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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 in$R$. (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!