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Let $k$ be a field. Suppose we have a conic in $\mathbb P^2_k$ given by a homogeneous polynomial $P\in k[x,y,z]$. Let $(a:b:c)\in\mathbb P^2_k$ be a point at which the partial derivatives $\frac{\partial P}{\partial x}$, $\frac{\partial P}{\partial y}$ and $\frac{\partial P}{\partial z}$ vanish. Is it true that $P$ also vanishes at $(a:b:c)$?

I first suspected that it was false and was looking for a counter example, but I could not find any. So now I think it may be true. I am interested in the case where $k = \mathbb F_p$ but (counter)examples over $\mathbb C$ may also be instructive.

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    For characteristic zero you have the formula $n F = \sum x_i \partial_{x_i}F$ where $n$ is the degree of the polynomial. – 2017-01-23
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    Either you are speaking in terms of **conics** and you are looking for its center or you are speaking in terms of **quadratic forms** and you are speaking about a stationnary point. But the terms "vanishing of a conic" is not very adapted... – 2017-01-23

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1) The result is true if $\operatorname{ char.} k\neq 2$ because $x\frac{\partial P}{\partial x}+ y\frac{\partial P}{\partial y}+z\frac{\partial P}{\partial z} =2P$, by Euler's formula for homogeneous polynomials, so that the vanishing of the partial derivatives implies (after division by $2$) the vanishing of $P$.
2) The result may fail if $\operatorname{ char.} k= 2$: the point $a=[1,1,1]\in \mathbb P^2$ annihilates the three partial derivatives of $P(x,y,z)=x^2+y^2+z^2$ (since all three of them are the zero polynomial!) but that point $a$ does not lie on the conic $x^2+y^2+z^2=0$.

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    This is precisely what I needed. Thank you! – 2017-01-23
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    I'm happy to read this, dear Mandelbrot! – 2017-01-23