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Throughout, we take R \subset \mathbb{C} as a complex subring (with unity).  In this article, we’ll be interested in natural analogues of Euclid’s proof of the infinitude of the primes (i.e. the case R = \mathbb{Z}).  In particular, we’ll show that Euclid’s proof (and generalizations to this method) can be recast as a relationship between the size of the unit group R^\times of R and \# \mathrm{Spec}(R), the number of prime ideals inR. (Here, \mathrm{Spec}(R) denotes the spectrum of R.)

For the sake of explicit analogy, we include a (needlessly abstracted) version of Euclid’s result now:

Theorem 1 (Euclid):  The integers have infinite spectrum.

Proof: If not, let \mathfrak{p}_1,\ldots, \mathfrak{p}_m exhaust the list of prime ideals.  As the integers form a PID, (in fact, they form a Euclidean domain), we may associate to each given ideal a prime generator p_i such that (p_i) = \mathfrak{p}_i.  Let N:= \prod p_i; as N+1 is not a unit (replacing N with kN \gg 0 if necessary), it admits a prime factor p which equals some p_i.  But then p divides both N and N+1, whence 1 as well.  This is a contradiction, and our result follows. \square

It is worth remarking that the “PID-ness” of the integers is not needed in the derivation of Theorem 1.  Indeed, if \mathfrak{p}_i were not principal, we may take instead p_i to be any non-zero element of \mathfrak{p}_i.  Thus, we see that the PID-ness of \mathbb{Z} is not the crucial property of the integers (viewed as a subfield of \mathbb{C}) upon which our proof stands.  Rather — as we’ll see after the fold — Euclid’s proof frames the infinitude of primes as a consequence of the finiteness of the units group!

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