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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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