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 $K \neq \mathbf{Q}$?
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17$\begingroup$ 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 $K$. $\endgroup$– Noam D. ElkiesCommented Sep 23, 2011 at 13:50
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5$\begingroup$ 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. $\endgroup$– David E SpeyerCommented Sep 23, 2011 at 17:23
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5$\begingroup$ 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". $\endgroup$– Noam D. ElkiesCommented Sep 24, 2011 at 4:00
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$\begingroup$ 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. $\endgroup$– TamedCommented Sep 26, 2011 at 10:22
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$\begingroup$ @Tamed: OK; see below. $\endgroup$– Noam D. ElkiesCommented Oct 2, 2011 at 3:30
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1 Answer
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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 $u = x^n$.
answered Oct 2, 2011 at 3:29