gnatopoditw

2022-06-20

One forms ${\mathrm{\Omega}}^{1}(M)$ is a module over ${C}^{\mathrm{\infty}}(M)$, therefore does that make ${C}^{\mathrm{\infty}}(M)$ an algebra or a commutative ring?

marktje28

Beginner2022-06-21Added 22 answers

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 $x\in M$).

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 $x\in M$).