I'm trying to understand this proof:
Theorem
If $\nu_{n,d}:\mathbb P^n\to \mathbb P^N$ is a Veronese embedding and $X$ a hypersurface of degree $d$, then $\nu(X)$ is a hyperplane in $\nu (\mathbb P^n)$.
Proof
We write the homogeneous coordinates in $\mathbb P^N$ as $v_{i_0,\ldots,i_n}$, where $i_0,\ldots,i_n\ge 0$ and $i_0+\ldots,i_n=d$. The map $\nu:\mathbb P^n\to \mathbb P^N$ is given by
$\nu(x_0:\ldots:x_n)=(\ldots:v_{i_0,\ldots,i_n}:\ldots)=(\ldots:x_0^{i_0}\cdots x_n^{i_n}:\ldots)$.
We see that the image $\nu (\mathbb P^n)$ is the set of the zero of the polynomials
$V_{i_0,\ldots,i_n}\cdot V_{j_0,\ldots,j_n}=V_{k_0,\ldots,k_n}\cdot V_{l_0,\ldots,l_n} \ \ \ \ \ \ (*)$
Where $i_r+j_r=k_r+l_r$ for every $r=0,\ldots,n$. Obviously, by definition that the points of $\nu(\mathbb P^n)$ satisfy $(*)$.
Conversely, if $v=(\ldots:v_{i_0,\ldots,i_n}:\ldots)$ satisfy $(*)$, we have at least one of the coordinates of this kind $v_{0,\ldots,d,\ldots,0}$ nonzero, because in the contrary every coordinates of $v$ is zero by $(*)$. (WHY?)
Suppose wlog $v_{d,0,\ldots,0}\neq 0$. Given
$x_0=v_{d,0,\ldots,0},\ x_1=v_{d-1,1,0,\ldots,0},\ \ldots, \ x_n=v_{d-1,0,\ldots,1}$
and consider $x=(x_0:\ldots :x_n)\in \mathbb P^n$. we want to prove that $\nu(x)=v$. We have
$v(x)= (\ldots:(v_{d,0,\ldots,0})^{i_0}(v_{d-1,1,\ldots,0})^{i_1}\cdots(v_{d-1,0,\ldots,1})^{i_n}:\ldots)=^{**}(\ldots:(v_{d,0,\ldots,0})^{d-1}(v_{i_0,\cdots,i_n}):\ldots)=(\ldots:v_{i_0,\ldots,i_n}:\ldots)=v$
Where the equality $(**)$ follows from $(*)$, since $d\cdot i_0+(d-1)(i_1+\ldots +i_n)=i_0+(d-1)(i_0+\cdots +i_n)=i_0+d(d-1)$.(WHY?)
I really need help in this proof, I would be very grateful if any specialist could help me in these two doubts.
Thanks a lot!!!
