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Let $\chi$ be a character mod $m$. Consider the Dirichlet series

$$ L(s,\chi)=\sum_{n=1}^{\infty}\chi(n)/(n^s)$$ We know that if $\chi$ is no trivial: $$L(s,\chi)=\prod_{p\nmid m}(1-\chi(p)/(p^s))^{-1}$$

Now if $\chi'$ is a character mod $d$ such that $d|m$ and $\chi'$ induces $\chi$, prove that $$ L(1,\chi)=\prod_{p|m,p\nmid d}(1-\chi'(p)/p)L(1,\chi')$$

I tried using the fact that if $\chi'$ induces $\chi$, then they should have same results, but failing.

I thought something like this, let $f$ denote the trivial character mod $m$, then $L(1,\chi)=\prod_{p}(1-(\chi'(p)f(p))/p)^{-1}=\prod_{p\nmid d}(1-(\chi'(p)f(p)/p)^{-1}$ but I'm not sure and I can't see how to proceed.

1 Answers 1

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If $a(n)$ is multiplicative (that is $gcd(n,m) = 1 \implies a(n) a(m) = a(nm)$) then as formal Dirichlet series $$\sum_{n=1}^\infty a(n) n^{-s} = \prod_p (1+\sum_{k=1}^\infty a(p^k) p^{-sk})$$ (Euler product)

Now $a(n)$ is completely multiplicative (that is $\forall n,m , a(n) a(m) = a(nm)$) iff $$\sum_{n=1}^\infty a(n) n^{-s} = \prod_p (1+\sum_{k=1}^\infty a(p)^k p^{-sk}) = \prod_p \frac{1}{1-a(p) p^{-s}}$$

Since a Dirichlet character $\chi(n)$ is completely multiplicative and periodic, $|\chi(n)|$ is bounded and so $$L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s} = \prod_p \frac{1}{1-\chi(p) p^{-s}}$$ converges absolutely and is analytic for $Re(s) > 1$ (Note : an Euler product is said to converge iff the Dirichlet series for its logarithm converges)

From this, you easily get that if $\psi(n) = \chi(n) 1_{gcd(n,k)=1}$ is a non-primitive character induced by $\chi(n)$ a character modulo $m$ then $$L(s,\psi) = \prod_p \frac{1}{1-\psi(p) p^{-s}}=\prod_p \frac{1}{1-1_{gcd(p,k)=1}\chi(p) p^{-s}}$$ $$=\prod_{p\, {\displaystyle\nmid}\, k} \frac{1}{1-\chi(p) p^{-s}} = L(s,\chi) \prod_{p | k} (1-\chi(p)p^{-s})= L(s,\chi) \prod_{p | k, p \, {\displaystyle\nmid}\, m} (1-\chi(p)p^{-s})$$

Finally, using some non-trivial theorems (see a proof of the prime number theorem and the PNT in arithmetic progressions) you get that the Euler product for $L(s,\chi)$ converges and is analytic on $Re(s) = 1$ whenever $\chi$ is non-principal, and so $$L(1,\psi) = L(1,\chi) \prod_{p | k, p\, {\displaystyle\nmid}\, m} (1-\chi(p)p^{-1})$$

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    sorry why $$\prod_{p\, {\displaystyle\nmid}\, k} \frac{1}{1-\chi(p) p^{-s}} = L(s,\chi) \prod_{p | k} (1-\chi(p)p^{-s})$$ ?? – 2017-01-20
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    @Richard what is $L(s,\chi)$ ? .. – 2017-01-20
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    I tried substituting, but I do not know how to manage the double product that comes out. Any details please? – 2017-01-20
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    @Richard no really it is trivial, write what is $L(s,\chi) \prod_{p | k} (1-\chi(p)p^{-s})$ – 2017-01-20
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    it is $\prod_{p}1/((1-(\chi(p)p^{-s}))\prod_{p|k}(1-\chi(p)p^{-s})$ – 2017-01-20
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    @Richard Forget what you wrote in your question and on your paper, and use **what I wrote** – 2017-01-20
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    It is $\prod_{p}1/((1-(\chi(p)p^{-s}))\prod_{p|k}(1-\chi(p)p^{-s})$. Ok so in this way works because only $p\nmid k$ factors survive. Thank you. – 2017-01-20
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    @Richard $\displaystyle=\frac{\prod_{p|k}(1-\chi(p)p^{-s})}{\prod_{p}(1-\chi(p)p^{-s})} = \frac{1}{\prod_{p \ \nmid\ k} (1-\chi(p)p^{-s}}$. And the $\displaystyle\prod_{p \ \nmid \ m}$ simply comes from $\chi(p ) = 0$ if $p | m$.. – 2017-01-20