# Proof of Generalized Bézout theorem using Veronese Embedding map

I was wondering if it is possible to prove a sort of Bézout theorem from the Veronese Embedding Map:

Let $$f_1, \ldots, f_m \in \mathbb{C}[x_0, x_1, \ldots, x_n]$$ be $$m$$ homogeneous polynomials all of the same degree $$d$$.

Considering the Veronese map of degree $$d$$ from the projective space $$\mathbb{P}^n$$, $$V_{d, n} : \mathbb{P}^n \rightarrow \mathbb{P}^N$$, we have that the locus $$Z(f_i) = V_{d,n}^{-1}(Z(H_i))$$ where $$H_i$$ is a hyperplane in $$\mathbb{P}^N$$.

Then we have $$Z(f_1, \ldots, f_m) = \cap_{i} Z(f_i) = V_{d,n}^{-1}(\cap_{i} Z(H_i))$$. Then we can say something about the dimension of $$Z(f_1, \ldots, f_m)$$ based on the dimension of the intersection of hyperplanes (easy to deal with) and the inverse image of the Veronese map (or the intersection with the Veronese variety).

I am asking if there is some source where I can look up the following questions (or if someone is willing to explain them here):

1. Is there something known about the (hopefully nice) intersection of linear subspaces with the veronese variety?

2. Is there any method to bypass the restriction on the degree of the polynomials to be the same, except the following one (which has the downside of requiring too many polynomials to substitute just one)?

Let $$f$$ be a polynomial of degree $$k < d$$. One can substitute it with $$f \cdot x_0^{d-k}, f \cdot x_1^{d-k}, \ldots, f \cdot x_n^{d-k}$$. (If they all vanish in a point $$p$$ and $$f(p) \neq 0$$, then we get a contradiction since $$x_0 = x_1 = \ldots = x_n = 0$$)

3. I am hoping in particular that the following is true: Let $$K$$ be a linear subspace of codimension $$r$$ (in $$\mathbb{P}^N$$). Then $$V_{d,n}^{-1} (K)$$ has codimension $$r$$ (in $$\mathbb{P}^n$$). [Answered Negatively by Mohan]

4. More specifically, if we have $$m = n = 3$$ and we add the hyphothesis that the associated hyperplanes to $$Z(f_1), Z(f_2), Z(f_3)$$ are the zero loci of linearly independent equations, is it possible to prove by this argument that the intersection $$Z(f_1, f_2, f_3)$$ consist of a finite number of points?

For ease of notation, take $d=2$ and consider the Veronese map with $y_0,\ldots, y_N$ the co-ordinates of $\mathbb{P}^N$ and the Veronese map has $y_0=x_0x_1, y_1=x_0x_2$. Then let $K=\{y_0=y_1=0\}$. $K$ has codimension 2, but $V_{2,n}^{-1}(K)$ has codimension one, since it contains $x_0=0$. This answers your 3.

4 is unclear, are you assuming something about $n$? If $n$ is arbitrary, why should the common zeroes of 3 polynomials be a finite set of points even with any assumption on them?

• Thank you for the counterexample to (3). Regarding (4), I have edited the question (I forgot to mention $n = m$) Aug 9 '17 at 10:26