# de Rham cohomology of singular varieties: why completion?

If $X$ is a smooth variety over an algebraically closed field $k$ of characteristic zero one can define algebraic de Rham complex $$\mathcal{O}_X \to \Omega^1_X \to \ldots \to \Omega_X^n,$$ where $\Omega^i_X = \wedge^i \Omega^1_X$, $n = \dim X$, and differential is the universal derivative $d: \mathcal{O}_X \to \Omega_X^1$ uniquely extended to $\Omega^\bullet_X$ by the product rule.

Sheaves $\Omega^i_X$ can have non-trivial higher cohomology, this is a new phenomenon for algebraic varieties as compared to differential geometry, where the sheaves of differentials are acyclic. For this reason de Rham cohomology of $X$ are defined as the hypercohomology groups of the complex of sheaves $$H^i_{dR}(X) = \mathbb{H}^i(\Omega_X^\bullet).$$

Now let $Y$ be a singular variety. How to define $H^i_{dR}(Y)$? The answer I saw several times is the following. Consider only $Y$ embeddable in a smooth $X$ with the sheaf of ideals $I$. Then take completion of the complex $\Omega_X^\bullet$ along $Y$: $$\Omega_Y^\bullet = \varprojlim \Omega_X^\bullet /I^k \Omega_X^\bullet.$$ Hypercohomology of the completion are by definition the de Rham cohomology of $Y$.

I have two questions:

1. What is the idea behind taking the completion?
2. Could we find a resolution of $Y$ inside $X$ as a sheaf of dg algebras on $X$ and define de Rham cohomology of $Y$ as the de Rham cohomology of this sheaf of dg algebras?