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.


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.


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.

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    $\begingroup$ Small nitpick of language: "what's wrong with my argument?" is a better way to put things. (We have no way of answering the question "what's wrong with me?" ;) ) $\endgroup$ – Yemon Choi Mar 5 '12 at 8:19
  • $\begingroup$ It will be easier to see what is wrong with your argument if you show us your argument! In particular, what flat family are you using and how did you compute what you computed? $\endgroup$ – Mariano Suárez-Álvarez Mar 5 '12 at 8:23
  • $\begingroup$ Thanks for your kind comment. I should try to more specific next time. $\endgroup$ – Choa Mar 5 '12 at 8:50

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 $\chi(\mathscr O_{\mathscr X_0})=h^0-h^1=2-1=1$.

  • $\begingroup$ Thank you so much! One more question; is that nilpotent function are really "global"? I mean, when I considered that $x_{2}$, I concluded that it cannot be captured as a global section. $\endgroup$ – Choa Mar 6 '12 at 0:44
  • $\begingroup$ When we take $Proj$ operation, cutting out lower degree parts changes nothing, so I ignored $x_{2}$...... $\endgroup$ – Choa Mar 6 '12 at 0:48
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    $\begingroup$ Dear choa, the thing is, that this $x_2$ is a renegade on $X_0$, because $X_0$ is inside $x_2=0$. If you look at affine charts on $X_0$, you see that $x_2$ survives in all of them, so it glues to give a global section. On the entire $\mathbb P^4$ it does not give a section on the affine charts either, but in the $\mathbb P^3$ defined by $x_2=0$ it is independent of the variables. $\endgroup$ – Sándor Kovács Mar 6 '12 at 2:21
  • $\begingroup$ Thanks for yourntrue kindness and spending your time for such a tedious calculation to check everything for me:) Thank you very much. $\endgroup$ – Choa Mar 7 '12 at 1:20
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    $\begingroup$ Playing around with this in Macaulay, I'm skeptical that $x_2$ is actually a nilpotent. Shouldn't the special fiber just be the reduced quartic curve? The ideal choa provided for the special fiber is not saturated, and does not generate the ideal in affine charts. One way to "fix" the exercise would be to look at $H^i(t,I_t(1))$ for $i=0,1$; then its just the statement that the special fiber is contained in a plane. $\endgroup$ – Jack Huizenga Mar 8 '12 at 1:35

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.


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.




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



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)

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

o74 : List

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