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?