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?

  • 1
    $\begingroup$ Perhaps it may be better to use the notation $V(f)$ or $\mathbb{V}(f)$ or $Z(f)$ rather than $\mathrm{Zeros}(f)$. $\endgroup$ – 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.

  • $\begingroup$ Why is every possible $Z$ of the form $\nu_d(S)$ for some $S\subseteq\Bbb{P}^2$? $\endgroup$ – Jyrki Lahtonen Jul 1 at 9:30

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