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I'm having trouble understanding how to find the preimage of the degree $d$ Veronese map, following these steps:

(the projective space are the projectivization of $k^{n+1}$ and $k^{N+1}$, where $k$ is an algebraically closed field)

$v:\mathbb{P}^{n}\longrightarrow \mathbb{P}^{N}$

$(x_{0}:\dots:x_{n})\longmapsto (x_{0}^{d}:x_{0}^{d-1}x_{1}:\dots:x_{n}^{d})$

where $N=$${n+d}\choose{d}$$-1$

Constructing a multidenx $I={i_{0},\dots,i_{n}}$ such that $|I|=d$, we can put coordinates on $\mathbb{P}^{N}$, namely for each multindex $I$ we set $(z_{I})_{I}$ standing for the $I$-th coordinate. In this coordinate setting, our map becomes $z_{I}=x_{0}^{i_{0}}\cdots x_{n}^{i_{n}}$ where $I=i_{0},\dots,i_{n}$

Now the statement I found in my note is that the image of $\mathbb{P}^{n}$ under the Veronese map is a irreducible closed subset of $\mathbb{P}^{N}$ and that $v$ is a homeomorphism onto the image. The proof that $v$ is injective and a homeomorphism is clear. To prove that the image is closed, I prove that

$v(\mathbb{P}^{n})=V(\ker(\theta))$

where:

$\theta:k[z_{I}]\longrightarrow k[x_{0},\dots,x_{n}]$ is a $k$-algebra homomorphism, sending a polynomial in the $z_{I}$ to the polynomial calculated in $x_{0}^{i_{0}}\cdots x_{n}^{i_{n}}$

$V$ represents the closed set formed by the common zeros of the ideal $\ker(\theta)$

It is clear for me that $\subseteq$ holds.

For the converse, my notes follow this method: For an arbitrary multindex $J=j_{0},\dots,j_{n}$ such that $|J|=d-1$, and for $a\in {0,\dots,n}$, we put $e_{a}={0,\dots,0,a,0,\dots,0}$, we consider the multindex $J+e_{a}$ of module $d$.

For a point $Q\in V(\ker(\theta))$, $Q=(z_{I})_{I}$ we claimed that

$P=(z_{J+e_{0}}:\dots:z_{J+e_{n}})$ is such that $v(P)=Q$

So we showed that

$(z_{J+e_{0}}^{i_{0}}\cdots z_{J+e_{n}}^{i_{n}})=\lambda z_{I}$ where $\lambda\neq 0$, the equality holding for every $I$ multindex (i.e. for every coordinate of the point Q. Since it must hold for every $I$, by checking it for $I=J+e_{0}$ we find a suitable $\lambda=z_{J+e_{0}}^{j_{0}}\cdots z_{J+e_{n}}^{j_{n}}$ and we easily checked that this $\lambda$ works for every multindex $I$. But then, in order to complete the proof, we were supposed to solve some exercises left, that I don't fully understand:

1)We still have to show that there is a suitable multindex $J=j_{0},\dots,j_{n}$ for which if $j_{a}>0$ then $z_{J+e_{a}}\neq 0$, so that the point $P$ has some non-zero coordinate and it is well defined. My question: Since $\lambda$ was defined by the product of ALL the coordinate of P, don't we have to check that all the coordinates of P are non-zero? (By the way, it wouldn't be always possible because of the injectivity of $v$). HINT LEFT: since at least one coordinate of Q is non-zero, supposing $z_{H}\neq 0$, show that if $h_{i},h_{j}>0$ then $z_{H-e_{i}+e_{j}}\neq 0$. My question: I know how to prove the statement, but I don't understand the utility of checking this. Is it useful because, if this holds, then there exists at least one of the $z_{d,0,\dots,0}, \dots, z_{0,0,\dots,d}$ non zero? The exercise ended, saying that if $h_{0}>0$ we could put $J=H-e_{0}$, finding in this way the suitable coordinates for P.

Thank you all for having read this. Any hint would be appreciated!

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