# Generalizations of divided-power algebras over finite fields

In Andrews, Askey, and Roy's Special Functions, the authors state that Gauß sums are finite field analogs of the $\Gamma$-function as Jacobi sums are to B-function. The $\Gamma$-function is well-known as the analytic continuation for the factorial function and if $x$ is in a divided power algebra then $$x^{[m]}\cdot x^{[n]}={(m+n)!\over (n!)(m!)}x^{[m+n]}.$$ where $x^{[n]}$ is the nth power of x under the divided power meaning of $x^n$. For characteristic 0, this combinatorial coefficient can be written as $${\Gamma(1+m+n)\over \Gamma(1+m)\Gamma(1+n)}=B(1+x,1+y)^{-1}\cdot[(1+x)+(1+y)].\quad (*)$$ What I mean to ask is: "Is there value in considering these objects--i.e. divided power algebras and these coefficients-- from this point of view--i.e. use of Gauß and Jacobi sums? Is this formula just a coincidence or a pattern that I want to see, or is there more to it?"

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maybe this is dumb, but your coefficient is always an integer, so can't you just reduce it mod p? – Sean Tilson Nov 8 '10 at 13:29
Sometimes there are issues with dividing by p. I only put up the formula for multiplying powers of x. The "normal" $\gamma_n(x)$ is ${1\over n!}x^n$ where the $\{\gamma_i\}$ are structure morphisms for the algebra. It's this that I'm having issues with, especially because I have an interest in exponential maps on this algebra, and division is a standard obstruction to this. Sorry for not making things more clear from the outset. – Adam Hughes Nov 8 '10 at 14:46
You don't have to define your divided power algebra that way. The $\gamma_i(x)$ are a choice of elements of the ring that must satisfy some formulas that would be satisfied if you used your definition. In fact, the equations you assume should imply that $i!\gamma_i(x)=x^i$. – Sean Tilson Dec 1 '13 at 23:05
I recognize that it's not strictly necessary, but I'm asking for this particular choice, sorry if that was unclear. – Adam Hughes Dec 2 '13 at 1:21
That particular choice is the classical example. If your ring contains the rationals, then that is the unique system of divided powers on your ring. – Sean Tilson Dec 4 '13 at 18:51