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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 ).

In 1832, Galois introduced the concept of normal subgroups, and proved that the groups (for ) and (for ) 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

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 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 ?*

By convention, we shall refer to this maximal order as . Thus, for instance, Lagrange’s theorem gives the elementary bound . We may obtain sharper bounds by noting that 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 be a partition of n into positive integers. Then , with equality if and only if for all i.*

*Proof: *Let be a partition of *n* such that is maximized. If for any *i*, we have , and the partition

admits a strictly larger product. Thus, we may assume that for all *i*. Furthermore, the relation implies that in a maximal partition we may have for at most two such *i*. It now follows (by cases) that our maximal product takes the form

Our result follows from the inequalities .

*Note: buried under all of this is the fact that the function attains a global maximum at , 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 :

**Corollary: ***For all n, we have , with equality if and only if .*

*Proof: *We note that is the maximum value of , as varies over the partitions of *n*. But , with equality if and only if (1) for all *i *and (2) the are pairwise coprime. In particular, equality holds above if and only if (achieved with the partition ).

*Note: this result is stronger than the commonly-cited bound , 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 divides ), and to strengthen it significantly requires the Prime number theorem (PNT). In fact, Landau’s classical result

(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

in which denotes the probability of the outcome *i*. If corresponds to another such die, we note that the product

has coefficients which reflect the probabilities of certain dice sums for *p* and *q *(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?*