Exercise 6.11

By evaluating ∑ ⁡ t ( 1 + ( t ∕ p ) ) ζ t in two ways, prove that g = ∑ ⁡ t ζ t 2 .

Answers

Proof. For a ∈ 𝔽 p , Write N [ x 2 = a ] the number of solutions of the equation x 2 = a in 𝔽 p . We know from Ex. 5.2 that N [ x 2 = a ] = 1 + ( a ∕ p ) . Therefore

∑ t = 0 p − 1 ζ t 2 = ∑ t ¯ ∈ 𝔽 p ζ t 2 = ∑ a ¯ ∈ 𝔽 p N [ x 2 = a ] ζ a = ∑ t ¯ ∈ 𝔽 p ( 1 + ( t p ) ) ζ t = ∑ t ¯ ∈ 𝔽 p ζ t + ∑ t ¯ ∈ 𝔽 p ( t p ) ζ t = ∑ t = 0 p − 1 ( t p ) ζ t = g □
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2022-07-19 00:00
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