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I'm trying to understand this property of the Veronese surfaces which is an exercise in Hartshorne's book as well:

Question:

Let $Y$ be the image of the $2$-uple embedding of $\mathbf P^2$ in $\mathbf P^5$. This is the Veronese surface. If $Z\subseteq Y$ is a closed curve (a curve is a variety of dimension $1$), show that there exists a hypersurface $V\subseteq \mathbf P^5$ such that $V\cap Y=Z$.

After trying to solve this question without success, I've been looking in some AG sites why this property is true and every site has the same technique to solve this problem and prove this property:

$v_2:\mathbb P^2 \to \mathbb P^5$ is given by $(x_0,x_1,x_2) \mapsto (x_0^2,x_1^2,x_2^2,x_0x_1,x_0x_2,x_1x_2).$ Let $C\subset \mathbb P^2$ be a curve defined by the homogeneous function $f(x_0,x_1,x_2)=0$. Then $0=f^2\in k[x_0^2,x_1^2,x_2^2,x_0x_1,x_0x_2,x_1x_2]$ defines a hypersurface $V\subset \mathbb P^5$. So $Z=v_2(C)=V\cap Y$.

So, Why $f^2=0$ and $f^2\in k[x_ 0^2,x_1^2,x_2^2,x_0x_1,x_0x_2,x_1x_2]$?

I would appreciate if anyone can give me a hand here.

Thanks in advance.

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  • $\begingroup$ I suggest you just ignore that solution (as written it is just a mess) and think more about it yourself. It is rarely more productive to try to decipher someone's line of thought than to come up with one oneselves $\endgroup$ Nov 3, 2013 at 23:46
  • $\begingroup$ @MarianoSuárez-Alvarez thanks for the advice. $\endgroup$
    – user42912
    Nov 4, 2013 at 1:01

3 Answers 3

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Pull back the curve to $P^2$. There it has an equation: can you use it to construct $V$?

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  • $\begingroup$ How can I conclude that the curve in $P^2$ is of degree $2$? $\endgroup$
    – user371231
    Nov 18, 2021 at 9:22
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For one, let $f(x_0,x_1)$ be homogeneous of degree $d$ in $S(\mathbb{P}^2)$. Then

\begin{align*} f = \sum a_I x^I \text{ homogeneous of degree }d \implies f^2(x_0,x_1) = \sum a_Ia_J x^I x^J \\ \text{ is homogeneous of degree }2d \text{ in }x_0,x_1 \\ \implies f^2 \in k[x_0^2, x_0x_1, x_1^2, x_1x_2, \dots, x_2x_0] \end{align*} since $f^2$ is degree $d$ in $x_0^2, x_0x_1, \dots, x_2x_0$.

This choice of how to write $f^2$ in terms of $x_0^2, ...x_2x_0$ is not unique, but will be unique in $\nu_2(\mathbb{P}^2)$, where $\nu_2$ is the 2-uple embedding.

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  • $\begingroup$ Can you explain the last line of your solution in more detail ? I understood that we can write $f^2$ in more than one way, but why it is unique after considering that embedding? $\endgroup$ Jul 11, 2023 at 12:41
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I have the feeling that this could be illuminated by an example. Consider the quadric $$C = \{x_0^2 + x_1^2 + x_2^2 = 0\} \subset \mathbf P^2,$$ so that $f = x_0^2 + x_1^2 + x_2^2$. Now take it's square $$f^2 = x_0^4 + x_1^4 + x_2^4 + (x_0 x_1)^2 + (x_1x_2)^2 + (x_2x_0)^2. \tag{1}$$ Clearly set-theoretically¹ $V(f) = V(f^2)$. Let $y_0, \dotsc, y_5$ be the coordinates on $\mathbf P^5$, and consider the polynomial $$g = y_0^1 + y_1^2 + y_2^2 + y_3^2 + y_4^2 + y_5^2 \tag{2},$$ so that $g(x_0^2, x_1^2, x_2^2, x_0 x_1, x_1, x_2, x_2 x_0) = f^2$. This equation means exactly, that $$v(\mathbf P^2) \cap V(g) = V(f^2),$$ because the Veronese embedding $v: \mathbf P^2 \to \mathbf P^5$ is defined by $$[x_0:x_1:x_2] \mapsto [x_0^2: x_1^2:x_2^2, x_0x_1, x_1x_2,x_2x_0].$$


¹ If you don't know about schemes yet, ignore this.

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    $\begingroup$ Maybe this isn't the most instructive example because the curve is already $V(y_0+y_1+y_2)$ intersected with the Veronese surface (no squaring necessary). $\endgroup$
    – KReiser
    Apr 19, 2022 at 19:49

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