There is a proof of the infiniteness of prime numbers using Topology. I was only informed of the existence of this proof. They say it's very elegant. One could show how this proof?


4 Answers 4


Here is a variant of Fürstenberg's proof that does not use topological notions (which obscure the main idea): We are arguing about periodic subsets of ${\mathbb Z}$. The set of integers not divisible by $p$ is periodic for any $p>0$, and the intersection of two periodic sets is periodic. If there were only finitely many primes the set $\{-1,1\}$ would be periodic.

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    $\begingroup$ It would help the reader to precisely define what "periodic" means above. $\endgroup$ Dec 20, 2011 at 18:02
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    $\begingroup$ I'd say that a subset of $\mathbb Z$ is periodic iff its characteristic function is. --- I find the answer brilliant! $\endgroup$ Dec 20, 2011 at 18:14
  • $\begingroup$ I am afraid to make this last remark true you have to define the characteristic function of a set as: 0 for elements in the set, and 1 for elements not in the set. In other words, this is the dual of the common way of defining characteristic function. $\endgroup$
    – boumol
    Dec 20, 2011 at 19:49
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    $\begingroup$ That's a GREAT proof-but to me,it defeats the entire point of the topological proof.The whole point of the proof to me was to use basic point set topology to prove a result in a seemingly unrelated field: number theory.Furstenberg's proof is not only a remarkable result,it has great significance for me personally as it was the first "theorem' I ever came up with and proved on my own without any references. I was disappointed when I found out I was beaten to it by almost 60 years by Furstenberg and Golumb, but not really surprised. Best of all-it motivated my first original research. $\endgroup$ Dec 20, 2011 at 19:51
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    $\begingroup$ @buomol: Why? Periodicity of $f$ is the same as periodicity of $1-f$. $\endgroup$
    – sdcvvc
    Dec 20, 2011 at 22:48

Eliminating the (unneeded) topological language from Fürstenberg's proof shows that it is simply the following trivial variation on Euclid's proof. If there are only finitely many primes $\rm\:p_1,\ldots,p_n\:$ then there are infinitely many units $\rm\:1+p_1\:\cdots\:p_n\ \mathbb Z,\:$ contra $\:\mathbb Z\:$ has only finitely many units $\pm1\:.\:$

For a much less trivial reinterpretation of Euclid's proof see my fewunits generalization.

THEOREM $\ $ An infinite ring $\rm R$ has infinitely many max ideals if it has fewer units $\rm U = U(R)$ than it has elements, i.e. $\rm\:|U| < |R|$.

The marvelous thing about this proof is that it preserves the constructivity of Euclid's proof. The key idea is that Euclid's construction of a new prime generalizes from elements to ideals, i.e. given some maximal ideals $\rm P_1,\ldots,P_k$ then a simple pigeonhole argument employing $\rm CRT$ implies that $\rm 1 + P_1\cdots P_k$ contains a nonunit, which lies in some maximal ideal $\rm P$ which, by construction, is comaximal (so distinct) from the prior max ideals $\rm P_i\:.\:$ Follow the above link for full details.

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    $\begingroup$ @Brian Don't let the topological language deceive you. As stated F's result is only syntactically topological - not semantically. If you deem F's proof innately topological then you may as well deem any trivial use of Boolean algebra topological, since that's what it amounts to when examined closely. Syntactic reformulations are a dime a dozen without significant semantic consequences. Of course it goes without saying that such evaluations are highly subjective. But some other mathematicians agree, e.g. B. Conrad in said MO thread. $\endgroup$ Dec 20, 2011 at 21:20
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    $\begingroup$ @BrianM.Scott : As evidence that there is less to Furstenberg's proof than meets the eye, it is worth pointing out that the "topological ideas" in it have led to no further serious mathematical results. One might have used the words "possibly productive" when the proof was released, but the succeeding history has shown that it was a dead end. And the reason for this is, as Bill pointed out, that it is just a pretentious way of describing Euclid's original proof. $\endgroup$
    – Adam Smith
    Dec 20, 2011 at 21:52
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    $\begingroup$ @Brian Please do elaborate on the "different possibilities and different modes of thought" that this suggests to you and, more importantly, the interesting number theory resulting from such. There can be no doubt that some topological notions do play nontrivial conceptual roles in modern number theory (these ideas were known long before F's proof). The same cannot be said for F's trivial reformulation of Euclid's proof (as presented here) $\endgroup$ Dec 20, 2011 at 22:35
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    $\begingroup$ +1 to Bill! I have no idea why Brian thinks this is "obviously" not a dead end. Ignoring the wild claims of a certain crackpot on mathoverflow, there are zero applications of Furstenberg's topology to number theory, and more generally zero applications of this style of point set topology to number theory. The "topological notions" that have been useful in number theory (eg ideles and adeles) are different and long predated Furstenberg. $\endgroup$
    – Adam Smith
    Dec 20, 2011 at 23:01
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    $\begingroup$ I think the crux of the matter is that when people speak of a proof of theorem X via methods from the field Y, then they generally mean that the proof uses theorems from Y. But in this case, the proof only uses the definition of a topological space, which only involves two short axioms! And the proof only uses one of them!! So calling it a "topological" proof is a bit of a misnomer. One would expect a genuinely topological proof to involve ideas of homotopy, homology, knots, the classification of surfaces -- anything which is actually a theorem. $\endgroup$ Dec 22, 2011 at 12:51

You may also like the paper On the exotic topology of the integers by Mezö and Lovas: Fürstenberg's topology turns $\mathbb Z$ into a metrizable, totally disconnected space, and $(\mathbb Z,+,\cdot)$ is a topological ring with respect to this topology (this last is in fact quite elementary!).

  • $\begingroup$ Great read. Thanks! $\endgroup$ Dec 21, 2011 at 0:04
  • $\begingroup$ I'm using this paper in some research I'm finally getting back to after a year off for health reasons.It's definitely worth people noticing it. $\endgroup$ Mar 5, 2012 at 5:30

By a domain I will mean a commutative ring without zero divisors. A Furstenberg domain is a domain in which every nonzero nonunit element is divisible by an irreducible element. This is a very weak factorization condition: one has Noetherian $\implies$ ascending chain condition on principal ideals $\implies$ all nonzero nonunits factor into products of irreducibles (factorization domain or atomic domain) $\implies$ Furstenberg domain, and in fact no implications can be reversed.

For $\alpha \in R$ we put $(\alpha) = \{x \alpha \mid x \in R\}$.

Furstenberg Lemma: a) A domain which is not a field is a factorization domain iff

$$R^{\times} = \bigcap (R \setminus (f)),$$

where the intersection is over a maximal set of mutually nonassociate irreducibles.

b) In particular, in a Furstenberg domain which is not a field and which has only finitely many nonassociate irreducibles $f_1,\ldots,f_n$ we have

$$R^{\times} = \bigcap_{i=1}^n (R \setminus (f_i)).$$

Proof: This is immediate from the definition.

Theorem: Let $R$ be a Furstenberg domain, which is not a field, and which has only finitely many nonassociate irreducibles. Then there is a nonzero element $\alpha \in R$ such that for all $x \in R$, $x \alpha + 1 \in R^{\times}$. In particular, $\# R^{\times} = \# R$.

Proof: Call a subset $X \subset R$ open if for all $x \in X$ there if a nonzero $\alpha \in X$ such that $x + (\alpha) = \{x+y\alpha \mid y \in R\} \subset X$. For nonzero $\alpha$, the complement $X = R \setminus (\alpha)$ is open: if $x \in X$ then for all $y = b\alpha \in (\alpha)$, if $x+y \in (\alpha)$ then $x+b\alpha = c\alpha$, so $x = (c-b)\alpha \in (\alpha)$. If $X,Y \subset R$ are both open, so is $X \cap Y$: if $x \in X \cap Y$, then there are $\alpha,\beta \neq 0$ such that $x+(\alpha) \subset X$ and $x+ (\beta) \subset Y$, and then $x + (\alpha \beta) \subset X \cap Y$. Of course it follows immediately that finite intersections of open sets are open.

Thus by Furstenberg's Lemma, $R^{\times}$ is open. Since $1 \in R^{\times}$, there is a nonzero $\alpha \in R$ such that $1 + (\alpha) \subset R^{\times}$. Since $R$ is a domain, the map $R \rightarrow 1+(\alpha)$ given by $x \mapsto x \alpha + 1$ is a bijection, so $\# R = \# (1+(\alpha)) \leq \# R^{\times}$. Certainly $\# R^{\times} \leq \# R$, so $\# R^{\times} = \# R$.

(0) If you read carefully Furstenberg's note, you see that he claims a "topological" proof rather than a topological proof. Is this proof topological? Answer: it is iff you want it to be! By calling something "open", anyone who knows what a topological space is will suspect that we are defining a topology on $R$, and the later arguments confirm that suspicion. (We do not explicitly say that arbitrary unions of open sets are open because we didn't need it in the proof, but that is immediate from the definition: if each $X_i$ is open and $x \in \bigcup X_i$, then there is some nonzero $\alpha$ such that $x + (\alpha) \subset X_i \subset \bigcup X_i$. Or: we can just remark that a set is open iff it is a union of cosets of nonzero principal ideals.) However, some years ago on Math Overflow Brian Conrad insisted that there was no actual topological theory in Furstenberg's proof, just topological language. This version of the argument, which is in part a response to a recent Monthly note of Idris Mercer, confirms that one can construe the argument that way: at no point does the argument make reference, either explicitly or implicitly in any logically necessary implicit way, to topological spaces! Still, though there is no necessary topology in Furstenberg's argument, there is still something interesting in it. I hope I've isolated what that something is.

(i) This is not a proof by contradiction.

(ii) Since $\mathbb{Z}$ is an infinite Furstenberg domain (indeed, certainly a factorization domain) with finite unit group, $\mathbb{Z}$ has infinitely many primes. This application of the result is a proof by contradiction.

(iii) The contrapositive form of this statement is: in a Furstenberg domain $R$ which is not a field, if for all nonzero $\alpha \in R$ there is $x \in R$ such that $x \alpha + 1$ is not a unit, then there are infinitely many nonassociate irreducibles. Euclid's original argument adapts immediately to prove this. Indeed, I presented this result two weeks ago in the first meeting of a seminar on the process of mathematical research that I am co-leading, and the point is that it is very close to being what you get just by looking for the genereal context in which Euclid's proof applies. (Except: in Euclid's argument, you multiply all your positive primes together and add $1$. Even in $\mathbb{Z}$ if you do this with "irreducibles" it doesn't quite work, because $(-2) + 1$ is a unit. So to make this work with "both positive and negative primes" you see that you should feel free to adjust the sign of $p_1 \cdots p_n$ before adding $1$, and then you see that you can in fact multiply $p_1 \cdots p_n$ by any nonzero integer $N$.)

At the moment I confess that I am viewing much of the fuss about whether Euclid's argument is by contradiction (quick answer: as presented by Euclid himself, certainly not; in fact it clearly means to be an argument by induction and is interesting as a very early example of that; however, so many later mathematicians, including Kummer, phrased the argument that way, and I think it would be dishonest to deny that there is something natural and appealing in that version: in fact, probably a proof by contradiction is more broadly appealing and understandable than a proof by induction) as being about which of the two contrapositive forms of this argument one wants to take as basic.

(v) The conclusion on $R$ is precisely that its Jacobson radical $J(R)$ is nonzero. In fact the argument would be if anything a little easier if instead of principal ideals we worked throughout with all nonzero ideals. (Since every nonzero ideal contains a nonzero principal ideal, the definition of "open" would not change.) However I wanted to present (here) a version which uses no ideal theory whatsoever.

(vi) Actually the Euclidean version gives something slightly stronger: the infinite sequence of irreducibles $p_1,\ldots,p_n,\ldots$ that it gives are pairwise comaximal: for all $i \neq j$, there are $a,b \in R$ with $a p_i + b p_j = 1$. Thus if you choose a maximal ideal $\mathfrak{m}_i$ containing each $p_i$, you get an infinite sequence of distinct maximal ideals. In fact, if you apply Euclid's argument without factoring in any domain with $J(R) = (0)$, then you get an infinite sequence of parirwise comaximal elements, hence again infinitely many maximal ideals. In particular we recover Dubuque's "Fewunit Theorem" when $R$ is a domain in a slightly sharper form. (Maybe this is what Kaplansky had in mind?) Over a ring which is not a domain this type of argument does not work, and though one can see some family resemblance between this argument and Dubuque's, they seem to remain distinct.


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