Do rings of smooth functions differ from rings of continuous functions? Let $M$, $N$ be connected nondiscrete compact smooth manifolds. Can the ring of continuous functions on $M$ be isomorphic to the ring of smooth functions on $N$?
 A: No. In both the smooth function ring and the continuous function ring a maximal ideal $\frak m$ consists of the functions vanishing at some point. In the smooth case $\frak m/\frak m^2$ is the cotangent space of the manifold at that point, while in the continuous case $\frak m^2=\frak m$.
A: Here is a different proof which maybe clarifies a different aspect of the situation. The ring $C(X)$ of continuous functions on a compact Hausdorff space, as an abstract ring, actually knows its $C^{\ast}$-norm (the sup norm). We get this through the following sequence of steps:

*

*First, as an abstract ring we can recover the copy of $\mathbb{Q}$ inside it (the rational-valued constant functions) starting from the copy of $\mathbb{Z}$ inside it.

*Next, we define a partial order on $C(X)$ where $f \le g$ iff there exists $h$ such that $f + h^2 = g$. Consider the set of functions $r$ with the property that for any $\varepsilon > 0$ there exist $p, q \in \mathbb{Q}$ such that $|p - q| \le \varepsilon$ and $p \le r \le q$. This recovers precisely the copy of $\mathbb{R}$ inside $C(X)$ (the constant functions). So $C(X)$ as an abstract ring knows its $\mathbb{R}$-algebra structure.

*Next, given the $\mathbb{R}$-algebra structure we can define the spectrum $\sigma(f) = \{ \lambda \in \mathbb{R} : f - \lambda \text{ is not invertible} \}$; this recovers the image of $f$, and hence the spectral norm $\| f \| = \sup_{\lambda \in \sigma(f)} | \lambda |$ recovers the sup norm of $f$.

This construction works and produces the same result (the sup norm) for the ring $C^{\infty}(M)$ of smooth functions on a compact smooth manifold (so this ring, as an abstract ring, also knows its sup norm). (Edit: as Tobias Fritz observes, in the smooth case the above relation is no longer transitive. However, we don't need transitivity, so I don't even need to say "partial order" above, just "relation.") Now we can distinguish them: $C(X)$ is always complete with respect to this norm, whereas $C^{\infty}(M)$ never is (since its completion is the continuous functions) unless $M$ is discrete.
This is noticeably more complicated than the existing answers but I think it's nice that 1) we avoided the classification of maximal ideals, and relatedly 2) this argument works in more generality: it's an adaptation of the classic proof that $\mathbb{R}$ itself has no nontrivial automorphisms (and specializes to that statement), and it also successfully identifies the copy of $\mathbb{R}$ inside, for example, $\mathbb{R}[x]$.
A: In this answer, let's assume that all functions are real-valued. In the ring of continuous functions, the following are equivalent:

*

*$f\geq 0$


*there is some $g$ with $g^2=f$.


*For each $n>0$, there is some $g$ with $g^{2n}=f$.
Furthermore, in the ring of continuous functions, for all $n\geq 0$ and $f$, there is a $g$ with $g^{2n+1}=f$.
This is clearly not the case with smooth functions since $\sqrt[3]{x^2}$ and $\sqrt{|x|}=\sqrt[4]{x^2}$ are not smooth.
A: Another conceptually interesting way to see that the answer is negative is to make the following observations (related to Tom Goodwillie's answer):

*

*For any topological space $X$, the algebra of real-valued continuous functions on $X$ has no nonzero derivations.

*On the other hand, the derivations on $C^\infty(M)$ for a manifold $M$ correspond to the vector fields on $M$.

It now follows that there is no $\mathbb{R}$-algebra isomorphism $C(X) \cong C^\infty(M)$ for any space $X$ and any nondiscrete manifold $M$ upon noting that such $M$ has nonzero vector fields.

In order to show that there is not even an isomorphism of rings, it is enough to show to characterize the $\mathbb{R}$-algebra structure on both $C(X)$ and $C^\infty(M)$ in purely ring-theoretic terms. See Qiaochu Yuan's answer for how to do this in the case of $C(X)$. For $C^\infty(M)$, one can proceed in the exact same way, with the minor difference that one should put $f \le g$ iff there exist finitely many $h_1,\ldots,h_n$ such that $f + \sum_i h_i^2 = g$. Allowing sums of squares there is relevant for showing that $\le$ is transitive, but not necessary for $C(X)$ since there one can always take $h := \sqrt{\sum_i h_i^2}$.
