Exercise 2.18 in Hartshorne (chapter I) is:
Let $C \subset \Bbb P^2$ be a closed curve (a variety of dimension 1). Then there exists a hypersurface $H \subset \Bbb P^5$ such that $\nu_2 (C)=H \cap \nu _2 (\Bbb P^2)$.
This can be proved by writing $C=\Bbb V(f)$ for $f \in k[x_0,x_1,x_2]$ an homogeneous polynomial, noting $\Bbb V(f)=\Bbb V(f^2)$ and $f^2$ can be seen as a polynomial $g$ in the six variables $x_0^2,x_0x_1,...,x_2^2$, so taking $H=\Bbb V(g)$ works.
What is the analogous statement in the category of schemes? Does it hold?
My try:
To reformulate the claim in a scheme setting, we take:
$\nu_2 (\Bbb P^2)=\operatorname{Proj} \bigoplus \limits_{d=0}^\infty k[x_0,x_1,x_2]_{2d} = \operatorname{Proj} \frac{k[y_0,...,y_5]}{(y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$.
$H=\operatorname{Proj} \frac{k[y_0,...,y_5]}{(g)}$, for some homogeneous polynomial $g \in k[y_0,...,y_5]$.
And the intersection of these two corresponds to the sum of the ideals by which we divide.
- $\nu _2 (C)$ may be replaced by a closed subscheme of dimension 1 of $\nu_2 (\Bbb P^2)$, and a closed subscheme of $\operatorname{Proj} S$ is always of the form $\operatorname{Proj} S/I$. There are two other approaches that seem possible, though: it might be realized as $\operatorname{Proj} \bigoplus \limits_{d=0}^\infty \frac{k[x_0,x_1,x_2]_{2d}}{(f)_{2d}}$ for some $f$, or as $\operatorname{Proj} \frac{k[y_0,...,y_5]}{I(\nu_2(C))}$ where $\nu_2(C)$ is the familiar veronese embedding of a variety. I'm not sure what are the differences between these 3 approaches, so any of them is interesting.
Taking it all together, the question looks unwieldy:
Is there a scheme $D=\operatorname{Proj} \frac{k[y_0,...,y_5]}{I+(f,y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$ with dimension 1 such that there is no $g$ with $$D=\operatorname{Proj} \frac{k[y_0,...,y_5]}{(g,y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)} $$?
I don't know if the considerations in this generalization are true or if there is a simpler way of looking at things. A drawback of this method is that with it it's unclear even what corresponds to the choice of $C=\{x_0=0\} \subset \Bbb P^2$, to try a concrete simple case.
The proof for varieties uses $\Bbb V(f)=\Bbb V(f^2)$ which does not hold for schemes, since $\operatorname{Proj} \frac {k[x_0,...,x_n]}{(f)} \neq \operatorname{Proj} \frac {k[x_0,...,x_n]}{(f^2)}$. Thus I expect a counterexample.
I don't see why you need all this computations in coordinates for schemes. Map $\nu_2$ is given by complete linear system of quadrics on $\mathbb{P}^2$ and $\nu_2^* \mathcal{O}_{\mathbb{P}^5}(1)= \mathcal{O}_{\mathbb{P}^2}(2)$. Hypersurface $H$ is a divisor in the linear system $|\mathcal{O}_{\mathbb{P}^5}(d)|$, where $d= \deg H$. When you pull it back to $\mathbb{P}^2$ or in other words consider intersection $H \cap \nu _2 (\Bbb P^2)$ you get a divisor on $\Bbb P^2$ of degree $2d$. So, in this way you can only get curves of even degree.
Now, your construction with $f^2$ works for varieties and does not work for schemes. In the case of schemes you get first infinitesimal neighbourhood of your curve in $\Bbb P^2$.
In general, if a subscheme $Z$ is defined by a sheaf of ideals $\mathcal{I}_Z$ that is $\mathcal{O}_Z =\mathcal{O}_X/\mathcal{I}_Z$, then first infinitesimal neighbourhood is $\mathcal{O}_Z ^{(1)}=\mathcal{O}_X/\mathcal{I}_Z^2$. It is the same as topological subspace but the structure sheaf is different.
Finally, a sheaf of ideals on $\Bbb P^2$ defining a curve $C$ is $\mathcal{O}(-C)$ and the sheaf of ideals of the first infinitesimal neighbourhood is $\mathcal{O}(-2C)$. Therefore first infinitesimal neighbourhood is always of even degree.