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Let $k/\mathbb{Q}$ be a number field, i.e. a finite field extension to $\mathbb{Q}$.  We recall that the ring of integers in k, denoted $\mathcal{O}_k$, is the ring

$\mathcal{O}_k := \{\alpha \in k : f(\alpha)=0 \text{ for some monic } f(t) \in \mathbb{Z}[t]\}.$

For $k= \mathbb{Q}$, the ring of integers is just the integers $\mathbb{Z}$, in which case we recall the  Fundamental Theorem of Arithmetic: that every integer $n \neq 0$ may be written as a finite product

$n=\pm 1\cdot p_1 \cdots p_k,$

in which the $\{p_i\}$ are prime and uniquely determined (up to permutation). Domains for which this holds are known in general as unique factorization domains (UFDs). For $k = \mathbb{Q}(\sqrt{d})$ — with $d \neq 0,1$ square-free — the ring of integers $\mathcal{O}_k$ will in general not be a UFD.  In fact, for $d <0$, the integers $\mathcal{O}_k$ have unique factorization only in the 9 cases

$d=-1, -2, -3, -7, -11, -19, -43, -67, -163.$

Far less is known in the case $d >0$ (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 $\mathbb{Q}(\sqrt{d})$ should be a UFD with probability $\approx 0.75446$ as $d \to \infty$ 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 $\tau(n)$ of distinct real quadratic fields $k=\mathbb{Q}(\sqrt{d})$ with $d\leq n$ for which $\mathcal{O}_k$ 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 $n \to \infty$, we have

$\tau(n) \gg \sqrt{n}\log n,$

in which the implied constant is made effective (e.g. greater than $1/11$).

In 1832, Galois introduced the concept of normal subgroups, and proved that the groups $A_n$ (for $n>4$) and $\mathrm{PSL}_2(\mathbb{F}_q)$ (for $q>4$) were simple, i.e. admit no non-trivial (proper) normal subgroups.  In 1892, Hölder asked for a classification of all finite simple groups, and the final classification of finite simple groups in 2004 by Aschbacher and Smith’s resolution of the quasithin case thus resolves an open problem over a century in the making. (Note: the great majority of the classification theorem concluded in the 1980’s.  A computational error was identified and resolved in 2008.)

One immediate consequence of the classification of finite simple groups (CFSG) is that the set S given by

$S:=\{n \in \mathbb{N} : \text{there exists a simple group of order } n\}$

has natural density 0.  Yet it seems unlikely that this result requires the tens of thousands of pages currently involved in the proof of the CFSG, and for the rest of this article we seek to bound the density of S using arithmetic methods.

Let $S_n$ denote the permutation group on n letters.  Herein, we consider the following question, brought to light by Landau in 1902:

Question: What is the maximal order of an element in $S_n$?

By convention, we shall refer to this maximal order as $g(n)$.  Thus, for instance, Lagrange’s theorem gives the elementary bound $g(n) \leq n!$.  We may obtain sharper bounds by noting that $S_n$ is centerless, but tweaks like this will fail to give more than a gain of a multiplicative constant.  A far more productive route comes from the following Proposition:

Proposition: Let $\{a_1,\ldots, a_k\}$ be a partition of n into positive integers.  Then $\prod_i a_i \leq 3^{n/3}$, with equality if and only if $a_i = 3$ for all i.

Proof: Let $\{a_1,\ldots, a_k\}$ be a partition of n such that $\prod a_i$ is maximized.  If $a_i >3$ for any i, we have $2(a_i-2)>a_i$, and the partition

$\{a_1,\ldots, a_{i-1}, 2, a_i-2,a_{i+1},\ldots, a_k\}$

admits a strictly larger product.  Thus, we may assume that $a_i =2,3$ for all i. Furthermore, the relation $2^3 <3^2$ implies that in a maximal partition we may have $a_i =2$ for at most two such i.  It now follows (by cases) that our maximal product $A(n)$ takes the form

$A(n) = \begin{cases} 3^{n/3}, \hspace{15.5 mm} n \equiv 0 \mod 3 \\ 4 \cdot 3^{(n-4)/3}, \quad n \equiv 1 \mod 3 \\ 2 \cdot 3^{(n-2)/3}, \quad n \equiv 2 \mod 3 \end{cases}.$

Our result follows from the inequalities $4 \cdot 3^{-4/3} < 2 \cdot 3^{-2/3} <1$. $\square$

Note: buried under all of this is the fact that the function $f(x) = x^{1/x}$ attains a global maximum at $x=e$, and that 3 is the closest integer to e (this also underlies the marginal appearance of 2, as the second closest integer approximation to e).  This will become more evident as we continue.

As a Corollary, we obtain a new upper bound for the function $g(n)$:

Corollary: For all n, we have $g(n) \leq 3^{n/3}$, with equality if and only if $n=3$.

Proof: We note that $g(n)$ is the maximum value of $\mathrm{lcm}(a_1,\ldots a_k)$, as $\{a_i\}$ varies over the partitions of n.  But $\mathrm{lcm}(a_1\ldots, a_k) \leq \prod a_i \leq 3^{n/3}$, with equality if and only if (1) $a_i =3$ for all and (2) the $\{a_i\}$ are pairwise coprime.  In particular, equality holds above if and only if $n=3$ (achieved with the partition $\{3\}$). $\square$

Note: this result is stronger than the commonly-cited bound $g(n) < e^{n/e}$, which holds for all n.

The bound arising from our Proposition is nevertheless weak in method (in that we have only used the fact that $\mathrm{lcm}(a_1,\ldots,a_k)$ divides $a_1\cdots a_k$), and to strengthen it significantly requires the Prime number theorem (PNT).  In fact, Landau’s classical result

$\log g(n) \sim \sqrt{n \log n}$

(which we will derive after the fold) is equivalent to the PNT by means of elementary methods.

For what follows, we define a loaded die as a discrete probability distribution with six outcomes (labelled {1,…,6}), each of which has positive probability. To each such die, we associate a generating polynomial, given by

$p(t):=p_1t+p_2t^2+\ldots +p_6 t^6$

in which $p_i$ denotes the probability of the outcome i.  If $q(t)$ corresponds to another such die, we note that the product

$p(t)q(t) =\sum_i a_i t^i:=\sum_{i=2}^{12}\big( \sum_{k=1}^{i-1} p_k q_{i-k}\big) t^i$

has coefficients which reflect the probabilities of  certain dice sums for p and (and this is the utility of generating polynomials).  We are now ready to ask the following question:

Question: Does there exist a pair of loaded dice such that the probability of rolling any dice sum ({2,…12}) is equally likely?