Notes on Speyer’s paper titled “Some Sums over Irreducible Polynomials”

by ayushkhaitan3437

Let \mathcal{P} be the set of irreducible polynomials over F_2[T]. Then \sum\limits_{P\in \mathcal{P}}\frac{1}{1-P}=0. The paper lists certain examples of \frac{1}{1-P} below. These are all expanded as geometric series. As one can see only P=T, T+1 contribute to the coefficient of T^{-1} in the sum \sum\limits_{P\in \mathcal{P}}\frac{1}{1-P}=0. Why don’t the other irreducible polynomials do the same? This is because these are the only two linear polynomials in F_2[T]. All other polynomials are of higher degree. Moreover, all other irreducible polynomials have the constant term 1; otherwise they would be reducible, as T would be a common factor. Hence \frac{1}{P-1} would be of the form \frac{1}{T^{a_1}+T^{a_2}+\dots+T^{a_n}}, where a_1>1. Now divide both the numerator and denominator by T^{a_1}. So we get an expression of the form \frac{1}{T^{a_1}}(\frac{1}{1+T^{a_2-a_1}+T^{a_3-a_1}+\dots+T^{a_n-a_1}}). As a_i-a_1<0 for all i\neq 1, this is a power series expansion in negative powers of T. Also, as a_1\geq 2, all such negative exponents will be less than -1. This proves that only the polynomials T and T+1 contribute to the coefficient of T^{-1} in \sum\limits_{P\in \mathcal{P}}\frac{1}{1-P}=0.

We now try and understand Theorem 1.1 in this paper. Let \mathcal{P_1} be the set of monic irreducible polynomials in F_{2^n}[T]. Then \sum\limits_{P\in \mathcal{P_1}}\frac{1}{P^k-1}\in F_{2^n}(T) for any k\equiv 0(\mod 2^n-1).

A corollary of this is that \sum\limits_{P\in \mathcal{P}}\frac{1}{P^k-1} is in F_{2^n}(T)

Proof of corollary: We have rewritten \sum\limits_{P\in \mathcal{P}}\frac{1}{P^k-1} as \sum\limits_{P\in \mathcal{P}_1}\sum\limits_{a\in \Bbb{F}_q^\times}\frac{1}{(aP)^k-1}, where q=2^n. Why can we do that? This is because for any a\in\Bbb{F}_q^\times, a^{q-1}=1. Hence, we’re essentially counting the same thing as before. Aren’t we counting each term |\Bbb{F}_q^\times| times? Also, every irreducible polynomial is of the form aP for some P\in\mathcal{P_1}. Now consider the identity \sum\limits_{a\in \Bbb{F}_q^\times}\frac{1}{(aX)^k-1}=\frac{1}{(X)^{lcm(k,q-1)}-1} in \Bbb{F}_q(U). Why is this true? This is because \frac{1}{(aX)^k-1} can be written as \sum\limits_{j=1}^\infty\frac{1}{(ax)^{kj}} (just multiply and divide \frac{1}{(aX)^k-1} by \frac{1}{({aX})^k}).

Now, as \sum\limits_{a\in\Bbb{F}_q}a^m=1 if m\equiv 0 \mod q-1, and \sum\limits_{a\in\Bbb{F}_q}a^m=0 otherwise. This is because if m\equiv 0 \mod q-1, then \sum a^m is essentially adding 1 to itself q-1 times. As the characteristic of the field is 2, and as q-1 is essentially 2^m-1, this sum is equal to the inverse of 1, which is exactly 1. When q\not\equiv 0\mod q-1, then \sum a^m=0. This can be verified independently.