$\newcommand{\Hdr}{H_{\mathrm{dRh}}}$ $\newcommand{\spec}[1]{\mathrm{spec}(#1)}$ $\require{amsmath}$

Let $A = k[x_1,\ldots,x_n]$ the polynomial ring over a field $k$ of characteristic zero and $I \subseteq A$ an ideal of $A$. Let $Y= V(I) = \spec{A/I}$ and $X = \spec{A}$.

According to R. Hartshorne, On the de Rham cohomology of algebraic varieties, Publications mathématiques de l’I.H.É.S., tome 45 (1975), p. 5-99 the algebraic de Rham cohomology of $Y$ is defined as follows:

Let $\hat{X}$ be the ringed space with $Y$ as a topological space and the sheaf derived from $\varprojlim_r A/I^r = \hat{A}$ as structure sheaf. Additionally let $\widehat{\Omega_X^\bullet}$ be the completion of $\Omega_{X|k}^\bullet$ (the exterior powers of the ordinary Kähler-differentials) at $I$, so that $\widehat{\Omega_X^\bullet} = \Omega_{X|k}^\bullet \otimes_A \hat{A}$. The exterior derivative extends to $d^p:\widehat{\Omega_X^p} \to \widehat{\Omega_X^{p+1}}$.

Now Hartshorne defines

$$\Hdr^p(Y) = \mathbb{H}^p(\hat{X}, \widehat{\Omega_X^\bullet})$$

with $\mathbb{H}$ standing for the hypercohomology.

If $Y$ is a smooth $k$-scheme, then by Hartshorne's Proposition 1.1, one can replace $X$ by $Y$ and get

$$(*) \quad \quad \Hdr^p(Y) = \mathbb{H}^p(Y, \Omega_{Y|k}^\bullet) = H^p(Y, \Omega_{Y|k}^\bullet)$$

where $\Omega_{Y|k}^\bullet$ is the complex of exterior powers of the ordinary Kähler-differentials of $Y/k$ and at the right stands ordinary cohomology, because of affine-ness of the situtation.

QUESTION: Does this equality (*) still hold, when $Y$ is no longer a smooth scheme over $k$?

The answers to

algebraic de Rham cohomology of singular varieties

seem to suppose so, but I could not find a proof by searching with google or on my own.

  • $\begingroup$ This question seems to be a duplicate of mathoverflow.net/questions/256379/… which in fact has an answer in the statement (a counterexample due to Arapura-Kang: $Y : x^5 + y^5 + x^2 y^2 = 0 $). $\endgroup$ Commented Jul 29, 2018 at 18:21
  • $\begingroup$ The cited question asks for an example where the algebraic de Rham cohomology does not coincide with singular cohomology. My question is about two different definitions of algebraic de Rham cohomology in the singular case. $\endgroup$ Commented Jul 29, 2018 at 18:26
  • 1
    $\begingroup$ I thought that the algebraic de Rham cohomology always agrees with singular cohomology over $\mathbf{C}$, by some form of $h$-descent of the usual isomorphism for smooth varieties. Unfortunately I cannot find a reference, so maybe it is false? Intuitively it should be true if we think of $\hat X$ as a tubular neighborhood of $Y$. $\endgroup$ Commented Jul 29, 2018 at 19:07
  • 3
    $\begingroup$ A reference for the isomorphism between algebraic de Rham cohomology and singular cohomology is the book Periods and Nori motives by Huber-Klawitter and Müller-Stach. The isomorphism is described in Definition 5.4.1. They use a different construction of algebraic de Rham cohomology, but Theorem 3.3.13 says their definition agrees with Hartshorne's. $\endgroup$ Commented Jul 29, 2018 at 20:05
  • 4
    $\begingroup$ @PiotrAchinger It's not recent. The comparison between alg. de Rham coh and singular coh is in Hartshorne's work in the 1970's. See Thm 1.6 of Hartshorne's survey article in Manuscripta Math 7 (1972), or Chapter IV of his longer paper in PMIHES 45 (1975). $\endgroup$ Commented Jul 29, 2018 at 21:14

1 Answer 1


(Synthesis of answers from comments, posted as community-wiki answer for convenience.)

  1. If $k = \mathbb{C}$ then algebraic de Rham cohomology, defined a la Hartshorne using the completion of $X$ along $Y$, is always isomorphic to singular cohomology. This is a theorem of Hartshorne, stated as Theorem 1.6 of his 1972 Manuscripta Math survey article Algebraic de Rham cohomology, and the proof is in Chapter IV of his 1975 PMIHES paper On the de Rham cohomology of algebraic varieties.

  2. There are examples, as in this question, of (non-smooth) affine $\mathbb{C}$-varieties $Y$ for which the "naive de Rham cohomology" $\mathbb{H}^\bullet(Y, \Omega^\bullet_Y)$ does not agree with singular cohomology. The Arapura--Kang curve $x^5 + y^5 + x^2 y^2 = 0$ is one such example.

  3. Hence the Arapura--Kang curve, or any other example as in (2), must also be an example where naive de Rham cohomology disagrees with Hartshorne's definition.

In particular, when this question uses the notation $H^\bullet_{\mathrm{dR}}(Y)$ to mean $\mathbb{H}^\bullet(Y, \Omega^\bullet_Y)$, this is misleading and non-standard, since this definition is not equivalent to the standard definition due to Hartshorne.


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