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

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.

Periods and Nori motivesby 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$1more comment