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!

Where $R \subset \mathbb{C}$ is a subring (with unity) as before, we define

$J_R:= \displaystyle\bigcap_{\mathfrak{p} \neq (0)} \mathfrak{p},$

in which the sum runs over all non-zero prime ideals in $R$.  This is not quite the nilradical of $R$, as our intersection omits the prime ideal $(0)$.  Nor is $J_R$ the Jacobson radical (the intersection of all maximal ideals), except when the Krull dimension of $R$ is $1$ (e.g. for $R$ a PID or a Dedekind domain).

Proposition 2: Let $R \subset \mathbb{C}$ be a subring (with unity).  Then $R^\times +J_R=R^\times$.

Proof: Let $j \in J_R$ and $u \in R^\times$.  If $j+u \in \mathfrak{p}$ for some $\mathfrak{p} \in \mathrm{Spec}(R)$, then$u \in \mathfrak{p}$ as $j \in J_R \subset \mathfrak{p}$.  Yet $u$ is a unit, so this contradiction forces $j+u \notin \mathfrak{p}$ for all prime ideals $\mathfrak{p}$.  Yet all non-units of $R$ lie in maximal ideals, so this implies that $j+u$ is a unit.  The reverse inclusion is obvious. $\square$

Corollary 3: If $R \subset \mathbb{C}$ has finitely many units, then $R$ has infinite spectrum.

Proof: If $R$ has finite unit group, then the identity $R^\times +J_R = R^\times$ forces$J_R=(0)$.  Yet if $J_R=(0)$, then $\mathrm{Spec}(R)$ must be infinite, as the intersection of finitely many non-zero ideals (in a domain) is itself a non-zero ideal. $\square$

As an application of Corollary 3, we may reestablish Euclid’s result on the infinitude of primes.  Moreover — if we assume the Dirichlet unit theorem — it follows that $\# \mathrm{Spec}(R)=\infty$  for all imaginary quadratic fields.  Yet, as the following Theorem shows, we have only begun to tap the power of our Proposition:

Theorem 4: If $R$ has finite spectrum, then $R^\times$ is dense in $\mathrm{cl}(\mathrm{Frac}(R))$, the topological closure of the fraction field of $R$.

Proof: Suppose that $R$ has finite spectrum, so that $J_R \neq (0)$ as in Corollary 3. If $R \subset \mathbb{R}$, then any non-zero $j \in J_R$ generates a lattice over $\mathbb{R}$.  If $R \not\subset \mathbb{R}$, let $k_1 \in J_R$ be non-zero and choose $z \in R$ not real.  Then $k_2:=zk_1$ and $k_1$ are independent over $\mathbb{R}$ and so generate a lattice in $\mathbb{C}$.  In each case, we shall denote this lattice by $\mathscr{L}$.

Fix $\ell \in \mathrm{cl}(\mathrm{Frac}(R))$, and choose a sequence $\{a_n\} \subset \mathscr{L}$ tending to infinity. There exists some $\eta \in (0,\infty)$ dependent only on $\mathscr{L}$ such that $\vert \ell a_n -b_n \vert < \eta$ for some choice of $\{b_n\} \subset \mathscr{L}$.  Then

$\displaystyle\bigg\vert \frac{b_n+1}{a_n+1} - \ell \bigg\vert \leq \bigg\vert \frac{b_n+1}{a_n+1}-\frac{b_n}{a_n}\bigg\vert + \bigg\vert \frac{b_n}{a_n} - \ell \bigg\vert < \bigg\vert \frac{a_n-b_n}{a_n(a_n+1)} \bigg\vert+\frac{\eta}{\vert a_n \vert},$

and it’s not hard to show that this rightmost sum is $O(\vert a_n \vert^{-1})$.  Thus the sequence $\{(b_n+1)/(a_n+1)\}$ tends to $\ell$, while Proposition 2 implies that each term is unital.  $\square$

Once again, we can use the Dirichlet unit theorem to furnish examples of number fields with infinite spectrum:

2. Cubic and quartic fields that are not totally real.
3. Select fields of degree five and six.

Actually — and in some sense with a lot less work — we can show that all number fields have infinite spectrum.  To do so, we’ll need just one more Proposition:

Proposition 5: Let $k$ be a number field of degree $n$.  Then the length of the longest arithmetic progression in $\mathcal{O}_k^\times$ is uniformly bounded as a function of $n$.

Proof: Following Newman in [1], suppose that the units

$\alpha, \alpha + \beta, \ldots, \alpha+m\beta \in \mathcal{O}_k^\times$

form an arithmetic progression.  Let $\tau = \beta/\alpha$, which lies in $\mathcal{O}_k$ as $\alpha$ is a unit. Thus $N(r\tau+1)=\pm 1$, for $r \in [0,k]$.  If we take $r>2$ as well, then$N(r\tau+1)=1$, a consequence of the fact that $r\tau+1 \equiv 1 \mod (r)$ (whereby $N(r\tau+1)\equiv 1 \mod r$).  Thus the polynomial $N(x\tau+1) \in \mathbb{Z}[x]$ (as a degree $n$ norm form) has roots at $r=3,\ldots, k$.  Then $k-2 \leq n$, which gives our uniform bound. $\square$

Coupled with Proposition 2, this implies the following:

Corollary 6: Let $k$ be a number field, with ring of integers $\mathcal{O}_k$.  Then $\mathcal{O}_k$ has infinite spectrum.

Proof: If our result does not hold, then $J_{\mathcal{O}_k}$ is non-zero; let $j \in J_{\mathcal{O}_k}$ be non-zero, and consider the (infinite) arithmetic progression $\{1+mj\}_{m=0}^\infty$.  Each element lies in $\mathcal{O}_k^\times$ by Proposition 2, but this contradicts Proposition 5 and so our result must hold. $\square$

(In particular, our previous appeals to the Dirichlet unit theorem have been unnecessary.)

— PART II —

Of all proofs of the infinitude of the primes, it is without a doubt that of Euler that has led to the greatest generalizations.  In summary, the fundamental theorem of arithmetic (that the integers form a UFD) allows us to write

$\displaystyle\sum_{n=1}^\infty n^{-s} = \prod_p \left(1-p^{-s}\right)^{-1},$

i.e. an expression of the zeta function $\zeta(s)$ as an Euler product.  In this way, the divergence of the harmonic series implies that the right-hand product contains infinitely many terms.  Yet this method (as written above) fails to generalize to number fields that are not UFDs.  Yet this setback admits a simple remedy: if one shows that all number fields are Dedekind domains (i.e. that the ideal group of $\mathcal{O}_k$ is always a UFD), then the factorization

$\displaystyle\sum_{\mathfrak{a} \neq (0)} N(\mathfrak{a})^{-s} = \prod_{\mathfrak{p}\neq 0} \left(1-N(\mathfrak{p})^{-s}\right)^{-1}$

holds, in which $\mathfrak{a}$ (resp. $\mathfrak{p}$) exhausts the non-zero ideals (resp. prime ideals) in $\mathcal{O}_k$.  Once again, we find that our infinite sum diverges at $s=1$ (it is minorized by the harmonic series) which implies that the right-hand sum constitutes an infinite product.  Yet this is to say that $\mathcal{O}_k$ has infinite spectrum (our result in Corollary 6).

Nevertheless, I contend that our generalizations to Euclid’s method have interest in their own right.  For one, they highlight the tension between the size of the units group and the cardinality of the spectrum in any ring of characteristic 0.  While our best example of this phenomenon — Theorem 4 — relies heavily on the topology of $\mathbb{C}$, it is not difficult to imagine other results that capture this sentiment.  In particular, I conjecture (but have not been able to prove) the following:

Conjecture 7: Let $R$ be a domain with characteristic 0, and suppose that $R^\times$ has finite rank as an abelian group.  Then

1. the arithmetic progressions in $R^\times$ have uniformly bounded length, so;
2. $R$ has infinite spectrum (à la Corollary 6).

— EXERCISES —

Exercise: Show that the ring

$R:= \{a/b \in \mathbb{Q} : (p,b)=1 \text{ for all } p \equiv 1 \mod 4\}$

satisfies $\mathrm{cl}(R^\times)=\mathbb{R}$, so the converse to Theorem 4 is false.

The following Exercise gives conditions under which our methodology is sharp.

Exercise: Let $R$ be a domain (with unity).  We have already shown that$J_R =(0)$ implies that $R$ has infinite spectrum.

1. Prove that these conditions are equivalent if $R$ is a Noetherian domain with finite Krull dimension.
2. We call $R$ a $k$PID if all ideals in $R$ can be written as modules over $R$ with at most $k$ generators.  Use the Krull height theorem to show that $k$-PIDs are Noetherian domains of finite Krull dimension, so that $J_R =(0)$ if and only if $\#\mathrm{Spec}(R)=\infty$ in a $k$-PID.
3. Show that the ring of integers in a number field (of degree $n$) is an $n$-PID.

(Hard.)

— REFERENCES —

[1] M. Newman, Units in Arithmetic Progression in an Algebraic Number Field, Proc. Amer. Math. Soc, 43(2), 1974.