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Let $K$ be a finite type field extension of $\mathbf{Q}$ and let $Y$ be a smooth quasi-projective variety over $K$. Let $\Omega_{Y/K}^{\bullet}$ denote the complex of sheaves of (algebraic) regular differential forms on $Y$. Recall that the $k$-th De Rham cohomology group of $Y$ is defined as $$ H^k(Y):=\mathbb{H}^k(Y/K,\Omega_{Y/K}^{\bullet}), $$ where $\mathbb{H}$ stands for the hypercohomology of a complex.

Q: How do you prove in a purely algebraic way that $H^k(Y)$ is a finite dimensional $K$-vector space?

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  • $\begingroup$ Am I allowed to use resolution of singularities? $\endgroup$ Commented Aug 10, 2011 at 19:05
  • $\begingroup$ Well, if you use Hironaka's desingularization theorem then I guess that you can express (using the Leray spectral sequence) the cohomology of $Y$ in terms of the cohomology of a compactification of $Y$ and of the normal crossing divisors. Is it what you have in mind ? $\endgroup$ Commented Aug 10, 2011 at 19:09
  • $\begingroup$ Exactly. I decided to write it out below. Without a normal crossing compactification, I really don't know how you would do this, even for $Y$ affine. So I wil be curious to see what answers people come up with. $\endgroup$ Commented Aug 10, 2011 at 19:28

2 Answers 2

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$\def\HH{\mathbb{H}}$The following sketch of an argument is taken from Grothendieck's On the de Rham cohomology of algebraic varieties. You can also find a good discussion in Voisin's Hodge Theory book, volume 1, chapter 8. I talked about it a little on the last day of my Hodge theory class.

Note: Every step here is meant to be nontrivial.

Step 1: Using resolution of singularities, compactify $Y$ as $X \setminus \bigcup D_i$, where $\bigcup D_i$ is a simple normal crossings divisor and $X$ is smooth and projective.

Step 2: Let $M \Omega^{\bullet}$ be the complex of sheaves on $X$ obtained by pushing forward $\Omega^{\bullet}$ from $Y$. Argue that $\HH(X, M\Omega^{\bullet}) = \HH(Y, \Omega^{\bullet})$.

Step 3: Let $\Omega^j(\log D)$ be the following subsheaf of $M \Omega^j$: Near a point $x \in X$, with local coordinates $(x_1, \ldots, x_n)$, and where $\bigcup D_i$ is locally cut out by $x_1 x_2 \cdots x_k=0$, a differential form is in $\Omega^j(\log D)$ if it is locally of the form: $$\sum f_{i_1 \ldots i_r j_1 \ldots j_s} \bigwedge_{1 \leq i_1 < \cdots < i_r \leq k} \frac{d x_{i_a}}{x_{i_a}} \wedge \bigwedge_{k+1 \leq j_1 < \cdots < j_s \leq n} d x_{j_b}.$$ where the $f_{i_1 \ldots i_r j_1 \ldots j_s}$'s are in the local ring at $x$.

So $\Omega^{\bullet}(\log D)$ is a subcomplex of $M \Omega^{\bullet}$. Define $Z^j$ to be the quotient sheaf $M \Omega^j/\Omega^j(\log D)$.

Show that $Z^{\bullet}$ is acyclic. Deduce that $\HH(M \Omega^{\bullet}) = \HH(\Omega^{\bullet}(\log D))$.

Step 4: Write out the spectral sequence for hypercohomology for $\Omega^{\bullet}(\log D)$. On page $2$ (or is it $1$?), all of the terms look like $H^q(X, \Omega^p(\log D))$. Observe that these terms are all finite dimensional, because they are the cohomology of a coherent sheaf on a projective scheme. Deduce that $\HH(\Omega^{\bullet}(\log D))$ is finite dimensional.

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I'm quite happy with David's answer, but perhaps I can also mention the following references:

Hartshorne, "On the algebraic de Rham cohomology of algebraic varieties" IHES (1975), chap III theorem 2.1.

Monsky, "Finiteness of de Rham cohomology", Amer. J Math (1972)

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  • $\begingroup$ Do any of these papers work in characteristic p? can we use alterations instead of resolution for instance? $\endgroup$
    – anon
    Commented Aug 10, 2011 at 21:40
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    $\begingroup$ I'm not sure how much of it works, but certainly the result in the strongest form is false in char p, e.g. $H^1_{DR}(\mathbb{A}^1)$ is infinite dimensional! $\endgroup$ Commented Aug 10, 2011 at 22:11
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    $\begingroup$ There are finiteness results for various "p-adic cohomologies",related to De Rham cohomology, for characteristic p varieties. I'm told that Mebkhout used some of the tools from my paper in giving a "local" proof of such finiteness theorems. I think Kedlaya is the person who can tell you about this. $\endgroup$ Commented Aug 11, 2011 at 0:26
  • $\begingroup$ It's nice to (virtually) meet you Prof. Monsky. $\endgroup$ Commented Aug 11, 2011 at 0:40

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