On the $\mathbb R$-algebra structure on $C^\infty(M)$. As it is more or less well-know, and as it has come up on MO a couple of times, the $\mathbb R$-algebra $C^\infty(M)$ of smooth functions on a (say) compact manifold contains essentially everything there is to know about $M$ itself.

Does one really need to know the $\mathbb R$-algebra structure, though? Can we manage with $C^\infty(M)$ given as an abstract ring?

For comparison, if I recall correctly one can reconstruct (as an analytic manifold) an open subset $U\subseteq\mathbb C$ from the field of meromorphic functions on it, given as an abstract field. (It follows, of course, that the ring of holomorphic functions also can be used to reconstruct $U$.)
 A: You can determine the R-algebra structure of $C^\infty(M)$ purely from its ring structure. As Robin Chapman mentions, the constant function 1M 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.
A: If $M$ is connected then one can determine $\mathbb{R}$ 
(the constant functions) within
$C^\infty(M)$ (the ring of smooth real-valued functions on $M$).
One can certainly determine $\mathbb{Q}$ within $C^\infty(M)$.
If $f$ is not in $\mathbb{Q}$ then it's a constant function iff
$f-a$ is a unit in $C^\infty(M)$ for all $a\in\mathbb{Q}$. The key
is that all nonconstant functions must take a rational value.
This doesn't work for non-connected $M$, since a function $f$
may take distinct constant values on a clopen partition on $M$.
Maybe one can use idempotents...
