Exercise 7.19

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $F$ be a finite field with $q$ elements . If $f(x) \in F[x]$ has degree $t$, put $|f| = q^t$. Verify the formal identity $\sum_f |f|^{-s} = (1-q^{1-s})^{-1}$. The sum is over all monic polynomials.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $U$ be the set of monic polynomials in $\mathbb{F}_q[x]$, and $U_t$ the set of monic polynomials of degree $t$, and $s\in \C$. Then $U = \bigcup_{t \in \N} U_t$, so
\begin{align*}
\sum_{f \in U} \vert f \vert^{-s }&= \sum_{t=0}^\infty \sum_{f \in U_t} \vert f \vert^{-s}\
&=\sum_{t=0}^\infty \frac{1}{q^{ts}} \sum_{f\in U_t} 1.
\end{align*}
As $\sum_{f\in U_t} 1 = \mathrm{Card} \, (U_t) = q^t$, then, for $\mathrm{Re}(s) >1$

\begin{align*}
\sum_{f \in U} \vert f \vert^{-s } &= \sum_{t=0}^\infty \frac{1}{q^{t(s-1)}}\
&=\frac{1}{1-\frac{1}{q^{s-1}}}\
&= (1-q^{1-s})^{-1}.
\end{align*}
As $\left | \frac{1}{q^{t(s-1)}} ight | = \frac{1}{q^{t(\mathrm{Re}(s)-1)}}$, the sum is absolutely convergent for $\mathrm{Re}(s)>1$. This justifies the grouping of terms in this sum.

\bigskip

Conclusion:  if $\mathrm{Re}(s)>1$, 
$$\sum_{f \in U} \vert f \vert^{-s } =(1-q^{1-s})^{-1},$$
where $U$ is the set of monic polynomials in  $\mathbb{F}_q[x]$.
\end{proof}

Exercise 7.19

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Exercise 7.19

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