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 $K \neq \mathbf{Q}$?



Is there a section disjoint from 0, 1 and infinity on the projective line
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 $u = x^n$. 


