This post serves to resolve some questions posed at the end of a previous article, Units Group and the Infinitude of Primes, 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. In a bit 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.