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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: 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 $v_n(\mathbb{P^1}) \subset \mathbb{P}^n$.)

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up vote 6 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.

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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

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