So if we have a differentiable manifold $M$, the set of smooth functions $C^{\infty}(M)$ on $M$ can be equipped with point-wise addition and multiplication.
Apparently it is important that this object will be a ring but not a field, however I am having trouble seeing how it fails to be field?
I know point wise addition on this set will be an abelian group. So $C^{\infty}(M)$ without ${f=0}$ and with pointwise multiplication must not have one (or more) of the following: commutativity, associativity, a neutral element and an inverse?
Or maybe the distributive rule of multiplication over addition?
But I’m just not sure which property does not hold and why; surely all do?!
Best Answer
It's not a field because not every non-zero smooth function has a multiplicative inverse.
For instance, for some open $U\subsetneq M$, pick $f$ to be any smooth function with compact support which vanishes outside $U$. Then, $f|_{M\setminus U}$ is identically $0$ and hence, there is no smooth function $g$ such that $g(x)f(x)=1$ for all $x\in M\setminus U$. In particular, there is no smooth $g$ such that $g(x)f(x)=1$ for all $x\in M$.