Suppose $F$ is a finite field of order $q$ a prime power. If $f\in F[x]$ of degree $t$, set $|f|=q^t$. Let $\sigma(f)=\sum_{g\mid f}|g|$ where the sum is over the monic divisors of $f$.
Why does $$ \sum_f \sigma(f)|f|^{-s}=(1-q^{1-s})^{-1}(1-q^{2-s})^{-1} $$ where the sum on the left is over all monic polynomials?
I have been able to verify that $\sum_f |f|^{-s}=(1-q^{1-s})^{-1}$ and that $\sum_f d(f)|f|^{-s}=(1-q^{1-s})^{-2}$ where $d(f)$ is the number of monic divisors of $f$. I can include proofs if needed. I started by writing $$ \sum_f \sigma(f)|f|^{-s}=\sum_f\left(\sum_{g\mid f}|g|\right)|f|^{-s} $$ and I know $\sum_{g\mid f}|g|$ consists of $d(f)$ terms, but I don't see a clever manipulation to tie it together. This sum is found in Ireland & Rosen. Thanks.
\begin{eqnarray*} \sum_f \sigma(f)|f|^{-s}&=&\sum_f \sum_{g: \ g|f} |g| |f|^{-s}\\ &=& \sum_g \sum_{f:\ g|f} |g| |f|^{-s}\\ &=&\sum_{g}\sum_h |g| |gh|^{-s} \ \ ({\rm setting}\ f=gh)\\ &=& (\sum_h |h|^{-s}) (\sum_g |g|^{1-s}) \\ &=& (\sum_{n\ge 0} q^{n(1-s)})(\sum_{m\ge 0} q^{m(2-s)})\\ &=& (1-q^{1-s})^{-1} (1-q^{2-s})^{-1}. \end{eqnarray*}