# Why de Rham complex is a complex of D-modules?

Usually, by a complex of $R$-modules, where $R$ is a ring (or sheaf of rings), one means a sequences of $R$-modules $C_n$ and $R$-homomorphisms $d_n:C_n\to C_{n-1}$ such that $d_{n-1}\circ d_{n}=0$.

However, consider the de Rham complex $$\mathcal{O}\to\Omega^1\to\cdots\to\Omega^n$$ where each term is an $\mathcal{O}$-module and each differential is merely $\mathbb{C}$-linear. I have seen thousands of times that the de Rham complex is a complex of right $D$-modules. ($D$ is the sheaf of differential operators)

What does “complex of right $D$-modules” means here?

• The modules are right $D$-modules and the maps in the complex respect this action. It's important that we specify "right" here, as the sheaf of rings $D$ is noncommutative. – KReiser Jun 3 '18 at 21:00