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Unformatted text preview: Elizabeth Vidaurre Homework 1, Sept 16 1. Let l > 3 be a prime number, be a primitive l th root of unity, D = Z [ ] and k an integer 6 0 (mod l ). Claim. k 1  1 and k + 1 are units of D . Proof. Note that k 1  1 = k 1 + k 2 + ... + 1, so k 1  1 D . It remains to show  1 k 1 D . Since l is prime, k is a primitive l th root of unity and j such that ( k ) j = . We have that  1 k 1 = ( k ) j 1 k 1 = ( k ) j 1 + ( k ) j 2 + ... + 1 Z [ k ] Since and k are primitive l th roots of unity, D = Z [ k ]. Therefore, k 1  1 is a unit. Note that k +1 = ( k ) 2 1 k 1 . Since l > 3, k 6 0 (mod l ). Therefore, we can apply the previous result and it follows that k + 1. 2. Let l > 3 be a prime number, be a primitive l th root of unity and k an integer 6 0 (mod l ). Claim. k + 1 is a not a root of unity. Proof. Since k is a root of unity,  k  = 1. Let k = a + bi for some a,b R . We have that 1 =  k  = 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 =  k +1  . In other words, k + 1 is not a root of unity. Claim. k 1  1 is not a root of unity except when k l 1 (mod l ) . Proof. Note that k 1  1 is a root of unity if and only if  k 1  1  = 1 if and only if  k 1  =   1  ....
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 Spring '11
 Meydup
 Algebra

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