Algebraic de Rham cohomology for singular varieties

Question

I would like to know to what extent the naive algebraic de Rham cohomology is a "bad" cohomology theory. If $X$ is smooth then there is a comparison theorem with singular cohomology. If $X$ is singular, then Hartshorne embeds $X$ in a smooth variety, generalises the definition and proves a similar comparison theorem singular cohomology, Poincaré duality and other things (http://www.numdam.org/article/PMIHES_1975__45__5_0.pdf).

In this thread (naive de Rham cohomology fails for singular varieties), we learn that the naive De Rham cohomology is, for sure, not the same as singular cohomology and does not satisfy Poincaré duality.

Despite these negative results, my question is the following.

If $X$ is singular, can we still prove that its naive de Rham cohomology is finitely generated? Can we prove that it is zero above its dimension?

Any known results or references are welcome.

Answer 1

Here is an answer for the naive de Rham cohomology $\mathbb{H}(X, \Omega_X^{\bullet})$ (not the more sophisticated one of the linked article by Hartshorne (which involves chooseing an embedding $X \subset Y$ in a smooth variety $Y$, completion along $X$ etc.)).

We can use the hypercohomology spectral sequence $$ E_1^{pq} = H^q(X, \Omega_X^p) \implies \mathbb{H}^{p+q}(X, \Omega_X^\bullet) $$ (a.k.a. the Hodge-to-de Rham spectral sequence). Assuming $X$ is proper all terms on this $E_1$-page are finite dimensional, hence so is the de Rham cohomology. Setting $n = \dim X$, by Grothendieck's vanishing theorem and the fact that the complex $\Omega_X^\bullet$ has length $n$ $$ H^q(X, \Omega_X^p) = 0 \text{ for } p > n \text{ or } q > n $$

and hence $\mathbb{H}^i(X, \Omega_X^\bullet) = 0$ for $i > 2 \dim X$.