Question

I've been playing around with some basic intersection theory, and I've wondered the following:

For every two integers $n$ and $m$, and complex numbers $a_1,...,a_n$, are there polynomials $f_1(x),...,f_n(x)$ with coefficients in $\mathbb{C}$ such that the following holds:

1. $f_i(0)=a_i$.
2. For every complex number $b$, $v_{(x-b)}(f_i(x)-f_j(x))$ is divisible by $m$ (in other words all of the intersection numbers away from $0$ are divisible by $m$).
3. $f_i\neq f_j$ for $i\neq j$.

(The $a_i$'s needn't be different from one another)

This is clearly true if $n\leq 2$ and every $m$ and $a_1,a_2$, but I can't think of a general way to do it for every $n$. Is it impossible?

Answer 1

There are no solutions for $n,m \geq 3$ via the Mason-Stothers Theorem, which is also called the polynomial abc-theorem. Only the hypothesis that all roots of $f_i - f_j$ have multiplicity $\geq m$ is needed. Let $c(p)$ be the number of distinct roots of a polynomial $p \in \mathbb{C}[x]$ and let $d(p)$ be its degree, so that $c(f_i- f_j) \leq d(f_i - f_j)/m$ under the above hypothesis. Mason-Stothers Theorem: If $p, q, p+q \in \mathbb{C}[x]$ have positive degrees $\leq d$, then $c(p) + c(q) + c(p+q) \geq d+1$. If $d = max{d(f_i)}$ and $n \geq 3$, applying the theorem with $p = f_1 - f_2, q = f_2 - f_3$ gives $d/m + d/m + d/m \geq d+1$, which is not possible if $m \geq 3$.

By the way, there have been many generalizations of the Mason-Stothers Theorem.