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?