Timeline for Hartshorne Exercise III.4.7 (cohomology of closed subschemes in $\mathbb{P}^2$)
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2010 at 7:45 | vote | accept | Martin Brandenburg | ||
| May 31, 2010 at 14:36 | comment | added | Frank | $dim H^1(X, O_X) = dim h^0(X, \Omega_X)$ by Serre duality (Hartshorne III.7.7). The geometric genus (=arithmetic genus in dimension 1) is given by the latter and is computed by what is known as the degree-genus formula. That's the exercise you stated and proved by Cech cohomology but can be proved in a couple of ways. For example there's a more general geometric argument which works for any non-singular curve on a surface which is given by the adjunction formula (Hartshorne V.1.5). See example 1.5.1 in loc. cit. | |
| May 31, 2010 at 13:17 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
added 36 characters in body
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| May 31, 2010 at 12:41 | answer | added | Sasha | timeline score: 4 | |
| May 31, 2010 at 12:39 | comment | added | Martin Brandenburg | What do you mean by adjunction here? | |
| May 31, 2010 at 10:17 | comment | added | Frank | Yeah. Also it's not surprising since h^1 is the genus which you could have computed using adjunction. | |
| May 31, 2010 at 9:11 | comment | added | Martin Brandenburg | Ah, the dimensions don't change because of the flat base change theorem? | |
| May 31, 2010 at 9:10 | comment | added | Martin Brandenburg | I suspected that this hypothesis was in the first chapter of Hartshorne. | |
| May 31, 2010 at 9:08 | comment | added | Robin Chapman | It's pretty much the default hypothesis in Hartshorne that $k$ is an algebraically closed field. If you are worried about the finite field case, consider the same variety over $k^{alg}$ and show that the dimensions of cohomology don't change. | |
| May 31, 2010 at 7:36 | history | asked | Martin Brandenburg | CC BY-SA 2.5 |