アルティン・ハッセの指数関数

アルティン・ハッセの指数関数は、1928年にアルティンとハッセによって、下の級数によって与えられた。

1. 歴史
-----

この級数を指数関数によって表す一つの動機は、無限積に由来する。 形式的冪級数環Q[[}}//x//において、この恒等式が成り立つ。

ここでμはメビウス関数である。これは両辺の対数微分を行うことで示すことができる。同様にして、アルティン・ハッセの指数関数の無限積は:

So passing from 積 over 全ての//n// to a product over only //n// 素数//p//, これは典型的な//p//進解析での操作であり、//ex// から //Ep// を導く。

The coefficients of //Ep// are rational. We can use either formula for //Ep// to prove that, unlike //ex//, all of its coefficients are //p//-integral; in other words, the denominators of the coefficients of //Ep// are not divisible by //p//. A first proof uses the definition of //Ep// and Dwork's lemma, which says that a power series //f// = 1   +   ... with rational coefficients has //p//-integral coefficients if and only if //f/////f////p// ≡ 1 mod //p//p////x//. When //f// = //Ep//, we have //f/////f////p// = //e//−//px//, whose constant term is 1 and all higher coefficients are in //p//p//. A second proof comes from the infinite product for //Ep//: each exponent -μ///n// for //n// not divisible by //p// is a //p//-integral, and when a rational number //a// is //p//-integral all coefficients in the binomial expansion of //a// are //p//-integral by //p//-adic
continuity of the binomial coefficient polynomials //t//...///k//! in //t// together with their obvious integrality when //t// is a nonnegative integer  . Thus each factor in the product of //Ep// has //p//-integral coefficients, so //Ep// itself has //p//-integral coefficients.

2. Combinatorial interpretation
-------------------------------

The Artin–Hasse exponential is the generating function for the probability a uniformly randomly selected element of //Sn//  has //p//-power order :

This gives a third proof that the coefficients of //Ep// are //p//-integral, using the theorem of Frobenius that in a finite group of order divisible by //d// the number of elements of order dividing //d// is also divisible by //d//. Apply this theorem to the //n//th symmetric group with //d// equal to the highest power of //p// dividing //n//!.

More generally, for any topologically finitely generated profinite group //G// there is an identity

where //H// runs over open subgroups of //G// with finite index  and //aG,n// is the number of continuous homomorphisms from //G// to //Sn//. Two special cases are worth noting.  If //G// is the //p//-adic integers, it has exactly one open subgroup of each //p//-power index and a continuous homomorphism from //G// to //Sn// is essentially the same thing as choosing an element of //p//-power order in //Sn//, so we have recovered the above combinatorial interpretation of the Taylor coefficients in the Artin–Hasse exponential series.  If //G// is a finite group then the sum in the exponential is a finite sum running over all subgroups of //G//, and continuous homomorphisms from //G// to //Sn// are simply homomorphisms from //G// to //Sn//. The result in this case is due to Wohlfahrt . The special case when //G// is a finite cyclic group is due to Chowla, Herstein, and Scott , and
takes the form

where //am,n// is the number of solutions to //gm// = 1 in //Sn//.

David Roberts provided a natural combinatorial link between the Artin–Hasse exponential and the regular exponential in the spirit of the ergodic perspective  by showing that the Artin–Hasse exponential is also the generating function for the probability that an element of the symmetric group is unipotent in characteristic //p//, whereas the regular exponential is the probability that an element of the same group is unipotent in characteristic zero.

3. Conjectures
--------------

At the 2002 PROMYS program, Keith Conrad conjectured that the coefficients of are uniformly distributed in the p-adic integers with respect to the normalized Haar measure, with supporting computational evidence. The problem is still open.

Dinesh Thakur has also posed the problem of whether the Artin–Hasse exponential reduced mod //p// is transcendental over .

4. See also
-----------

Witt vector
Formal group

5. References
-------------

//A course in p-adic analysis//, by Alain M. Robert