Exercise 7.20

With the notation of Exercise 19 let d ( f ) be the number of monic divisors of f and σ ( f ) = ∑ ⁡ g ∣ f | g | , where the sum is over the monic divisors of f . Verify the following identities :

(a)
∑ ⁡ f d ( f ) | f | − s = ( 1 − q 1 − s ) − 2 .
(b)
∑ ⁡ σ ( f ) | f | − s = ( 1 − q 1 − s ) − 1 ( 1 − q 2 − s ) − 1 .

Answers

Proof. (a) With the notation of 7.19, for s ∈ ℂ , Re ( s ) > 1 , ∑ ⁡ f ∈ U | f | − s is absolutely convergent and

( 1 − q 1 − s ) − 1 = ∑ f ∈ U | f | − s .

Then

( 1 − q 1 − s ) − 2 = ∑ f ∈ U | f | − s ∑ g ∈ U | g | − s = ∑ ( f , g ) ∈ U 2 | fg | − s = ∑ h ∈ U ∑ g ∈ U , g ∣ h | h | − s ,

indeed, the application

φ : { U × U → { ( h , g ) ∈ U × U , g ∣ h } ( f , g ) ↦ ( fg , g )

is a bijection.

So

( 1 − q 1 − s ) − 2 = ∑ h ∈ U | h | − s Card { g ∈ U , g ∣ h } = ∑ h ∈ U | h | − s d ( h ) = ∑ f ∈ U d ( f ) | f | − s .

(b) Similarly,

( 1 − q 1 − s ) − 1 ( 1 − q 2 − s ) − 1 = ∑ f ∈ U | f | − s ∑ g ∈ U | g | − s + 1 = ∑ ( f , g ) ∈ U 2 | g | | fg | − s = ∑ h ∈ U ∑ g ∈ U , g ∣ h | g | | h | − s = ∑ h ∈ U | h | − s ∑ g ∈ U , g ∣ h | g | = ∑ h ∈ U σ ( h ) | h | − s = ∑ f ∈ U σ ( f ) | f | − s . □
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2022-07-19 00:00
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