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Show that the image of the veronese map $[a : b] \mapsto [a^{2}: b^{2} : ab]$ is not contained in any hyperplane of $\mathbb{P}^{2}$.

Using the result from a previous question I asked:

Hypersurface becomes an hyperplane after embedding

Let Im denote the image of the $2$-Veronese map and suppose Im is contained in, say $V(cx+dy+ez)$, where not all the coefficients c,d,e are zero. Then this defines an hyperplane in $\mathbb{P}^{2}$ so the preimage under the Veronese map defines an hypersurface of degree $2$ in $\mathbb{P}^{1}$. Since Im $\subset V(cx+dy+ez)$ applying the inverse yields that $\mathbb{P}^{1}$ is contained in an hypersurface, which has dimension $1$ and this is impossible. Is this wrong?

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up vote 3 down vote accepted

The general result is that the image of the $d$-uple Veronese embedding $v:\mathbb P^n\to \mathbb P^N \; (N=\binom{n+d} {n}-1)$ is not contained in a hyperplane.

The proof is trivial: if some hyperplane of equation $\sum a_Iz_I=0$ ( not all $a_I$ zero) contained $v(\mathbb P^n)$, we would have $\sum a_Ix^I$ for all $x\in \mathbb P^n$.
This is only possible if all $a_I=0$: a non-zero polynomial over an infinite field cannot vanish for all values of its arguments.

As a trivial illustration consider the $2$-uple embedding $v:\mathbb P^1\to \mathbb P^2:(x:y)\mapsto (x^2:xy:y^2)$.
The result says that it is impossible to have a line $au+bv+cw=0$ ($a,b,c$ not all zero ) containing $v(\mathbb P^1)$: else we would have $ax^2+bxy+cy^2=0$ for all $x,y\in k$ ($k=$ base field) which is impossible [just take successively $(x,y)=(1,0), (0,1), (1,1) \:!]$.

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