# Property of Veronese Embedding

I am working on the following problem:

Let $$Y$$ be the image of $$\mathbb{P}^2$$ in $$\mathbb{P}^5$$ by the Veronese embedding. Let $$Z$$ be a closed subvariety of $$Y$$ of dimension 1. Show that there exists a hypersurface $$V$$ of $$\mathbb{P}^5$$ such that $$V\cap Y = Z$$

This is what I have done so far:

As $$Z$$ is a subvariety of $$Y$$ and the Veronese embedding is an injection, I can see the preimage of $$Z$$, noted $$X$$, as a closed subvariety of $$\mathbb{P}^2$$. This means that $$X=V(f)$$ where $$f$$ is an irreducible polynomial in $$k[X_0,X_1,X_2]$$.

Now, I have that $$f^2 = g \in k[X_0^2,X_1^2,X_2^2,X_0X_1,X_0X_2,X_1X_2]$$.

Then $$V(f) \subset V(g)$$. I can factor $$g$$ into irreducible $$g=g_1\dots g_n$$ where each $$g_i \in k[X_0^2,X_1^2,X_2^2,X_0X_1,X_0X_2,X_1X_2]$$.

I know there should be one $$g_i$$ such that $$Y\cap V(g_i)=Z$$. I don't know how to continue

1. Is this reasoning right? I have found "easier" ways to proof this but I can't see them clearly (Example Why this property holds in a Veronese surface)

2. How do I know there is one $$g_i$$ with such property?

• Perhaps it may be better to use the notation $V(f)$ or $\mathbb{V}(f)$ or $Z(f)$ rather than $\mathrm{Zeros}(f)$. – mathphys Jun 27 at 3:45

I'll outline the general situation. Firstly, note the general fact that we can write $$V(F) = V(x_0F, \dots, x_nF)$$ because the $$x_i$$ cannot all simultaneously vanish (we are working in projective space).
Now consider the Veronese embedding $$\nu_d : \mathbb{P}^n \rightarrow \mathbb{P}^m$$. If $$S \subset \mathbb{P}^n$$ is a projective variety, then $$S= V(F_1, \dots, F_N)$$ for some homogeneous $$F_i$$; their degrees may be different. By the fact mentioned above, $$S = V(G_1, \dots, G_M)$$ where the $$G_i$$ are homogeneous and their degrees are the same, in fact $$\deg(G_i) = c \cdot d$$ for some $$c$$. Therefore we can write $$G_i = H_i \circ \nu_d$$ for homogeneous $$H_i$$ where $$\deg(H_i) = c$$. Then we have the map $$\mathbb{P}^n \supset S = V(G_1, \dots, G_M) \xrightarrow{\nu_d} V(H_1, \dots, H_M) \subset \mathbb{P}^m.$$ Thus $$\nu_d(S) = \nu_d(\mathbb{P}^n) \cap V(H_1, \dots, H_M)$$.
In your case, $$Y= \nu_d(\mathbb{P}^n)$$, $$Z = \nu_d(S)$$ and $$V = V(H_1, \dots, H_M)$$ is the hypersurface you're looking for.
• Why is every possible $Z$ of the form $\nu_d(S)$ for some $S\subseteq\Bbb{P}^2$? – Jyrki Lahtonen Jul 1 at 9:30