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 $F\colon\mathbb{R}\to C^\infty(M)$ is unique, which also uniquely defines the R-algebra structure.

The positive elements $x\in\mathbb{R}$ are squares, so $F(x)$ must be a square in $C^\infty(M)$, hence nonnegative everywhere. Then, for any $x\in\mathbb{R}$ and rational numbers $a\le x\le b$ 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 $A\equiv C^{\infty}(M,\mathbb{C})$ is uniquely determined by its ring structure, for which the answer is no. For any $\sigma\in{\rm Gal}(\mathbb{C}/\mathbb{Q})$ it is not possible to distinguish a constant $f\in A$ 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 $f\in A$ are those for which $f-\lambda1_M-\mu i_M$ are units for all but at most one choice of $\lambda,\mu\in\mathbb{Q}$. 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 $\sigma\colon M\to{\rm Gal}(\mathbb{C}/\mathbb{Q})$, it is not possible to distinguish $f\in A$ from $f_\sigma(P)\equiv\sigma(P)(f(P))$ 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.

ring$C^\infty(M)$? – Donu Arapura Aug 28 '10 at 16:39kept. @Steve, I do not understand the relation between your comment and my question! – Mariano Suárez-Alvarez♦ Aug 28 '10 at 22:18