Let be a number field, i.e. a finite field extension to . We recall that the ring of integers in k, denoted , is the ring
For , the ring of integers is just the integers , in which case we recall the Fundamental Theorem of Arithmetic: that every integer may be written as a finite product
in which the are prime and uniquely determined (up to permutation). Domains for which this holds are known in general as unique factorization domains (UFDs). For — with square-free — the ring of integers will in general not be a UFD. In fact, for , the integers have unique factorization only in the 9 cases
Far less is known in the case (in which case k is known as a real quadratic field), although an unproven conjecture dating back to Gauss suggests that there should be infinitely many real quadratic fields. More recently, some heuristics stemming from Cohen suggests that the ring of integers in should be a UFD with probability as on the square-free integers.
Here, we’ll focus on a more tractable variant of this problem:
Question: What can be said about the number of distinct real quadratic fields with for which is not a UFD?
For a weak answer to the question above, we devote the rest of this article to the establishment of the following bound:
Theorem: As , we have
in which the implied constant is made effective (e.g. greater than ).
For a high-level perspective, our plan is to identify a “large” infinite family of d for which the (images of the) norm forms in are both uniformly well-behaved and restricted, in the sense that they can be chosen to uniformly avoid the norms of some select low-lying primes. To control the distribution of said primes (i.e. to ensure that they remain small), we trade power for control and restrict our study to the primes that ramify in as opposed to those that split (which are far more numerous).
— PART I —
As always, we require a few Lemmas:
Lemma 1: Let be of degree 2. Then the set
has natural density 0.
Proof: For any prime , we note that for some n iff admits a root in the finite field , which occurs precisely when
in which denotes the polynomial discriminant of f and denotes the Legendre symbol. Of course, this implies that for all integers k, and for it follows that is composite. Thus, on a set of density , and we have
in which A denotes the set of primes such that (1) holds. Now, let, in which is odd. (If , then is reducible over and our theorem holds trivially.) We have for all primes in a coset of , and quadratic reciprocity implies for all primes in a coset of . Thus A contains all primes in some arithmetic progression, and thus
which tends to 0 provided that the sum diverges. This fact follows from the Chebotarev density theorem (or a sufficiently strong version of Dirchlet’s theorem on arithmetic progressions), and the well-known estimate.
Our second lemma concerns the density of square-free values for the polynomial (as above). We define
If is assumed, we denote by the set of roots of over the ring .
Lemma 2: Let be of degree 2, with leading coefficient . Then
In particular, the density is positive iff the content of is square-free and .
Proof: Let be prime. As in Lemma 1, implies . If and , then Hensel’s Lifting Lemma implies that admits exactly two roots over . In particular, we have with probability as . If for some , we find (likewise) that with probability , and so the formula in (2) holds. As
wherein the last inequality follows from absolute convergence of the series, we have iff for one of the finite primes dividing . If , then admits roots over . For , this forces (as is a field and ). Then, and we repeat this argument for to show . In the case , finite computation gives the stated exception, and these conditions are clearly sufficient for to hold.
Our final Lemma can be viewed as a strengthening of Lemma 1:
Lemma 3: Let be of degree 2, and let be a finite set of primes. Let be given by
Proof: It suffices to show that the set
has density 1. Let be prime such that there exists an with divisble by . Define such that for all (which exists as). As ranges across the primes in (where is as in Lemma 1), we have
As in Lemma 1, it follows that .
— PART II —
In this section, we’ll begin to see how our Lemmas apply to the construction of real quadratic number fields without unique factorization.
Let be of degree 2. Then is a quadratic irrational for any fixed , and so
in which the expression at right denotes the (periodic!) continued fraction expansion of . If the terms can be taken as integer polynomials in , and if can be taken independent of (for sufficiently large), then we say that has a uniform root. For example, satisfies
for and so admits a uniform root.
Theorem 1: Suppose that of degree 2 admits a uniform root. As , the ring of integers in fails to be a UFD for all n in a set of density .
Proof: Take square-free and set . Let be a ramified prime ideal, lying over the rational prime . If is a UFD, then is principal (with generator of norm ). For , the norm form in is given by , hence there exist integers such that
If we suppose further that , then (3) has a solution if and only if with one of the “pre-periodic” approximants to (i.e. it suffices to check successive approximants up to ; see here for more information).
If admits a uniform root and , the approximants to appear as a rational function in , and the norm form of the i-th approximation to takes the form
Let denote the (finite) set of primes which arise when is constant. Otherwise, , and we have as . For sufficiently large, let be a prime divisor of . Then the norm form fails to surject onto , and will not be a UFD by the remarks above. This proves our Theorem in the case , and a proof for the case is outlined in the Exercises. The result for general follows by consideration of the four polynomials , for (each of which satisfies one of our conditions on ).
With this result in hand, we derive a (preliminary) lower bound on the function . This is presented in the following example, which moreover sets the stage for our work in Part III.
Example: For , define . As
for all , it follows that admits a uniform root. Let . As , we note that if and only is square-free. In either case, Theorem 1 implies that
To evaluate , we note that , so
after some simplification. If we take square-free, then for all (just consider the reduction into ). Now, if we define and , it follows that
For , we get the weak estimate . (With more care, this constant may be raised to .)
— PART III —
We now approach our final step in the proof that
which is achieved by (delicately!) adding the contributions of each to our lower bound on . We require one more Lemma:
Lemma 4: If for any two , then and.
Proof: We first note that for , since
for (and the lower bound is obvious). So implies that these values lie between the same perfect squares, i.e. . Yet for all , which implies (whence ).
This “injectivity” result implies that we need not worry about double-counting our contributions to our estimate on . We note that provided that , whereby
(This step requires some uniformity in the rate at which tends to, but this is not difficult, as for .) By introducing an error term (and slightly reducing our constants), this implies that
With this in hand, we are ready to prove our final estimate on :
Theorem 2: As , we have
Proof: To estimate the sum above in (4), we define the Dirichlet series
where the introduction of the Möbius function is used to restrict our sum to the square-free integers. For real, we see that
in which denotes the infinite product
We recognize the rightmost product in (5) as an Euler product relating to the Möbius function, and so
as along the positive axis. It follows that the N-th partial sum to satisfies
(by a consideration of the residue of at ), which yields to us the estimate
To get a handle on the constant appearing in the final estimate of this proof, we can unravel all of our infinite products. The resulting constant is then
Thus the implied constant in the statement of Theorem 2 may be taken in (slight) excess of , as earlier claimed. Without a doubt, this constant can be improved (perhaps with a more careful treatment of ).
— EXERCISES —
Exercise: As varies over the polynomials of degree 2, show that can be made both arbitrarily large (less than 1) and arbitrarily small (while greater than 0). Is it the case that
is dense in ?
Exercise: When is of degree 2, what are the possible values of ? Give necessary and sufficient conditions for each to occur. (Take care in the case .)
Exercise: Complete the proof of Theorem 1 in the case , using the fact that
Exercise: What is the bound on which follows from the observation
for (i.e. that admits a uniform root)? (This exceeds the bound which we established at the end of Part II.)