I'm trying to understand following Example in Hartshorne (Chapter III, Example 9.8.2 & Example 12.9.2):
Let $X_1 \subset \mathbb{P}^{3}$ be a twisted cubic curve not containing the point $(0:...:1) $ (will become important when we project $X_0$ to $\mathbb{P}^{2}$) and consider the automorphisms parametrized by $\sigma_t, t \in k$ of $\mathbb{P}^{3}$ defined by $(x_0:...:x_3) \mapsto (x_0:...:t \cdot x_3)$. Let $X_t:= \sigma_a(X_t)$. For $t \in k \neq 0$ all $X_t$ are isomorphic as abstract schemes to $X_1$. Then the $X_t$ form a flat family parametrized by $\mathbb{A}^1 - \{0\}$.
Now according to (9.8) this family extends uniquely to a flat family defined over all of $\mathbb{A}^1$, and with $X_0$ happen crazy things.
Following Hartshorne it can be proved (for details see book pages 259/260 ) that set theoretically $X_0$ coincides with $X_1$ but as a scheme $X_0$ admits an embedded component $(0:0:0)$, in this case it is a "double point" and the concern of this question is how does it effect on cohomology.
Recall that the twisted cubic curve in $\mathbb{P}^{3}$ was defined by the parametric equations $x_0 = t^3 , x_1 = t^2 u, x_2 = tu^2 , x_3 = u^3$ . In other words, it is just the $3$-uple embedding of $\mathbb{P}^{1}$ in $\mathbb{P}^{3}$ (Chap. II Example 7.8.5 and Chap I, Ex. 2.12).
Example 12.9.2 (page 289) contains some things about the cohomology of the members of the family $X_a$ I not pretty understand. It says:
Example 12.9.2. In the flat family of (9.8.4), we have $h^0 (X_t, O_{X_i}) = 1$ if $t \neq 0$, and $2$ if $t = 0$, because of the nilpotent elements. On the other hand, $h^1 (X_t, O_{X_t}) = 0$ for $t \neq 0$, since $X$, is rational, and $h^1 (X_t, O_{X_t}) =h^1 (X_t, O_{X_t})_{red})=1$ since $(X_0)_{red}$ is a plane cubic curve. So in this case the functions $h^0 ,h^1$ both jump up at $t = 0$.
Several caclulations I not completly understand and would like to discuss:
I) on $h^0 (X_t, O_{X_t}) = 1$
Since $X_1 \cong X_t$ for $t \neq 0$ we can assume $t=1$. $X_1$ is a twisted cubic curve of degree $3$ embedded by $3$-uple Veronese $\nu: \mathbb{P}^{1} \to \mathbb{P}^{3}$. The first idea came up to me is to use the overkiller $X_1 \cong \mathbb{P}^{1}$ and to think only about the cohomology of $\mathbb{P}^{1}$. I think that Hartshorne used at this point another argument to reach $h^0 (X_t, O_{X_t}) = 1$, because looking at the used argument for $h^1 (X_t, O_{X_t}) = 0$ where $t \neq 1$ he only used that $X_1$ is rational, that is birational equivalent to $\mathbb{P}^{1}$, not neccessary isomorphic. And indeed genus $g_1=h^1 (X_1, O_{X_1})$ is by Theorem 8.19 a birational invariant.
This leads me to suspicion that to establish $h^0 (X_1, O_{X_1}) = 1$ Hartshorne used another argument. Which one? I'm looking for an argument using cohomology theory that for a rational, non singular irreducible cubic curve of degree $3$ in $\mathbb{P}^3$ the space of global sections has dimension one?
II) on $h^0 (X_0, O_{X_0}) = 2$
How the fact that $X_0$ is not reduced anymore rases the the dimension to $2$? More specificaly: the $X_0$ difers from $X_1$ by the observation the point $(0:0:0) \in X_0$ in an embedded point, $O_{X_0, (0:0:0)}$ has nilpotent elements. How to see formally that these contribute exactly an additional one dimension to the global sections: that is $h^0 (X_0, O_{X_0})= h^0 (X_1, O_{X_1})+1$?
III) $h^1 (X_t, O_{X_0}) =h^1 (X_t, O_{X_t})_{red})$
Why passing from $X_0$ to associated reduced $(O_{X_0})_{red}$ not changes $h^1$? (recall that the birationality argument not work anymore since $X_0$ isn't smooth.)
