3
$\begingroup$

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 $f_1-f_2$.

Improve this question
$\endgroup$
26
  • 1
    $\begingroup$ It seems that if you make all your polynomials equal to 0, then you get the desired result (for the 1-variable case). $\endgroup$
    – Zatrapilla
    Commented Feb 12, 2012 at 18:05
  • $\begingroup$ In fact, if you make $f_i(x)=x^m$, then all the pair-wise intersections are $2m$. $\endgroup$
    – Zatrapilla
    Commented Feb 12, 2012 at 18:07
  • 1
    $\begingroup$ Hmmm, let me clarify in the body of the question. $\endgroup$
    – James D. Taylor
    Commented Feb 12, 2012 at 18:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – James D. Taylor
    Commented Feb 12, 2012 at 18:39
  • 1
    $\begingroup$ @Mahdi: Yes, that's indeed what I mean. $\endgroup$
    – Florian Eisele
    Commented Feb 12, 2012 at 22:57

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Just a small remark: we should only include the nonzero $a_i$'s, otherwise we will ruin the prescribed condition at zero. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Feb 12, 2012 at 23:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .