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I don't know whether this is known or not, but I was thinking of the following problem.

Let $n$ and $m$ be natural numbers. Are there $n$ polynomial $f_1,...,f_n\in \mathbb{C}[x]$, such that all of their intersection numbers are at least $m$?

It is also possible to state an even harder question in a more natural (albeit less basic) way: Are there $n$ homogeneous polynomials in 2 variables for which all of their intersection numbers in $\mathbb{P}^1_{\mathbb{C}}$ are at least $m$?

EDIT: Sorry, the comments made me realize I wanted a slightly different condition: that for every $j$ there's a unique $k\neq j$ such that $f_j(0)=f_k(0)$. (In particular, $n$ is assumed to be even.) I will allow the intersection at $x=0$ to not be of multiplicity $\geq m$, but I will ask that over $x\neq 0$ the intersection multiplicity will be $\geq m$.

Clarification

The question was posed in a way that algebraic geometers would understand, because I suspect they are most likely to come up with a solution to this question. I wanted to emphasize, however, that intersection multiplicity is something that every high-school student can understand: If $f_1$ and $f_2$ are polynomials with coefficients in $x$, and they meet at say $x=3$, then their intersection multiplicity at $x=3$ is the greatest natural number $l$ such that $(x-3)^l$ divides the polynomial $f_1-f_2$.

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It seems that if you make all your polynomials equal to 0, then you get the desired result (for the 1-variable case). –  Zatrapilla Feb 12 '12 at 18:05
    
In fact, if you make $f_i(x)=x^m$, then all the pair-wise intersections are $2m$. –  Zatrapilla Feb 12 '12 at 18:07
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Hmmm, let me clarify in the body of the question. –  James D. Taylor Feb 12 '12 at 18:09
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Let's do a quick example: $0,x,1,1+x$ have the desired property at 0. If $m=2$, you're saying to multiply by $(x-1)^2$, say. $1(x-1)^2$ and $x(x-1)^2$ indeed intersect with multiplicity $2$ at $x=1$, but they would also intersect with multiplicity $1$ at some $x\neq 0,1$. So that's undesirable. –  James D. Taylor Feb 12 '12 at 18:39
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@Mahdi: Yes, that's indeed what I mean. –  Florian Eisele Feb 12 '12 at 22:57

1 Answer 1

up vote 3 down vote accepted

Ok, so here's the long version of what I already said in the comments:

Start by picking $g_1,\ldots,g_n\in \mathbb C[x]$ such that $g_i\neq g_j$ whenever $i\neq j$ which satisfy the prescribed condition at $0$ (that's not hard). Since $g_i-g_j\neq 0$ for each $i\neq j$, there will be only finitely many $0\neq a\in\mathbb C$ such that $g_i(a)-g_j(a)=0$ for some pair $i,j$ with $i\neq j$. Call these values $a_1,\ldots,a_k$. Then define $$ f_i(x) := \left(\prod_{j=1}^k (x-a_j)^m\right)\cdot g_i(x) $$ Those $f_1,\ldots,f_n$ have pairwise intersection also in $a_1,\ldots,a_k$ (and possibly in $0$), but now with multiplicty at least $2m$ (except in $0$, where the intersection multiplicty is left as it was). The condition at zero is clearly preserved and we didn't add any further intersection points either.

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Just a small remark: we should only include the nonzero $a_i$'s, otherwise we will ruin the prescribed condition at zero. –  Mahdi Majidi-Zolbanin Feb 12 '12 at 23:40
    
@Mahdi: Thanks! Just edited. –  Florian Eisele Feb 12 '12 at 23:47

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