One forms Ω 1 ( M ) is a module over C ∞ ( M...

If $R$ is a commutative ring, we say that a commutative ring $S$ equipped with a ring homomorphism $R\to S$ is an "algebra" over $R$, or just an $R$-algebra for brevity.
Thus, there is no inconsistency in saying that $C^{\mathrm{\infty }}(M)$ is both a commutative ring, and an R-algebra (where $\mathbb{R}$ denotes the ring of real numbers, and the homomorphism $\mathbb{R}\to C^{\mathrm{\infty }}(M)$ sends a real number $a$ to the constant function $c_a:M\to \mathbb{R}$ defined by $c_a(x)=a$ for all $x\in M$).