# FInding pre-image of the Veronese map of degree d

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!