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Unformatted text preview: x 3 ≡ 2 + 3 ω (11) is diﬃcult to solve in R since there are 121 residue classes mod 11. By cubic reciprocity ³ 2 + 3 ω 11 ´ = ³ 11 2 + 3 ω ´ . Use this fact, together with your knowledge of the congruence x 3 ≡ 11 (7) to prove that the initial congruence mod 11 in R is not solvable. 6. Prove g (1 , p ) = ∑ t (mod p ) ζ t 2 p by evaluating the sum X t (mod p ) [1 + ³ t p ´ ] ζ t p (Recall that ζ p denotes a p th root of unity e 2 πi/p and the Gauss sum g ( a, p ) = X t (mod p ) ³ t p ´ ζ at p 7. Let f be a function from Z → C , the complex numbers. Suppose that for p prime, f ( n + p ) = f ( n ) for all integers n . Deﬁne ˆ f ( a ) = p1 X t (mod p ) f ( t ) ζat p . Prove that f ( t ) = X a (mod p ) ˆ f ( a ) ζ at p (Note the similarity with Fourier analysis for periodic functions on R .)...
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 Spring '09
 BRUBAKER
 Algebra, Number Theory, Algebraic number theory, initial congruence mod, trivial character mod

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