smooth manifolds - On the $mathbb R$-algebra structure on $C^infty(M)$.

You can determine the R-algebra structure of $C^infty(M)$ purely from its ring structure. As Robin Chapman mentions, the constant function 1_M is uniquely determined by the fact that it is the identity element, and multiplication by rationals is uniquely defined, so the functions equal to a constant rational value are uniquely determined.

Actually, the ring homomorphism $Fcolonmathbb{R}to C^infty(M)$ is unique, which also uniquely defines the R-algebra structure.

The positive elements $xinmathbb{R}$ are squares, so $F(x)$ must be a square in $C^infty(M)$, hence nonnegative everywhere. Then, for any $xinmathbb{R}$ and rational numbers $ale xle b$ we have $F(x)-a1_M=F(x-a)ge0$ and $b1_M-F(x)=F(b-x)ge0$, so $F(x)in C^infty(M)$ takes values in the interval $[a,b]$. This shows that, in fact, $F(x)=x1_M$.

Thinking about it, this works because $mathbb{R}/mathbb{Q}$ has trivial Galois group. You can see this by asking if the C-algebra structure on $Aequiv C^{infty}(M,mathbb{C})$ is uniquely determined by its ring structure, for which the answer is no. For any $sigmain{rm Gal}(mathbb{C}/mathbb{Q})$ it is not possible to distinguish a constant $fin A$ from $sigma(f)$ in terms of ring operations [edit: if $sigma$ is continuous, that is. So, only considering the identity element and complex conjugation]. Instead, you could ask if it is possible to determine the C-algebra structure up to the action of the Galois group. If the manifold is connected then the answer to this is yes. The constant function taking the value $pm i$ everywhere is given by $i_M^2+1_M=0$, and the constant functions $fin A$ are those for which $f-lambda1_M-mu i_M$ are units for all but at most one choice of $lambda,muinmathbb{Q}$. The constant functions are isomorphic to $mathbb{C}$, which is determined up to the action of the Galois group. If it is not connected, then we can't even say that much. For any locally constant map $sigmacolon Mto{rm Gal}(mathbb{C}/mathbb{Q})$, it is not possible to distinguish $fin A$ from $f_sigma(P)equivsigma(P)(f(P))$ using ring operations. The C-algebra structure is uniquely determined up to the action of such a locally constant $sigma$ though, which should still be enough to tell you everything about the manifold. Working over the reals, none of this matters, because of the triviality of the Galois group.