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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).

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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?

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