Why this property holds in a Veronese surface

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $$[x_0:x_1:x_2] \mapsto [x_0^2: x_1^2:x_2^2, x_0x_1, x_1x_2,x_2x_0].$$

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