This post serves to resolve some questions posed at the end of a previous article, Unit-Prime Dichotomy for Complex Subrings, in which we began to look at the tension between the relative sizes of the units group and the spectrum of  a complex subring (with unity).  As a disclaimer, I assume from this point familiarity with the topics presented therein.  To begin, we recall a Theorem from the prequel (Theorem 4):

Theorem: Let $R \subset \mathbb{C}$ be a subring (with unity), and suppose that $R$ has finite spectrum.  Then $R^\times$ is dense in $\mathrm{cl}(\mathrm{Frac}(R))$, the topological closure of the fraction field of $R$.

In particular — assuming a finite spectrum — it follows that the rank of $R^\times$ (considered as an abelian group) is bounded below by two. Unfortunately, this is as far as our previous methods will take us, even when $R \not\subset \mathbb{R}$ (see the Exercises).

The primary objective of this post is an extension to this result.  Specifically, we would like to capture in a purely algebraic way (e.g. without mention of the topology on $\mathbb{C}$) the fact that $R^\times$ becomes quite large in certain rings with finite spectrum.  After the fold we’ll accomplish exactly that, with the following Theorem and its generalizations to a wide class of commutative rings:

Theorem 1: Suppose that $R \subset \mathbb{C}$ is a subring (with unity).  If $R$ has finite spectrum, then $R^\times$ has infinite rank as an abelian group.

— PART I —

Before we proceed to our proof, some remarks are in order.  First off — if we continue our count from the previous article — this provides a third proof of the infinitude of prime ideals in the ring of integers over a number field. Indeed, Dirichlet’s unit theorem implies that the units in such a ring have finite rank.

Secondly, and perhaps more importantly, the conclusion of this theorem makes sense in any unital ring, so we may ask whether or not our theorem holds in greater generality (as will be done in Part II).

As one final reminder, we recall the definition of the ideal

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

in which the sum runs over the non-zero prime ideals in $R$.  Here, $J_R \neq (0)$ if$R$ is a domain with finite spectrum.  Having recalled this, we present the proof of Theorem 1:

Theorem 1: Suppose that $R \subset \mathbb{C}$ is a subring (with unity).  If $R$ has finite spectrum, then $R^\times$ has infinite rank as an abelian group.

Proof: This proof will proceed in two cases, dependent on whether or not $J_R$ contains transcendental elements.  In our first case, take $j \in J_R$ transcendental over $\mathbb{Q}$; then $R^\times +J_R =R^\times$ (an earlier Proposition) implies that $1+j\mathbb{Z}[j] \subset R^\times$.  Now, suppose that $\{\pi_i(j)\}_{i=1}^m \subset 1 + j \mathbb{Z}[j]$ is a set of irreducible polynomials.  If $\{\pi_i(j)\}_{i=1}^m$ is not independent (as a multiplicative set), then there exist integers $a_i,b_i \geq 0$ such that

$\displaystyle\prod \pi_i^{a_i} = \prod \pi_i^{b_i},$

at which point the fact that $\mathbb{Z}[j]$ is a UFD implies that $a_i=b_i$ for all $i$.  Thus$\{\pi_i(j)\}_{i=1}^m$ is an independent set, whence $R^\times$ has rank at least $m$.  It thus suffices (for our first case) to evince infinitely many irreducible polynomials in$1+j\mathbb{Z}[j]$, and such an example is provided by the cyclotomic polynomials$\Phi_n(j)$ with $n>1$ (or the linear polynomials).

Our second case is slightly more involved.  Take $j \in J_R$ algebraic; we may assume without loss of generality that $j \in J_R$ is an algebraic integer.  Now, suppose that the (multiplicative) abelian group $S$ generated by $\{1+kj\}_{k \in \mathbb{Z}}$ has finite rank.  The field norm $N: \mathbb{Q}(j) \to \mathbb{Q}$ restricts to a homomorphism$N:S \to \mathbb{Q}^\times$, and the image of this map has finite rank.  Let

$f(x):= N(1+xj) \in \mathbb{Z}[x].$

Finiteness of rank implies that the set of prime divisors of $\{f(k) : k \in \mathbb{Z}\}$ is finite.  Yet if $P:= \prod p_i$ represents the product of these primes, it follows that$f(1+kP) \equiv 1 \mod P$, so our hypothesis forces $f(1+kP) = \pm 1$ for all integers $k$.  (This is an extension of Euclid’s proof of the infinitude of the primes to a statement about the prime divisors of a polynomial.)  This gives a contradiction, and so $S$ has infinite rank.  As $S \subset R^\times$, it follows that $R^\times$ has infinite rank as well. $\square$

— PART II —

From here on, we direct our attention to extensions of our Theorem 1 to more general rings.  To begin, let $R$ be a ring with unity $1_R$.  Let $K$ be the kernel of the ring homomorphism $\mathbb{Z} \to R$ generated by $1 \mapsto 1_R$; then $K=m\mathbb{Z}$ for some $m \geq 0$, and we define the characteristic $\mathrm{char}(R)$ of $R$ to be $m$ (see here for more information).  In particular, a ring $R$ of characteristic $0$ contains a subring isomorphic to $\mathbb{Z}$, and if $R$ is a domain, then $\mathrm{Frac}(R)$ contains a subfield isomorphic to $\mathbb{Q}$.

The advantage of this definition is obvious when we realize that our purely algebraic proof of Theorem 1 cannot distinguish between $\mathbb{Z}$ (resp. $\mathbb{Q}$) and their embedded copies in $R$ (resp. $\mathrm{Frac}(R)$).  In other words,

Theorem 2: Let $R$ be a unital domain of characteristic $0$.  If $R$ has finite spectrum, then $R^\times$ has infinite rank as an abelian group.

Note: the shift from Theorem 1 to Theorem 2 represents the paradigmatic change between two characterizations of $R$: first as a subring of $\mathbb{C}$; then as a “super-ring” of $\mathbb{Z}$.  To emphasize our new-found generality, we present the following:

Corollary 3: Let $R$ be a discrete valuation ring of characteristic $0$.  Then $R^\times$ has infinite rank.

Proof: A discrete valuation ring is a PID with a unique non-zero prime ideal (hence finite spectrum). $\square$

It’s easy to construct counter-examples to Theorem 2 if we drop either of the hypotheses that

1. $R$ have characteristic $0$ (e.g. $R = \mathbb{Z}/p\mathbb{Z}$, with $p$ prime); or
2. $R$ have finite spectrum (e.g. $R=\mathbb{Z}$).

What is less clear is the dependence on $R$ being a domain, which runs deep in our assumptions (e.g. that the intersection of any two ideals be non-trivial). As some small consolation when $R$ is not a domain, we find that

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

the nilradical of $R$ (in which the intersections exhaust the prime ideals of $R$), which offers a nice reinterpretation of our earlier construction.

Non-Example: Take $R$ to be the dual numbers over the integers, i.e.$R = \mathbb{Z}[\epsilon]/(\epsilon^2)$.  Then $\mathrm{char}(R)=0$, and some algebra implies that the units group of $R$ is precisely

$R^\times = \{a+b\epsilon : a=\pm 1\} = \{\pm (1+\epsilon)^k : k \in \mathbb{Z}\}$

so that in particular $R^\times$ has finite rank.  Next, recall that $\mathfrak{p} \subset R$ is a prime ideal if and only if $R/\mathfrak{p}$ is an integral domain.  Thus $\mathfrak{p}=(\epsilon,p)$ is a prime ideal for any prime $p \in \mathbb{Z}$ so $R$ has infinite spectrum, which is consistent with the general form of Theorem 2. $\square$

This non-example gives some hope to the idea of extending Theorem 2 to certain rings that are not domains.  And indeed — such extensions do exist, under hypotheses that require us to recall one more definition: given a short exact sequence

$\Omega: 0 \to A \stackrel{\alpha}{\rightarrow} B \stackrel{\beta}{\rightarrow} C \to 0$

of modules over a commutative ring, $\Omega$ is said to split if $B \cong A \oplus C$ (under a natural isomorphism for which $\alpha = \mathrm{id}\vert_{A}$ and $\beta$ is projection to $C$).

With these preparations, we make our final refinement of Theorem 1:

Theorem 4: Let $R$ be a commutative ring of characteristic $0$ (with unity), such that $\mathrm{nil}(R)$ is prime and the short exact sequence

$\Omega: 0 \to \mathrm{nil}(R) \to R \to R/\mathrm{nil}(R) \to 0$

splits.  If $R$ has finite spectrum, then $R^\times$ has infinite rank.

Proof: Let $D:=R/\mathrm{nil}(R)$.  The lattice theorem for commutative rings gives a bijection $\varphi$ between the ideals of $R$ containing $\mathrm{nil}(R)$ and the ideals $D$, given by $\varphi(\mathfrak{n})=\mathfrak{n}/\mathrm{nil}(R)$.  For $\mathfrak{p} \in \mathrm{Spec}(R)$, we have

$R/\mathrm{nil}(R)/\mathfrak{p}/\mathrm{nil}(R) \cong R/\mathfrak{p},$

(a domain) and so $\varphi$ restricts to a map $\mathrm{Spec}(R) \twoheadrightarrow \mathrm{Spec}(D)$.  Moreover, for any $\mathfrak{q} \in \mathrm{Spec}(D)$, the preimage $\varphi^{-1}(\mathfrak{q})$ lies in $\mathrm{Spec}(R)$; this is a general property of (unital) ring homomorphisms.  Thus $\varphi$ induces a bijection

$\mathrm{Spec}(R) \leftrightarrow \mathrm{Spec}(D),$

and in particular $\mathrm{Spec}(D)$ is finite.  Next, we claim that $D$ has characteristic $0$. Indeed, if $\iota: \mathbb{Z} \to R$ denotes the canonical (characteristic) injection, suppose that $\iota(k) \in \mathrm{nil}(R)$.  Then $\iota(k^m)=\iota(k)^m=0$ for some $m \gg 0$, a contradiction to injectivity.  Thus $D$ is a unital domain of characteristic $0$ with finite spectrum, and it follows by Theorem 2 that $D^\times$ has infinite rank.

As $\Omega$ splits by hypothesis, we obtain an injective section $D \to R$, which restricts to an injective group homomorphism $D^\times \to R^\times$.  So $R^\times$ has infinite rank, as claimed. $\square$

Our work in generalizing Theorem 2 deserves a few remarks.  Firstly, while we have used Theorem 2 in the proof of Theorem 4, we note that Theorem 4 recovers a proof of Theorem 2 directly.  For if $R$ (taken as in Theorem 4) is a domain, then $\mathrm{nil}(R) =(0) \in \mathrm{Spec}(R)$, and the short exact sequence $\Omega$ splits trivially.

Secondly, we note that the condition $\mathrm{nil}(R) \in \mathrm{Spec}(R)$ implies that all zero divisors in $R$ are nilpotent (and these conditions are equivalent).  On the other hand, the condition on $\Omega$ is less transparent, but we can nevertheless force it through various tricks; e.g. if $R$ is a $\mathbb{Z}$-module and $R/\mathrm{nil}(R)$ is a free $\mathbb{Z}$-module (as in our Non-Example).

The obstruction to variants of Theorem 4 in more general rings is outlined in the Exercises.  Suffice to say that the relationship between $\mathrm{Spec}(R)$ and$\mathrm{rank}(R^\times)$ is less clear-cut in general.

— EXERCISES —

Exercise: Let $\alpha,\beta \in \mathbb{C}^\times$.  Prove that the multiplicative group $\langle \alpha,\beta \rangle$ is dense in $\mathbb{C}$ if and only if (1) $\vert\alpha\vert$ and $\vert\beta\vert$ are multiplicatively independent and (2) at least one of $\alpha,\beta$ has argument incommensurable with $2\pi$(Consider the lattice of logarithms.)

Exercise: Suppose that $R$ is a commutative ring with unity, such that the set

$Z:=\{z \in R : z=0 \text{ or } z \text{ is a zero-divisor}\}$

forms an ideal.  Give an example of a ring $R$ of characteristic $0$ such that the quotient $R/Z$ has positive characteristic (which is necessarily prime).

Exercise: If $R$ is a commutative ring (with unity) such that $Z$ is an ideal, prove that $R/Z$ is a domain and that

$\#\mathrm{Spec}(R) \geq \#\mathrm{Spec}(R/Z).$

Use this to prove that $R^\times$ has infinite rank provided that (1) $\mathrm{char}(R/Z)=0$, (2) $R$ has finite spectrum, and that (3) the short exact sequence

$0 \to Z \to R \to R/Z \to 0$

splits.