Let $p$ be a prime number, $m$ be any integer, $f(p)$ be the order of $p$ in $(Z/mZ)^*$, $i.e.$ $p^{f(p)} \equiv 1 \pmod m$ with $f(p)$ smallest.

Let $g(p)=\frac{\phi(m)}{f(p)}$ is a integer where $\phi$ is the euler $\phi$-function.

Then why is $\zeta_m(s)=\prod\limits_{p\nmid m} \frac{1}{(1-\frac{1}{p^{f(p)s}})^{g(p)}}$ a Dirichlet series with positive integral coefficients?

(The above formula is derived from $L$ functions, $\zeta_m(s)=\prod\limits_\chi L(s,\chi).$)

A Dirichlet series is a series of the form $\sum a_ne^{-\lambda_n z}$$\,$ $(a_n, z\in \mathbb C,$ $\lambda_n$ increasing to $\infty $ in $\mathbb R$) , if let $\lambda_n=\log n$, then get $\sum \frac{a_n}{n^s}$, the Ordinary Dirichlet Series.

It also has a formula $\sum\limits^\infty_1 \chi(n)n^{-s}=\prod\limits_{p}\frac{1}{1-\frac{\chi(p)}{p^s}} $, where $\chi$ is any multiplicative function--i.e. $\chi(xy)=\chi(x)\chi(y))$.

I attempted to apply the product formula above, then I need $\left(1-\frac{\chi(p)}{p^s}\right)= \left(1-\frac{1}{ p^{f(p)s} }\right)^{g(p)}$, but it doesn't solve to a multiplicative function $\chi$, so why is $\zeta_m(s)$ a Dirichlet series?

(All things above are found in Serre's A Course In Arithmetic.)

  • 3
    $\begingroup$ Use $1/(1-x)= 1+ x + x^2 + \cdots$ and multiply it through. $\endgroup$ – i707107 Feb 7 '15 at 15:43

Let's recall the basics of Dirichlet series: with $A(s)=\sum_n a_nn^{-s}, B(s) =\sum_n b_nn^{-s}$ we have

$$A(s)B(s)=\sum_n\left(\sum_{d|n}a_d b_{n/d}\right)n^{-s}$$

is the Dirichlet product. We conclude that if $a_n\ge 0, b_n\ge 0$ for all $n$, then $AB(s)$ has all positive coefficients.

But then we note that $\zeta_m(s)$ is exactly a series of this type, since if we remove the $g(p)$ power we have a $p$-local factor of

$$A_p={1\over 1-p^{-f(p)s}}=\sum_{n}p^{-nf(p)s}$$

so raising this to the $g(p)$ power gives us another Dirichlet series with non-negative coefficients. But then multiplying all the $p$ components together gives us a result with non-negative coefficients. One might complain that we have an infinite product, so I cannot use the finite products rule, but recall that each $n$ factors as a finite number of primes, so that it's coefficients are still only finite sums of non-negative numbers, hence we may still use the result.


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