I'm in trouble with the exercise problem iii.12.3 in Hartshorne's AG.

Let $X_{1}$ be a rational normal curve in $\mathbb{P^4}$ ( = image of 4th veronese embedding of $\mathbb{P^1}$).

$X_{0}$ be a rational quartic curve in $\mathbb{P^3}$ with parametrization $[t^4, t^3u, tu^3, u^4]$.

Construct flat family $X$ using projection $\pi: \mathbb{P^4}\to \mathbb{P^3}$ parametrized by $\mathbb{A^1}$,with the given fibers $X_{1}$ and $X_{0}$ for $t=1$ and $t=0$.

More precisely, $X_{a}$ is parametrized by $[t^4, t^3u, at^2u^2, tu^3, u^4]$.

Let $\mathcal{I} \subset$ $\mathcal{O_{\mathbb{P^4}\times\mathbb{A^1}}}$ be a total idael sheaf of $X$.

Show that the functions $h^0(t,\mathcal I_{t})$ and $h^1(t,\mathcal{I}_{t})$ are jump at $t=0$.

Still, my calculation doesn't lead to such answer. Rather, it seems to be that they are constant functions.

After some calculation, I found out that

$\mathcal{I}\otimes k(t\ne 0)$ $\simeq$ idael sheaf of rational normal curve

$\mathcal{I}\otimes k(0)$ $\simeq$ [$(x_{2}x_{0}, x_{2}x_{1}, x_{2}^2, x_{2}x_{3}, x_{2}x_{4})$ + ideals of quartic rational cuves in $\mathbb{P^3}$]

In any cases, there zeroth and first cohomology group on $\mathbb{P^4}$ vanish, which comes from the exact sequence

$0$ $\rightarrow$ $\mathcal{I}$ $\rightarrow$ $\mathcal{O_{P^4}}$ $\rightarrow$ $\mathcal{O}_{closed subscheme} \rightarrow 0$

(By the flat base change theorem, they do computes functions $h^0$ and $h^1$ above)

I think above result is more acceptable because all fibers of $X$ are just $\mathbb{P^1}$.

What's wrong with this calculation? I'll appriciate any comments.

(edited)

Thank you, Sandor-kovacs. I'm really appriciate about your kind explaination. But still, there are two things make me confuse.

- It seems to me that......according to your answer, calculation leads to $h^1=0$. (I forgot about flatness and just calculated it)
- Honestly, I still don't know why $x_{2}$ becomes independent. In my opinion, the relations $x_{2}x_{0} = x_{2}x_{1}=x_{2}^2= x_{2}x_{3}=x_{2}x_{4}=0$ are already in the construction of the structure sheaf. After sheafifying those relation, $x_{2}$ itself vanishes at any affine chart, because $\frac {x_{2}}{x_{i}}=\frac{x_{2}x_{i}}{x_{i}^2}=0$.

I'ii appreciate to any feedbacks.

(edited)

I realize that my first question is foolish (I think that the geometric genus is 1), as Sander-Kovacs pointed out. But there are some problems remain.

At first, I think Sander-Kovacs' answer is right. But if local sections $x_{2}$ were "renegades" and form a single global section, then it does not contribute to higher cohomology modules. This was a real reason of my confusing.

Second question still remains. Precisely,

$x_2 = \frac {x_2 x_0}{x_0}\in (x_2x_0, x_2x_1, x_2^2, x_2x_3, x_2x_4)_{(x_0)}\subset \mathscr {I} (U_{0})$

(I assumed $x_0$ has degree 0, as Sander-kovacs implicitly do. But my argument is same even if it has degree 1)

There are some related problems. I'm continuously recalling that when one defines closed subscheme of $Proj S$ using its homogeneous ideal $I$, cutting off some lower degree part changes nothing, because saturation of ideal(not the ideal it self) determines scheme structure. (see Hartshorne's book ch ii, ex 3.12) Now, note that

$(x_0x_2,x_1x_2,x_2^2,x_3x_2,x_4x_2)$ = $\bigoplus_{d \geq 2} (x_2)_d$

If I were wrong, then there are very serious gaps and errors of my whole undegraduate AG study.

I'll waiting for any comments

p.s Thank you Yemon Choi. I did not know about MO's policy, and I'm not good at English..... Apologize for those mistakes.