Definition of the ring of differentiable maps $C^\infty(M)$, where $M$ is a differentiable manifold

Question

Let $M$ be a differentiable manifold. We define the set $C^{\infty}(M) := \{f : M \to \mathbb{R} \text{ differentiable}\}$.

We can endow this set with a ring structure considering the common sum of functions:

$$(f+g)(p) := f(p)+g(p)$$

and the product (instead of the composition):

$$(fg)(p) := f(p)g(p)$$

So, the neutral element for $+$ is the constant map zero, and the neutral element for the product is the constant map $1$ or the identity map? (then is a domain?)

Or we consider $(C^{\infty}(M),+,\circ)$ with $\circ$ the composition of differentiable maps?

Thanks.

Answer 1

$M$ is an arbitrary differentiable manifold, so the "identity map" does not make sense unless $M$ itself is $\mathbb{R}$. So, the multiplicative identity is the constant map $1$.

Again, since $M$ is an arbitrary differentiable manifold, composition does not necessarily make sense. It only makes sense if $M = \mathbb{R}$. So, composition cannot in general define a multiplication on $C^\infty(M)$.

The question

(then is a domain?)

does not make sense to me. Are you perhaps asking if $C^\infty(M)$ is an integral domain? If so, the answer is no, and it is a good exercise to find two nonzero smooth functions on $M$ whose product is the zero function.