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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.

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Have you used LHS is multiplicative? –  user27126 Jan 29 '13 at 8:25
    
@Sanchez Sorry, what do you mean? I realize that $|h|^{-s}\cdot |g|^{-s}=|hg|^{-s}$, but I don't know if that's what you're getting at. –  Noomi Holloway Jan 29 '13 at 8:32
    
I meant to say that $\sigma(fg) = \sigma(f)\sigma(g)$ if $f$,$g$ are relatively prime. This, together with $F[x]$ is UFD, gives that LHS = $\prod_f \sum_{n=o}^{\infty} \sigma (f^n) |f|^{-ns}$. Simplifying this would probably give you what you want, but the answer below is definitely simpler. –  user27126 Jan 29 '13 at 9:17

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\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*}

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Thanks, this is very clear. –  Noomi Holloway Jan 29 '13 at 9:18

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