Trouble with semicontinuity

Question

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.

1. It seems to me that......according to your answer, calculation leads to $h^1=0$. (I forgot about flatness and just calculated it)
2. 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.

Answer 1

It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$ (i.e., the arithmetic genus is $0$), it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

The corresponding calculation for $t\neq 0$ gives $h^1(\mathscr X_t,\mathscr O_{\mathscr X_t})=0$ and so the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ in this case implies that then $h^2(\mathbb P^4,\mathscr I_t)=0$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.

Addendum Here is an argument to prove that $x_2$ is indeed a global regular function on $\mathscr X_0$:

Using choa's description of the ideal we have that $$ \mathscr I_0=\mathscr J + (x_0x_2,x_1x_2,x_2^2,x_3x_2,x_4x_2). $$ where $\mathscr J$ is the ideal sheaf of a rational normal quartic curve in $\mathbb P^3_{x_0,x_1,x_3,x_4}$.

Consider the affine charts $U_i=(x_i\neq 0)\subset \mathbb P^3$ for $i=0,1,3,4$. Observe that $U_i\simeq \mathbb A^3$ with coordinates $y_j=\dfrac{x_j}{x_i}$ ($j\neq i$). In these coordinates $\mathscr J$ becomes very simple. For simplifying the notation I will work on $U_0$, but the other charts work the exact same way. So, $\mathscr J|_{U_0}=(y_3-y_1^3, y_4-y_1^4)$ and hence the affine coordinate ring of $\mathscr X_0\cap U_0$ is isomorphic to $k[y_1,x_2]/(y_1x_2,x_2^2)$. In particular, $x_2\in \Gamma(U_0,\mathscr O_{\mathscr X_0})$. Similarly, the affine coordinate ring of $\mathscr X_0\cap U_1$ is isomorphic to $k[y_0, y_0^{-1},x_2]/(y_0x_2,x_2^2)$ and so $x_2\in \Gamma(U_1,\mathscr O_{\mathscr X_0})$, the affine coordinate ring of $\mathscr X_0\cap U_3$ is isomorphic to $k[y_4, y_4^{-1},x_2]/(y_4x_2,x_2^2)$ and so $x_2\in \Gamma(U_3,\mathscr O_{\mathscr X_0})$ and the affine coordinate ring of $\mathscr X_0\cap U_4$ is isomorphic to $k[y_3,x_2]/(y_3x_2,x_2^2)$ and so $x_2\in \Gamma(U_4,\mathscr O_{\mathscr X_0})$.
Therefore $x_2$ is regular on each affine chart of a covering and hence it is a global section.

Note that the arithmetic genus of $\mathscr X_0$ is still $0$ since $\chi(\mathscr O_{\mathscr X_0})=h^0-h^1=2-1=1$.

Answer 2

Nothing, as far as I can tell*. That is, I think you're right. Congratulations! You've found a mistake in H, which is not that easy to do**. (I suspect it was more of a typo. He probably left out a twist or something, but I'm not really sure what he intended.)

• Actually, I only checked your $h^0$. It does seem that there is a jump in $h^1$ and $h^2$ as Sándor points out. So his explanation 1 of the typo seems highly plausible.

** I have to confess that I'm amazed by just how few errors there are.

Answer 3

I have problems to verify your calculations with Macaulay 2, concretely the fiber over $t = 0$ (resp. $a=0$ as its called by me).

I used the description given at the OP for the ideal $\mathcal{I}$ and computed it as id1 by elimination. With the map phi I set $a=0$ and got the ideal id11 in S2=QQ[w_0..w_4]. This is the homogeneous coordinate ring of the $\mathbb{P}^4_{QQ}$ which is the fiber of $\mathbb{P}^4_{QQ[a]}$ over $a=0$.

Calculating cohomology in $\mathbb{P}^4_{QQ}$ I get all cohomologies zero for id11.

I would be very happy if someone could reconcile my calculation with the results above and find a possible mistake that I have made.

A=QQ[a,Degrees=>{1:{}}];


S=A[x_0..x_4];

T=A[t,u];

phi = map(T,S, {t^4, t^3*u, a* t^2 * u^2, t * u^3, u^4});

id1 = ker phi;

                                        2                             2          3      2     2    2     3    2       2    2           2        2     2        2
ideal (x x  - x x , - x x  + a*x x , a*x  - x x , - x x  + a*x x , a*x  - x x , x  - x x , x x  - x x , x  - x x , - x  + a x x , - x x  + a*x x , - x x  + a*x x )
        1 3    0 4     2 3      1 4     3    2 4     1 2      0 3     1    0 2   3    1 4   0 3    1 4   1    0 3     2      0 4     2 3      0 4     1 2      0 4

S2=QQ[w_0..w_4]

phi=map(S2,S,{w_0,w_1,w_2,w_3,w_4})


id11 = phi(id1)

                                                 3      2     2    2     3    2      2      2    2
ideal (w w  - w w , -w w , -w w , -w w , -w w , w  - w w , w w  - w w , w  - w w , -w , -w w , -w w )
        1 3    0 4    2 3    2 4    1 2    0 2   3    1 4   0 3    1 4   1    0 3    2    2 3    1 2

i73 :  for i from 0 to 4 list prune HH^i(sheaf module id11)

o73 = {0, 0, 0, 0, 0}

o73 : List

i74 : for i from 0 to 4 list prune HH^i(sheaf S2^1/id11)

         1
o74 = {QQ , 0, 0, 0, 0}

o74 : List