u' = 1$ in $O_K^*$. This is indeed impossible for $K = {\bf Q}$, when $O_K = {\bf Z}$ and the only units are $\pm 1$; but there can be such solutions for other number fields $K$, though it is known that in each $K$ there are only finitely many solutions.

A "section of ${\bf P}^1_{O_K}$ over $O_K^{\phantom|}$" is a $K$-point of the projective line, i.e. either $\infty$ or a field element. The "sections" $u$ disjoint from $\infty$ are precisely the algebraic integers, because $u$ "intersects $\infty$ at the prime $\wp$" iff $u$ has negative valuation at $\wp$, and the algebraic integers are precisely the field elements none of whose valuations are negative. Likewise $u$ is disjoint from $0$ iff $u$ has no positive valuation at any $\wp$, and disjoint from $1$ iff $u-1$ has no positive valuations. Therefore, $u$ is disjoint from $0$, $1$, and $\infty$ iff both $u$ and $u-1$ are units, which is to say iff $(u,1-u)$ is a solution of the unit equation.

One easy way to get such $(K,u)$ is to make $u(1-u)$ a unit in some number field $F$, say $\epsilon$, because then $u$ and $1-u$ are themselves algebraic integers in a number field containing $F$ with degree at most $2$ (they're the roots of the monic quadratic polynomial $x^2-x+\epsilon$ over $O_F$), and thus units because they divide the unit $\epsilon$. For example, taking $F={\bf Q}$ and $\epsilon = 1$ or $-1$ we recover the simplest solutions of the unit equation: the sixth roots of unity $(1 \pm \sqrt{-3})/2$, and the golden ratio $(1 \pm \sqrt{5})/2$. More generally, if $x^{m+n} - x^m = \epsilon$ for some unit $\epsilon$ and positive integers $m,n$ then both $x$ and $1-x^n$ are units, so we may take $u = x^n$.

share|cite|improve this answer

Your Answer


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.