On a topological proof of the infinitude of prime numbers.

Solution 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.

Solution 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.