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 be a subring (with unity), and suppose that has finite spectrum. Then is dense in , the topological closure of the fraction field of .*

In particular — assuming a finite spectrum — it follows that the rank of (considered as an abelian group) is bounded below by two. Unfortunately, this is as far as our previous methods will take us, even when (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 ) the fact that 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 is a subring (with unity). If has finite spectrum, then 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

in which the sum runs over the non-zero prime ideals in . Here, if is a domain with finite spectrum. Having recalled this, we present the proof of Theorem 1:

**Theorem 1: ***Suppose that is a subring (with unity). If has finite spectrum, then has infinite rank as an abelian group.*

*Proof: *This proof will proceed in two cases, dependent on whether or not contains transcendental elements. In our first case, take transcendental over ; then (an earlier Proposition) implies that . Now, suppose that is a set of irreducible polynomials. If is not independent (as a multiplicative set), then there exist integers such that

at which point the fact that is a UFD implies that for all . Thus is an independent set, whence has rank at least . It thus suffices (for our first case) to evince infinitely many irreducible polynomials in, and such an example is provided by the cyclotomic polynomials with (or the linear polynomials).

Our second case is slightly more involved. Take algebraic; we may assume without loss of generality that is an algebraic integer. Now, suppose that the (multiplicative) abelian group generated by has finite rank. The field norm restricts to a homomorphism, and the image of this map has finite rank. Let

Finiteness of rank implies that the set of prime divisors of is finite. Yet if represents the product of these primes, it follows that, so our hypothesis forces for all integers . *(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 has infinite rank. As , it follows that has infinite rank as well.

**— PART II —**

From here on, we direct our attention to extensions of our Theorem 1 to more general rings. To begin, let be a ring with unity . Let be the kernel of the ring homomorphism generated by ; then for some , and we define the * characteristic* of to be (see here for more information). In particular, a ring of characteristic contains a subring isomorphic to , and if is a domain, then contains a subfield isomorphic to .

The advantage of this definition is obvious when we realize that our *purely algebraic *proof of Theorem 1 cannot distinguish between (resp. ) and their embedded copies in (resp. ). In other words,

**Theorem 2: ***Let be a unital domain of characteristic . If has finite spectrum, then has infinite rank as an abelian group.*

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

**Corollary 3: ***Let be a discrete valuation ring of characteristic . Then has infinite rank.*

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

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

- have characteristic (e.g. , with prime); or
- have finite spectrum (e.g. ).

What is less clear is the dependence on 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 is *not* a domain, we find that

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

**Non-Example: **Take to be the dual numbers over the integers, i.e.. Then , and some algebra implies that the units group of is precisely

so that in particular has finite rank. Next, recall that is a prime ideal if and only if is an integral domain. Thus is a prime ideal for any prime so has infinite spectrum, which is consistent with the general form of Theorem 2.

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

of modules over a commutative ring, is said to ** split** if (under a natural isomorphism for which and is projection to ).

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

**Theorem 4: ***Let be a commutative ring of characteristic (with unity), such that is prime and the short exact sequence*

*splits. If has finite spectrum, then has infinite rank.*

*Proof: *Let . The lattice theorem for commutative rings gives a bijection between the ideals of containing and the ideals , given by . For , we have

(a domain) and so restricts to a map . Moreover, for any , the preimage lies in ; this is a general property of (unital) ring homomorphisms. Thus induces a bijection

and in particular is finite. Next, we claim that has characteristic . Indeed, if denotes the canonical (characteristic) injection, suppose that . Then for some , a contradiction to injectivity. Thus is a unital domain of characteristic with finite spectrum, and it follows by Theorem 2 that has infinite rank.

As splits by hypothesis, we obtain an injective section , which restricts to an injective group homomorphism . So has infinite rank, as claimed.

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 (taken as in Theorem 4) is a domain, then , and the short exact sequence splits trivially.

Secondly, we note that the condition implies that all zero divisors in are nilpotent (and these conditions are equivalent). On the other hand, the condition on is less transparent, but we can nevertheless force it through various tricks; e.g. if is a -module and is a free -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 and is less clear-cut in general.

**— EXERCISES —**

**Exercise: **Let . Prove that the multiplicative group is dense in if and only if (1) and are multiplicatively independent and (2) at least one of has argument incommensurable with . *(Consider the lattice of logarithms.)*

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

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

**Exercise: **If is a commutative ring (with unity) such that is an ideal, prove that is a domain and that

Use this to prove that has infinite rank provided that (1) , (2) has finite spectrum, and that (3) the short exact sequence

splits.

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