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Unformatted text preview: Elizabeth Vidaurre Homework 1, Sept 16 1. Let l > 3 be a prime number, Î¶ be a primitive l th root of unity, D = Z [ Î¶ ] and k an integer 6â‰¡ 0 (mod l ). Claim. Î¶ k 1 Î¶ 1 and Î¶ k + 1 are units of D . Proof. Note that Î¶ k 1 Î¶ 1 = Î¶ k 1 + Î¶ k 2 + ... + 1, so Î¶ k 1 Î¶ 1 âˆˆ D . It remains to show Î¶ 1 Î¶ k 1 âˆˆ D . Since l is prime, Î¶ k is a primitive l th root of unity and âˆƒ j such that ( Î¶ k ) j = Î¶ . We have that Î¶ 1 Î¶ k 1 = ( Î¶ k ) j 1 Î¶ k 1 = ( Î¶ k ) j 1 + ( Î¶ k ) j 2 + ... + 1 âˆˆ Z [ Î¶ k ] Since Î¶ and Î¶ k are primitive l th roots of unity, D = Z [ Î¶ k ]. Therefore, Î¶ k 1 Î¶ 1 is a unit. Note that Î¶ k +1 = ( Î¶ k ) 2 1 Î¶ k 1 . Since l > 3, k 6â‰¡ 0 (mod l ). Therefore, we can apply the previous result and it follows that Î¶ k + 1. 2. Let l > 3 be a prime number, Î¶ be a primitive l th root of unity and k an integer 6â‰¡ 0 (mod l ). Claim. Î¶ k + 1 is a not a root of unity. Proof. Since Î¶ k is a root of unity,  Î¶ k  = 1. Let Î¶ k = a + bi for some a,b âˆˆ R . We have that 1 =  Î¶ k  = âˆš a 2 + b 2 . Next, notice that a 2 + b 2 = a 2 +2 a +1+ b 2 = ( a +1) 2 + b 2 if and only if a = 1 2 . Since the only root of unity with real part 1 2 is a cubed root of unity and l 6 = 3, we have that a 6 = 1 2 and a 2 + b 2 6 = ( a +1) 2 + b 2 . Therefore, 1 = a 2 + b 2 6 = ( a +1) 2 + b 2 =  Î¶ k +1  . In other words, Î¶ k + 1 is not a root of unity. Claim. Î¶ k 1 Î¶ 1 is not a root of unity except when k â‰¡ l 1 (mod l ) . Proof. Note that Î¶ k 1 Î¶ 1 is a root of unity if and only if  Î¶ k 1 Î¶ 1  = 1 if and only if  Î¶ k 1  =  Î¶ 1  ....
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This note was uploaded on 02/11/2011 for the course MATH 1111 taught by Professor Meydup during the Spring '11 term at Hartford.
 Spring '11
 Meydup
 Algebra

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