Any projective variety can be seen as a plane section of the Veronese variety (Harris 2.9) I am trying to solve this excercise from Harris. Our instructor did not spend much time on excercises and problem solving and I literally do not know if my ideas are naive or have some meaning. 
As in the title we have to show that given a projective variety $Y\subset\mathbb{P}^n$ it is isomorphic to the intersection of a Veronese variety of some deegre $d$ and an hyperplane, hence to $v_d(\mathbb{P}^n)\cap H\ \subset \mathbb{P}^N$, where $H$ is an hyperplane (and $N=$${n+d}\choose{d}$$-1$ the usual dimension for the codomain of the Veronese map $v_d$). 
What I had in mind is to somehow use the following facts
Fact 1 Given a variety $Y\subset\mathbb{P}^n$ its image through the Veronese map  is a subvariety of $\mathbb{P}^N$ isomorphic to $Y$
Fact 2 Given an hypersurface of deegre $d$ in $\mathbb{P}^n$ this is isomorphic to  a plane section of $v_d(\mathbb{P}^n)$
Anyhow I am not quite sure about how to ''compose'' them.
Thanks in advance 
 A: Fix your favourite projective variety $X \subseteq \mathbb{P}^n$, and let $X = V(F_1, \dots F_r)$ where all $F_i$s are homogeneous polynomials of, say, degree $d$. Now fix the corresponding Veronese embedding $v_d: \mathbb{P}^n \to \mathbb{P}^N$, where $N$ is $n+d \choose d$. Let $(x_0: \dots : x_n)$ be homogeneous coordinates on $\mathbb{P}^n$, and let $(z_{I})$ be homogeneous coordinates on $\mathbb{P}^N$ (indexed by multi-indices $I$ of length $d$). In coordinates, $v_d$ can be written as $z_I = x^I$. Suppose $F_j = \sum_{\vert I \vert = d} b^{j}_I x^I$, then define $G_j = \sum_{\vert I \vert = d} b^{j}_I z_I$. All the $G_j$s are linear polynomials in the homogeneous coordinates of $\mathbb{P}^N$. Let $\Lambda = V(G_1, \dots G_r)$ be the associated projective subspace of $\mathbb{P}^N$. One can easily check that $p = (a_0: \dots :a_n) \in X$ if and only if $v_d(p) \in \Lambda \cap v_d(\mathbb{P}^n)$, hence $v_d$ gives an isomorphism between $X$ and a linear section of $v_d(\mathbb{P}^n)$.
