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I'm reading about algebraic de Rham cohomology over characteristic zero which is constructed using hypercohomology. Already, constructing injective resolutions is difficult, and coupling this with finding an injective resolution of the de Rham complex makes this extraordinarily difficult to figure out how to compute an example. Where can I learn about computational techniques?

For example, how can I compute the algebraic de rham cohomology of the cubic surface $$ \textbf{Proj}\left(\frac{\mathbb{C}[x,y,z,w]}{(x^3 + y^3 + z^3 + w^3)}\right) $$ ?

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    $\begingroup$ I've seen explicit calculations of algebraic de Rham cohomology for curves (via an explicit injective resolution involving big direct sums of skyscraper sheaves). Of course you realise that algebraic de Rham cohomology is isomorphic via GAGA to the hypercoh of the de Rham complex in the analytic category and hence to usual de Rham cohomology? So that's a way to compute it but probably not what you want. $\endgroup$
    – znt
    Commented Jul 12, 2016 at 20:31
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    $\begingroup$ I realize this, but I the main point I care about is the question in the title. $\endgroup$
    – 54321user
    Commented Jul 12, 2016 at 20:43
  • $\begingroup$ The discussions/exercises/examples in Eisenbud's Commutative Algebra book might be helpful, although I admit that I can't think of precise references off the top of y head $\endgroup$
    – Yemon Choi
    Commented Jul 12, 2016 at 21:34
  • $\begingroup$ In your given example, I think the Kaehler module shouldn't be two hard to work out (see my earlier comments about Eisenbud's book) and then the de Rham complex is, IIRC, given by the exterior powers of this module. Perhaps the terms in the de Rham complex are then sufficiently nice for one to spot ad hoc injective resolutions? $\endgroup$
    – Yemon Choi
    Commented Jul 12, 2016 at 21:36
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    $\begingroup$ I cannot remember the details, but someone once showed me how to prove that hypercohomology is closed differentials of the second or third kind (I can't even remember the terminology!) modulo exact meromorphic differentials using this. You resolve the structure sheaf by embedding it into something like the sheaf of meromorphic functions, and then onto something like the direct sum over all x of (K_x/O_x) where K_x is the field of fractions of O_x and these are local rings concentrated at one point. You do the same with the 1-forms and then follow your nose. I'm sure this must be a std calcn? $\endgroup$
    – znt
    Commented Jul 14, 2016 at 12:45

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Since you are asking about computing algebraic de Rham, don't use injective resolutions, since they are not really constructive. You can use a Cech complex: If $\{U_i\}$ is an affine open cover of your variety $X$, form a double complex $C^{\bullet\bullet}=C^\bullet(\{U_i\}, \Omega_X^\bullet)$ with Cech coboundary in one direction, and exterior derivative in the other direction. Then $$H^i_{dR}(X) \cong H^i(Tot(C^{\bullet\bullet}))$$ Depending on the example, this may not be a bad way to proceed.

If you want a quick and dirty approach for your specific example: use Grothendieck's algebraic de Rham theorem to rewrite it in terms of singular cohomology. Then you have access to all the standard exact sequences. You can also look at Hartshorne's paper in IHES on de Rham cohomology, where he reproduces most of these standard results by algebraic arguments. Using the fact that a smooth cubic surface is the blow up of $\mathbb{P}^2$ at $6$ points, the Betti numbers are 1,0,7,0,1,0...

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