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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?

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That should give that the intersection numbers are at least $m$, not a multiple of $m$. Do you mean to say that $f_i(x)-f_j(x)$ is an $m$th power? – Will Sawin Feb 18 '12 at 1:29
I fixed it in the body. – Makhalan Duff Feb 18 '12 at 1:36
Why do you say that? That's not true... – Makhalan Duff Feb 18 '12 at 2:29
Apologies, I was still thinking in terms of the previous setup. – Jack Huizenga Feb 18 '12 at 3:01
When $m\geq 4$ this should be impossible by Mason-Stothers. – Gjergji Zaimi Feb 18 '12 at 4:56

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.

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Nice! But a minor correction: $d$ should be $\max\lbrace d(f_i - f_j): i \neq j \rbrace$. – auniket Feb 22 '12 at 22:57

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