This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 1 Solutions 1. Case 2 of the Division Algorithm follows almost identically to Case 1. The big change is replacing q + 1 in the existence proof with q 1 since a is now negative. 2. Problem 1.51: Let I = { k ∈ Z : ζ k = 1 } ⊆ Z . We claim that I meets the hypotheses of Corollary 1.37. (1) follows clearly since ζ = 1. (2): Suppose a,b ∈ I then ζ a b = ζ a ζ b = 1 1 = 1 which implies a b ∈ I . (3): Suppose a ∈ I and q ∈ Z . Then ζ qa = ( ζ a ) q = 1 q = 1 which implies qa ∈ I . Since I meets the hypotheses of Corollary 1.37 we know that there exists a nonnegative integer d such that I consists precisely of all the multiples of d . Furthermore, d 6 = 0 since ζ is a root of unity and thus ∃ k 6 = 0 such that ζ k = 1. Hence we have a positive d as desired. 3. Problem 1.56: Let a,b ∈ Z such that as + bt = 1 for some s,t ∈ Z . From the proof of Theorem 1.35 we know that ( a,b ) is the smallest positive linear combination of a and b . Since as + bt = 1, 1 must be the smallest positive linear combination (there are no smaller positive integers). Thereforethe smallest positive linear combination (there are no smaller positive integers)....
View
Full
Document
This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra, Division

Click to edit the document details