# naive de Rham cohomology fails for singular varieties

Let $X$ be a variety over a field $k$ of characteristic zero. If $X$ is smooth, algebraic de Rham cohomology defined as $$H^n_{dR}(X / k)=\mathbb{H}^n(X, \Omega^\bullet_{X/k})\qquad (\star)$$ is a good cohomological theory, for instance, is isomorphic tensor $\mathbb{C}$ to the singular cohomology of $X(\mathbb{C})$ once one chooses and embedding $k \hookrightarrow \mathbb{C}$.

It is "well known" that this fails for singular $X$ unless one considers more sophisticated definitions of de Rham cohomology, but I had difficulties finding a counterexample of, say, a singular variety $X$ such that the dimension of $(\star)$ is different from the dimension of singular cohomology. After a while, I found a paper by Arapura-Kang where the example of the plane curve $$x^5+y^5+x^2y^2=0$$ is presented. The proof is not so easy. Does anybody know a simpler example?

• We found ourselves in the same position as you a few years ago, and played around until we found this. But I agree that there ought to be something easier. – Donu Arapura Dec 4 '16 at 14:14
• Great, a comment from the source himself! Thanks for writing that paper, I learned a lot from it – dr91 Dec 4 '16 at 14:20
• See also this question – Julian Rosen Sep 14 '17 at 0:54