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I'm trying to solve the exercise problem iii.7.4 in Hartshorne's AG book. It says that

"first chern class of the line bindle associated to a smooth subvarirety of codimen 1" = "cohomology class of that subvariety"

Suddenly it comes to me that I have NO! ideas about what Serre dualty looks like! There are no differential forms and integration. Everything is just so abstract, very hard to manipulate. Moreover, there are no exponential sequence, which makes the problem much easier in complex analytic case.

So my question is, is there any "explicit" formulation of the Serre duality? Of course, over arbitary (algebraically closed) field.

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  • $\begingroup$ The duality comes from a very explicit pairing $\mathrm{Ext}^i(F,\omega) \otimes \mathrm{H}^{n-k}(X,F) \to \mathrm{H}^n(X,\omega)$. $\endgroup$ Oct 28, 2011 at 12:21
  • $\begingroup$ I think that's a little unfair, Martin. Except when $X$ is a curve, it's pretty hard to determine when a class in $H^n(X,\omega)$ is zero, which makes the explicitness of that pairing less useful than you'd hope. $\endgroup$ Oct 28, 2011 at 12:37
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    $\begingroup$ Various people, including my colleague Lipman, have thought about how to make this (and more generally Grothendieck duality) more explicit. His book "Dualizing sheaves, differentials..." Asterisque (1984) might be a good place to start. The main issue as David says is to make the isomorphism $H^n(X,\omega)\cong k$ explicit in the algebraic setting. $\endgroup$ Oct 28, 2011 at 13:00
  • $\begingroup$ Amazon says that this book is out of print...... $\endgroup$
    – choa
    Oct 28, 2011 at 13:30
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    $\begingroup$ It's not too hard to make explicit on $\mathbb P^n$, at least. If you cover it by $n+1$ copies of $\mathbb A^{n+1}$, there is a natural basis for the tangent bundle or cotangent bundle which you can use to find a natural element of the canonical bundle, which will give a natural (nonzero) cohomology class. $\endgroup$
    – Will Sawin
    Oct 28, 2011 at 17:43

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