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For any topological space $X$, we can define $C(X)$ to be the commutative ring of continuous functions $f\,:\,X\rightarrow \mathbb{R}$ under pointwise addition and multiplication. Then $C(-)$ becomes a contravariant functor $C(-)\,:\,\bf{Top}\rightarrow \text{ComRing}$.

A theorem due to Gelfand and Kolmogorov states the following:

Let $X$ and $Y$ be compact Hausdorff spaces. If $C(X)$ and $C(Y)$ are isomorphic as rings, then $X$ and $Y$ are homeomorphic.

I encountered this theorem as an example in a book on homological algebra, without proof. I have searched for the proof, but have been unable to find it.

If anyone has an idea of how to prove this, or a reference to a proof, I would appreciate it greatly.

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    $\begingroup$ You might already know this, but there is a similar statement which is known as Gelfand duality. It deals with functions to $\mathbb{C}$ and commutative $C^*$-algebras though. $\endgroup$
    – Sebastian
    Nov 1 '12 at 12:37
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    $\begingroup$ Hist. ref.: И. М. Гельфанд, А. Н. Колмогоров, “О кольцах непрерывных функций на топологических пространствах”, Докл. АН СССР, 22:1 (1939), 11–15 $\endgroup$
    – Grigory M
    Nov 1 '12 at 12:39
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    $\begingroup$ An english translation of (part of) the paper referenced by @Grigory's (I hope...). $\endgroup$ Nov 1 '12 at 12:47
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I found a proof in these lecture notes by Garrido and Jaramillo. See Theorem 18. They also have historical references.

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  • $\begingroup$ Note that the theorem in those notes uses the $\mathbb{R}$-algebra structure on $C(X)$, not just the ring structure. $\endgroup$ Aug 6 '19 at 5:54
  • $\begingroup$ @EricWofsey: The R-algebra structure can be canonically reconstructed from the ring structure in this case. $\endgroup$
    – Dmitri P.
    Nov 24 '19 at 16:22
  • $\begingroup$ @DmitriPavlov: I'm aware of that; just pointing out that the reference given in this answer does not fully answer the question. $\endgroup$ Nov 24 '19 at 16:27
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Dugundji's Topology has a very short, readable proof.

you can find the proof on page 289. its very readable.

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    $\begingroup$ Can you provide a more specific reference? $\endgroup$
    – Ludolila
    Jan 24 '14 at 12:29
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The key to the proof is the following fact.

Lemma: Let $X$ be a compact space and let $\varphi:C(X)\to\mathbb{R}$ be a ring-homomorphism. Then there exists $x\in X$ such that $\varphi(f)=f(x)$ for all $f\in C(X)$.

Proof: Note that if $f\in C(X)$ is such that $f\geq 0$ everywhere, then $\varphi(f)=\varphi(\sqrt{f})^2\geq 0$. I claim furthermore that $\varphi$ is an $\mathbb{R}$-algebra homomorphism, so $\varphi(r)=r$ for any $r\in\mathbb{R}$ (thinking of it in $C(X)$ as a constant function on $X$). Indeed, we know this must be true if $r\in\mathbb{Q}$; for arbitrary $r\in\mathbb{R}$, now use the fact that $\varphi(r-q)\geq0$ if $q\leq r$ and $q\in\mathbb{Q}$ and $\varphi(q-r)\geq 0$ if $q\geq r$ and $q\in\mathbb{Q}$.

Now suppose that $\varphi$ is not given by evaluation at any point. Then for each $x\in X$, there is a function $f_x\in C(X)$ such that $\varphi(f_x)\neq f_x(x)$. Letting $r=\varphi(f_x)$ and replacing $f_x$ with $f_x-r$, we may assume $f_x(x)\neq 0$ and $\varphi(f_x)=0$. Replacing $f_x$ with its square, we may further assume that $f_x\geq 0$ everywhere. By compactness of $X$, finitely many of the sets $\{y:f_x(y)>0\}$ cover $X$, and so adding together the corresponding $f_x$'s, we get a function $f\in C(X)$ such that $f>0$ everywhere and $\varphi(f)=0$. But then $1/f$ is continuous so $f$ is a unit and so $\varphi(f)$ cannot be $0$, so this is a contradiction.

Given this fact, the result you ask for follows easily. If $X$ is compact Hausdorff, then we can recover the set $X$ from $C(X)$ (up to canonical bijection) as the set of ring-homomorphisms $C(X)\to\mathbb{R}$. We can moreover recover the topology on $X$ since it is the coarsest topology that makes each element of $C(X)$ continuous, by Urysohn's lemma. (Here if we are identifying $X$ with homomorphisms $C(X)\to\mathbb{R}$, we can think of an element of $C(X)$ as a function on $X$ by evaluation.) So we can recover the space $X$ up to homeomorphism from the ring $C(X)$.

(In fact, it similarly follows from the Lemma that if $X$ and $Y$ are compact Hausdorff, then ring-homomorphisms $C(X)\to C(Y)$ are naturally in bijection with continuous maps $Y\to X$, and this preserves composition. So this gives a contravariant equivalence of categories between compact Hausdorff spaces and rings of the form $C(X)$.)

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  • $\begingroup$ Hey Eric, can this approach be made to work when one considers complex-valued functions instead? $\endgroup$ Nov 17 '20 at 1:55
  • $\begingroup$ To prove what result, exactly? That if $C(X,\mathbb{C})\cong C(Y,\mathbb{C})$ as rings then $X\cong Y$ (for compact Hausdorff spaces $X,Y$)? $\endgroup$ Nov 17 '20 at 2:08
  • $\begingroup$ The issue with that is that the direct analogue of my Lemma is horribly false over $\mathbb{C}$ (for instance, even in the case where $X$ is a single point). But you can replace the Lemma with a different Lemma that says every maximal ideal is the set of functions that vanish at some point, and the proof is very similar. $\endgroup$ Nov 17 '20 at 2:16
  • $\begingroup$ The remainder of the proof using the Lemma is then similar, except that you have to determine the topology on $X$ (identified with the maximal ideals of $C(X,\mathbb{C})$) differently, since you don't know what the evaluation maps are. However, the closed sets of a compact Hausdorff space are generated by the zero sets of elements of $C(X,\mathbb{C})$, so you can recover $X$ topologically as the maximal spectrum of $C(X,\mathbb{C})$ with the Zariski topology. $\endgroup$ Nov 17 '20 at 2:18
  • $\begingroup$ That's great, thanks! $\endgroup$ Nov 17 '20 at 2:23
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Gillman-Jerison, Rings of continuous functions, Theorem 7.3.

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