# 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 $$N$$?

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 $$\frak m^2=\frak m$$.

• 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) Oct 2, 2022 at 21:43
• For example, if $f$ is continuous, then $g=\sqrt[3]{f}$ and $f/g=g^2$ are continuous as well. Oct 2, 2022 at 22:45
• @OlegEroshkin's argument seems to assume, like the other answers, that the functions are real-valued; but that can be avoided by something like a partition of unity that allows us still to take roots on the support of $f$, right? Oct 3, 2022 at 3:25
• I wonder whether this argument works for open manifolds? I don't know the classification theorem of maximal ideals in this case.
– Z. M
Oct 3, 2022 at 13:02
• @LSpice rather write a complex-valued function $f$ as the linear combination $f=f_1+if_2$ of real-valued 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.
– YCor
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:

1. 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.
2. 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.
3. 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, $$\mathbb{R}[x]$$.

• I guess your functions are real-valued, right? Oct 3, 2022 at 3:21
• @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. Oct 3, 2022 at 6:25

In this answer, let's assume that all functions are real-valued. In the ring of continuous functions, the following are equivalent:

1. $$f\geq 0$$

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

3. 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.

• I guess your functions are real-valued, right? Oct 3, 2022 at 3:24
• I assumed everything was real-valued. Oct 3, 2022 at 3:59
• Note that this shows that for every positive-dimensional 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 $A_1$.
– YCor
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 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 $$h := \sqrt{\sum_i h_i^2}$$.

• 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.
– Z. M
Oct 3, 2022 at 12:05
• @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. Oct 3, 2022 at 13:17
• 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. Nov 7, 2022 at 16:33
• @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 infinite-dimensional, e.g. as a consequence of the infinite-dimensionality of $\Omega_{\mathbb{R}/\mathbb{Q}}$ per Theorem 16.14 in Eisenbud.) Nov 7, 2022 at 20:18
• 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. Nov 7, 2022 at 21:10