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Unformatted text preview: SOLUTIONS TO EXERCISES 1.3, 1.12, 1.14, 1.16 JOHN DOYLE Exercise 1.3: Let  Â·  be a norm on a field k . a) If x âˆˆ k is a root of unity, then  x  = 1. b) For a field k , TFAE: i) Every nonzero element of k is a root of unity. ii) char k = p > 0, and k/ F p is algebraic. c) If k/ F p is algebraic, then the only absolute value on k is the trivial absolute value. Solution: a) Let x be an n th root of unity. x n = 1 = â‡’  x n  =  1  = â‡’  x  n =  1  = â‡’  x  n = 1 = â‡’  x  is a positive real n th root of 1. Therefore  x  = 1. b) ( (i) = â‡’ (ii): ) Suppose every nonzero element of k is a root of unity. Then every nonzero element satisfies the polynomial x m 1 for some positive in teger m . Let n = (1 + 1 + Â·Â·Â· + 1), the nfold sum of 1. Then either 2 = 0 (hence k has characteristic 2), or there exists m âˆˆ N such that 2 m = 1. Then 2 m 1 = 0; since k has no zero divisors, it must be the case that one of the prime factors p of 2 m 1 is zero. Therefore k has characteristic p > 0. Finally, because every element of k satisfies the polynomial x m 1 âˆˆ F p [ x ] for some m âˆˆ N , it follows that k is algebraic over F p . ( (ii) = â‡’ (i): ) Suppose k has characteristic p and is algebraic over F p , and let 0 6 = Î± âˆˆ k . Since k is algebraic, there exists a polynomial f âˆˆ F p [ x ] such that f ( Î± ) = 0. Let n := deg f . Then Î± is contained a subfield of k that is a degree n extension of F p , i.e., Î± âˆˆ F p n . Therefore Î± satisfies x p n = x ; since Î± is nonzero, it has an inverse, so we conclude that Î± p n 1 = 1. Thus Î± is a root of unity. c) By part (b), we have that every nonzero element of k is a root of unity, and by (a) we have that every root of unity has absolute value 1. Therefore it is determined that any absolute value must take the value zero at zero, and 1 at every other element; i.e., every absolute value is trivial....
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 Fall '11
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 Number Theory, Valuation, Abelian group, John Doyle

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