781hw9 - x 3 2 3 ω(11 is difficult to solve in R since...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
18.781, Fall 2007 Problem Set 9 Due: WEDNESDAY, November 14 This set of problems concludes our work on reciprocity laws and algebraic number theory. 1. Show that p - 1 p - 1 X t =0 ζ t ( x - y ) p = δ ( x, y ) where δ ( x, y ) = ± 1 x y ( p ) 0 x 6≡ y ( p ) 2. Let χ be a character on Z /p Z . (That is, a multiplicative homomor- phism from ( Z /p Z ) × to the non-zero complex numbers.) Define, as in class, g ( a, χ ) = X t (mod p ) χ ( t ) ζ at p . Prove that if a 6 = 0 and χ 6 = 1 p , the trivial character mod p , that g ( a, χ ) = χ ( a - 1 ) g (1 , χ ) 3. The Jacobi sum (used in the cubic case for our proof of reciprocity) is defined by J ( χ, λ ) = X a,b ( p ) a + b =1 χ ( a ) λ ( b ) where χ and λ are characters of Z /p Z . Prove the following: (a) J (1 p , 1 p ) = p (again 1 p is the trivial character which is always 1 mod p . By convention, we take 1 p (0) = 1 for this problem, though that is not always the standard convention.) (b) J (1 p , χ ) = 0 (c) J ( χ, χ - 1 ) = - χ ( - 1) (d) If χ · λ 6 = 1 p , then J ( χ, λ ) = g ( χ ) g ( λ ) g ( χλ )
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4. Let p 1 (3), and set p = π ¯ π , with π a primary prime in R . Show that x 3 a ( p ) is solvable in the integers if and only if ± a π ² = 1 . 5. The congruence
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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 ) = p-1 X t (mod p ) f ( t ) ζ-at 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 .)...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

781hw9 - x 3 2 3 ω(11 is difficult to solve in R since...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online