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 $N$?
4 Answers
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 $\frak m^2=\frak m$.

4$\begingroup$ Thank you! Can you suggest how to prove that in the continuous case $m^2 = m$? (it is approximately clear why this is so, but so far I do not see an accurate proof) $\endgroup$ Commented Oct 2, 2022 at 21:43

12$\begingroup$ For example, if $f$ is continuous, then $g=\sqrt[3]{f}$ and $f/g=g^2$ are continuous as well. $\endgroup$ Commented Oct 2, 2022 at 22:45

6

1$\begingroup$ I wonder whether this argument works for open manifolds? I don't know the classification theorem of maximal ideals in this case. $\endgroup$– Z. MCommented Oct 3, 2022 at 13:02

4$\begingroup$ @LSpice rather write a complexvalued function $f$ as the linear combination $f=f_1+if_2$ of realvalued functions (if $f$ vanishes at $x$ observe that so do $f_1$ and $f_2$), and then $\mathfrak{m}^2=\mathfrak{m}$ follows from the real case. $\endgroup$– YCorCommented Oct 3, 2022 at 13:56
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 rationalvalued 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, $\mathbb{R}[x]$.

1$\begingroup$ I guess your functions are realvalued, right? $\endgroup$– LSpiceCommented Oct 3, 2022 at 3:21

3$\begingroup$ @LSpice: yes, we need this. Otherwise there are automorphisms of $C(X, \mathbb{C})$ given by applying a wild automorphism of $\mathbb{C}$ pointwise... in the complex case we would need to either be given complex conjugation or be given the $\mathbb{C}$algebra structure. $\endgroup$ Commented Oct 3, 2022 at 6:25
In this answer, let's assume that all functions are realvalued. 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.

1$\begingroup$ I guess your functions are realvalued, right? $\endgroup$– LSpiceCommented Oct 3, 2022 at 3:24

2$\begingroup$ I assumed everything was realvalued. $\endgroup$ Commented Oct 3, 2022 at 3:59

4$\begingroup$ Note that this shows that for every positivedimensional manifold $M$ and every topological space $X$, the rings (and even the underlying multiplicative monoids) $A_1=C^\infty(M,\mathbf{R})$ and $A_2=C^0(X,\mathbf{R})$ are not elementary equivalent. Indeed, the formula ($\forall f\exists g: f^2=g^4$) is true in $A_2$ but false in $A_1$. $\endgroup$– YCorCommented Oct 3, 2022 at 14:01
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 realvalued 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 ringtheoretic 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 $h := \sqrt{\sum_i h_i^2}$.

$\begingroup$ Sorry, I failed to understand this argument. It seems that this proves that $C^0(M)$ and $C^\infty(M)$ are not isomorphic as $\mathbb R$algebras, but the question is whether they are isomorphic as rings. $\endgroup$– Z. MCommented Oct 3, 2022 at 12:05

$\begingroup$ @Z.M, ah, right, thanks! Well, then one still complete the argument by showing that both $C(X)$ and $C^\infty(M)$ "know" their $\mathbb{R}$algebra structure. I'll revise my answer accordingly. $\endgroup$ Commented Oct 3, 2022 at 13:17

1$\begingroup$ Could it be that if $A,B$ are $\mathbb{R}$algebras whose underlying rings are isomorphic, then $A$ and $B$ are isomorphic? What might help is that $\mathbb{R}$ has only the identity as ring endomorphism. $\endgroup$ Commented Nov 7, 2022 at 16:33

$\begingroup$ @MartinBrandenburg: Interesting question. No, such $A$ and $B$ are not necessarily isomorphic. For example if $\Delta : \mathbb{R} \to \mathbb{R}$ is any $\mathbb{Q}$linear derivation, then $\mathbb{R}[x]/(x^2)$ has a ring automorphism given by $a + bx \mapsto a + (b + \Delta(a))x$, but this automorphism is clearly not an $\mathbb{R}$algebra automorphism for $\Delta \neq 0$. (And the vector space of such derivations is infinitedimensional, e.g. as a consequence of the infinitedimensionality of $\Omega_{\mathbb{R}/\mathbb{Q}}$ per Theorem 16.14 in Eisenbud.) $\endgroup$ Commented Nov 7, 2022 at 20:18

1$\begingroup$ Thanks. But this shows that the stronger statement, that isomorphisms are reflected (by the forgetful functor), is not true. I was wondering if the property of being isomorphic is reflected. $\endgroup$ Commented Nov 7, 2022 at 21:10