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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?

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The proof is presented in Wikipedia: en.wikipedia.org/wiki/… –  Brad Dec 20 '11 at 16:00
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If you want to see a lengthy discussion on the originality and interest of this proof, head over to the comments at mathoverflow.net/questions/34699/… –  David Speyer Dec 20 '11 at 17:53
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The proof also appears in Aigner and Ziegler's Proofs from THE BOOK, which I highly recommend. –  Shaun Ault Jul 4 '13 at 16:02

4 Answers 4

up vote 9 down vote accepted

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!).

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Great read. Thanks! –  Isaac Solomon Dec 21 '11 at 0:04
    
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. –  Mathemagician1234 Mar 5 '12 at 5:30

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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It would help the reader to precisely define what "periodic" means above. –  Bill Dubuque Dec 20 '11 at 18:02
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I'd say that a subset of $\mathbb Z$ is periodic iff its characteristic function is. --- I find the answer brilliant! –  Pierre-Yves Gaillard Dec 20 '11 at 18:14
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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. –  Mathemagician1234 Dec 20 '11 at 19:51
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@Mathemagician1234 Please see my comment to Brian in my answer. –  Bill Dubuque Dec 20 '11 at 21:05
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@buomol: Why? Periodicity of $f$ is the same as periodicity of $1-f$. –  sdcvvc Dec 20 '11 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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The objection in the first paragraph rather misses the real point of Fürstenberg’s proof, which is that topological language/ideas can offer a possibly productive way of thinking about something apparently non-topological. And the restatement in terms of units seems to me rather less perspicuous than the topological version. –  Brian M. Scott Dec 20 '11 at 20:14
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@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. –  Bill Dubuque Dec 20 '11 at 21:20
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@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) –  Bill Dubuque Dec 20 '11 at 22:35
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That was a general comment about the importance of language and mathematical context: you appeared to think that syntactic reformulations are necessarily insignificant, which is nonsense. With ‘not a dead end’ I was referring to the application of topological dynamics in ergodic Ramsey theory and combinatorial number theory generally, not to the specific topological elements of this pretty but elementary result (which I don’t consider a trivial reformulation of Euclid’s proof). –  Brian M. Scott Dec 20 '11 at 23:07
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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. –  Dan Petersen Dec 22 '11 at 12:51

This is Fürstenberg's proof (follow the link). It is indeed beautiful.

PS: Oh, sorry, I didn’t see Brad’s answer before answering (in fact curiously I think it was not visible when I answered).

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