gnatopoditw

## Answered question

2022-06-20

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

### Answer & Explanation

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 ${C}^{\mathrm{\infty }}\left(M\right)$ is both a commutative ring, and an R-algebra (where $\mathbb{R}$ denotes the ring of real numbers, and the homomorphism $\mathbb{R}\to {C}^{\mathrm{\infty }}\left(M\right)$ sends a real number $a$ to the constant function ${c}_{a}:M\to \mathbb{R}$ defined by ${c}_{a}\left(x\right)=a$ for all $x\in M$).

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