Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is asserted in Exercise 1.15 of Joe Harris's algebraic geometry book (Algebraic Geometry: A First Course, Pg. 11 in my copy). This result struck my fancy but I'm unable to solve it myself or find references to it elsewhere. The closest reference I've found is the following document, which asserts something weaker: http://www.mast.queensu.ca/~tony/kn+1.ps Does anyone know of any references or see a solution off the top of their head?

(A rational normal curve is any curve equivalent to the Veronese image $v_n(\mathbb{P^1}) \subset \mathbb{P}^n$.)

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

Parameterize the curve with $[s:t]\to [s^d:s^{d-1}t:\ldots :t^d]$ and plug it into the polynomial $F$ of degree $k$. The result is a homogeneous polynomial of degree $kd$ in $s,t$. But any homogeneous polynomial in two variables factors into linear factors over an algebraically closed field, and each linear factor corresponds to a point. So, there are exactly $kd$ points in the curve that vanish on $F$ counting multiplicities, unless the polynomial vanishes identically which just means that $F$ vanishes on the whole curve.

share|improve this answer
add comment

Apply Veronese in degree k to $\mathbb{P}^d$. In this way the zero locus of your polynomial becomes an hyperplane $H$. Your rational normal curve becomes a curve of degree $kd$. If the polynomial vanish on $kd+1$ points this means that $H$ intersects your curve in $dk+1$ points, but your curve has degree $kd$ so this means that it lives on $H$. Coming back you have that the polynomial vanishes on the curve

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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