Exercise 11.20

If in Theorem 2 we consider the base field to be F s instead of F , we get a different zeta function, Z f ( s ) ( u ) . Show that Z f ( s ) ( u ) and Z f ( u ) are related by the equation Z f ( s ) ( u s ) = Z f ( u ) Z f ( ρu ) ⋯ Z f ( ρ s − 1 u ) , where ρ = e 2 iπ ∕ s .

Answers

Proof. Let Ω be an algebraic closure of 𝔽 p and write 𝔽 q for the unique subfield of Ω with cardinality q , if q is a power of p . Here F = 𝔽 q , and F s = 𝔽 q s . Recall that the function zeta only depends on the cardinality of the finite field, not on the choice of this field (see Exercise 3).

Then

Z f ( s ) ( u ) = exp ⁡ ( ∑ t = 1 ∞ N t ( s ) u t t ) ,

where N t ( s ) is the number of points of H ¯ f ( 𝔽 q st ) , because the degree of 𝔽 q st over 𝔽 q s is

[ 𝔽 q st : 𝔽 q s ] = [ 𝔽 q st : 𝔽 q ] [ 𝔽 q s : 𝔽 q ] = st s = t .

Therefore N t ( s ) = N st , where as usual N s is the number of points of H ¯ f ( 𝔽 q ) . This gives

Z f ( s ) ( u ) = exp ⁡ ( ∑ t = 1 ∞ N st u t t ) .

Now, since ln ⁡ ( Z f ( u ) ) = ∑ ⁡ k = 0 ∞ N k u k k , we obtain

ln ⁡ ( Z f ( ρ j u ) ) = ∑ k = 0 ∞ N k ρ kj u k k , j = 0 , 1 , … , s − 1 .

The sum of these s equalities gives

∑ j = 0 s − 1 ln ⁡ ( Z f ( ρ j u ) ) = ∑ k = 0 ∞ N k ( ∑ j = 0 s − 1 ρ kj ) u k k .

Moreover,

∑ j = 0 s − 1 ρ kj = { 1 − ρ ks 1 − ρ k = 0 if  s ∤ k , s otherwise.

Therefore

∑ j = 0 s − 1 ln ⁡ ( Z f ( ρ j u ) ) = ∑ s ∣ k N k s u k k = ∑ t = 1 ∞ N st u st t ( k = st ) = ln ⁡ ( Z f ( s ) ( u s ) ) .

To conclude,

Z f ( u s ) = Z f ( u ) Z f ( ρu ) ⋯ Z f ( ρ s − 1 u ) , ( ρ = e 2 iπ ∕ s ) .

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2022-07-19 00:00
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