On a topological proof of the infinitude of prime numbers.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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!).

Answer 4

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 $\# R^{\times} = \# R$.

Comments:
(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.