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 $f_1-f_2$.

$\endgroup$
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
    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
    Feb 12, 2012 at 18:07
  • 1
    $\begingroup$ Hmmm, let me clarify in the body of the question. $\endgroup$ 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. $\endgroup$ Feb 12, 2012 at 18:39
  • 1
    $\begingroup$ @Mahdi: Yes, that's indeed what I mean. $\endgroup$ Feb 12, 2012 at 22:57

1 Answer 1

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.

$\endgroup$
1
  • $\begingroup$ Just a small remark: we should only include the nonzero $a_i$'s, otherwise we will ruin the prescribed condition at zero. $\endgroup$ Feb 12, 2012 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.