Throughout, we take 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 . 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 of and , the number of prime ideals in. (Here, denotes the spectrum of .)
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 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 such that . Let ; as is not a unit (replacing with if necessary), it admits a prime factor which equals some . But then divides both and , whence as well. This is a contradiction, and our result follows.
It is worth remarking that the “PID-ness” of the integers is not needed in the derivation of Theorem 1. Indeed, if were not principal, we may take instead to be any non-zero element of . Thus, we see that the PID-ness of is not the crucial property of the integers (viewed as a subfield of ) 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 is a subring (with unity) as before, we define
in which the sum runs over all non-zero prime ideals in . This is not quite the nilradical of , as our intersection omits the prime ideal . Nor is the Jacobson radical (the intersection of all maximal ideals), except when the Krull dimension of is (e.g. for a PID or a Dedekind domain).
Proposition 2: Let be a subring (with unity). Then .
Proof: Let and . If for some , then as . Yet is a unit, so this contradiction forces for all prime ideals . Yet all non-units of lie in maximal ideals, so this implies that is a unit. The reverse inclusion is obvious.
Corollary 3: If has finitely many units, then has infinite spectrum.
Proof: If has finite unit group, then the identity forces. Yet if , then must be infinite, as the intersection of finitely many non-zero ideals (in a domain) is itself a non-zero ideal.
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 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 has finite spectrum, then is dense in , the topological closure of the fraction field of .
Proof: Suppose that has finite spectrum, so that as in Corollary 3. If , then any non-zero generates a lattice over . If , let be non-zero and choose not real. Then and are independent over and so generate a lattice in . In each case, we shall denote this lattice by .
Fix , and choose a sequence tending to infinity. There exists some dependent only on such that for some choice of . Then
and it’s not hard to show that this rightmost sum is . Thus the sequence tends to , while Proposition 2 implies that each term is unital.
Once again, we can use the Dirichlet unit theorem to furnish examples of number fields with infinite spectrum:
- All quadratic fields.
- Cubic and quartic fields that are not totally real.
- 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 be a number field of degree . Then the length of the longest arithmetic progression in is uniformly bounded as a function of .
Proof: Following Newman in , suppose that the units
form an arithmetic progression. Let , which lies in as is a unit. Thus , for . If we take as well, then, a consequence of the fact that (whereby ). Thus the polynomial (as a degree norm form) has roots at . Then , which gives our uniform bound.
Coupled with Proposition 2, this implies the following:
Corollary 6: Let be a number field, with ring of integers . Then has infinite spectrum.
Proof: If our result does not hold, then is non-zero; let be non-zero, and consider the (infinite) arithmetic progression . Each element lies in by Proposition 2, but this contradicts Proposition 5 and so our result must hold.
(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
i.e. an expression of the zeta function 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 is always a UFD), then the factorization
holds, in which (resp. ) exhausts the non-zero ideals (resp. prime ideals) in . Once again, we find that our infinite sum diverges at (it is minorized by the harmonic series) which implies that the right-hand sum constitutes an infinite product. Yet this is to say that 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 , 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 be a domain with characteristic 0, and suppose that has finite rank as an abelian group. Then
- the arithmetic progressions in have uniformly bounded length, so;
- has infinite spectrum (à la Corollary 6).
— EXERCISES —
Exercise: Show that the ring
satisfies , so the converse to Theorem 4 is false.
The following Exercise gives conditions under which our methodology is sharp.
Exercise: Let be a domain (with unity). We have already shown that implies that has infinite spectrum.
- Prove that these conditions are equivalent if is a Noetherian domain with finite Krull dimension.
- We call a –PID if all ideals in can be written as modules over with at most generators. Use the Krull height theorem to show that -PIDs are Noetherian domains of finite Krull dimension, so that if and only if in a -PID.
- Show that the ring of integers in a number field (of degree ) is an -PID.
— REFERENCES —
 M. Newman, Units in Arithmetic Progression in an Algebraic Number Field, Proc. Amer. Math. Soc, 43(2), 1974.
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