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

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.