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Martin Brandenburg
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I have some questions about the following exercise in Hartshorne (III.4.7):

Let $f \in k[x_0,x_1,x_2]$ be a homogeneous polynomial of degree $d \geq 1$ and $f \neq 0$ and let $X$ be the closed subscheme of $\mathbb{P}^2_k$ defined by $f$. Then $\dim H^0(X,\mathcal{O}_X) = 1, \dim H^1(X,\mathcal{O}_X) = (d-1)(d-2)/2$. This is done using Cech cohomology.

1 - Hartshorne makes the assumption $f(1,0,0) \neq 0$. Is this necessary?

This implies that $f$ is monic in $x_0$ and yields a very nice description of the Cech complex (if necessary, I'll add this), which makes the computation possible. But what about the general case?

It's not hard to see that $f$ is mapped by a graded isomorphism of $k[x_0,x_1,x_2]$ to a polynomial, which does not vanish in $(1,0,0)$, if and only if $f$ does not vanish on $k^3$. Thus if $k$ is infinite, you're done. But what happens when $k$ is finite? For example

$f = xy \prod_{\alpha \in k} (x - \alpha y)$

is a nontrivial homogeneous polynomial of degree $|k|+2$ and vanishes on $k^2$ (and thus on $k^3$).

2 - Is the finite case important for some applications (for example in arithmetic geometry)?

3 - Is it surprising that the cohomology only depends on $d$?

I have some questions about the following exercise in Hartshorne (III.4.7):

Let $f \in k[x_0,x_1,x_2]$ be a homogeneous polynomial of degree $d \geq 1$ and $f \neq 0$ and let $X$ be the closed subscheme of $\mathbb{P}^2_k$ defined by $f$. Then $\dim H^0(X,\mathcal{O}_X) = 1, \dim H^1(X,\mathcal{O}_X) = (d-1)(d-2)/2$.

1 - Hartshorne makes the assumption $f(1,0,0) \neq 0$. Is this necessary?

This implies that $f$ is monic in $x_0$ and yields a very nice description of the Cech complex (if necessary, I'll add this), which makes the computation possible. But what about the general case?

It's not hard to see that $f$ is mapped by a graded isomorphism of $k[x_0,x_1,x_2]$ to a polynomial, which does not vanish in $(1,0,0)$, if and only if $f$ does not vanish on $k^3$. Thus if $k$ is infinite, you're done. But what happens when $k$ is finite? For example

$f = xy \prod_{\alpha \in k} (x - \alpha y)$

is a nontrivial homogeneous polynomial of degree $|k|+2$ and vanishes on $k^2$ (and thus on $k^3$).

2 - Is the finite case important for some applications (for example in arithmetic geometry)?

3 - Is it surprising that the cohomology only depends on $d$?

I have some questions about the following exercise in Hartshorne (III.4.7):

Let $f \in k[x_0,x_1,x_2]$ be a homogeneous polynomial of degree $d \geq 1$ and $f \neq 0$ and let $X$ be the closed subscheme of $\mathbb{P}^2_k$ defined by $f$. Then $\dim H^0(X,\mathcal{O}_X) = 1, \dim H^1(X,\mathcal{O}_X) = (d-1)(d-2)/2$. This is done using Cech cohomology.

1 - Hartshorne makes the assumption $f(1,0,0) \neq 0$. Is this necessary?

This implies that $f$ is monic in $x_0$ and yields a very nice description of the Cech complex (if necessary, I'll add this), which makes the computation possible. But what about the general case?

It's not hard to see that $f$ is mapped by a graded isomorphism of $k[x_0,x_1,x_2]$ to a polynomial, which does not vanish in $(1,0,0)$, if and only if $f$ does not vanish on $k^3$. Thus if $k$ is infinite, you're done. But what happens when $k$ is finite? For example

$f = xy \prod_{\alpha \in k} (x - \alpha y)$

is a nontrivial homogeneous polynomial of degree $|k|+2$ and vanishes on $k^2$ (and thus on $k^3$).

2 - Is the finite case important for some applications (for example in arithmetic geometry)?

3 - Is it surprising that the cohomology only depends on $d$?

Source Link
Martin Brandenburg
  • 63.1k
  • 12
  • 207
  • 424

Hartshorne Exercise III.4.7 (cohomology of closed subschemes in $\mathbb{P}^2$)

I have some questions about the following exercise in Hartshorne (III.4.7):

Let $f \in k[x_0,x_1,x_2]$ be a homogeneous polynomial of degree $d \geq 1$ and $f \neq 0$ and let $X$ be the closed subscheme of $\mathbb{P}^2_k$ defined by $f$. Then $\dim H^0(X,\mathcal{O}_X) = 1, \dim H^1(X,\mathcal{O}_X) = (d-1)(d-2)/2$.

1 - Hartshorne makes the assumption $f(1,0,0) \neq 0$. Is this necessary?

This implies that $f$ is monic in $x_0$ and yields a very nice description of the Cech complex (if necessary, I'll add this), which makes the computation possible. But what about the general case?

It's not hard to see that $f$ is mapped by a graded isomorphism of $k[x_0,x_1,x_2]$ to a polynomial, which does not vanish in $(1,0,0)$, if and only if $f$ does not vanish on $k^3$. Thus if $k$ is infinite, you're done. But what happens when $k$ is finite? For example

$f = xy \prod_{\alpha \in k} (x - \alpha y)$

is a nontrivial homogeneous polynomial of degree $|k|+2$ and vanishes on $k^2$ (and thus on $k^3$).

2 - Is the finite case important for some applications (for example in arithmetic geometry)?

3 - Is it surprising that the cohomology only depends on $d$?