8410BestofJohnDoyle - SOLUTIONS TO EXERCISES 1.3, 1.12,...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTIONS TO EXERCISES 1.3, 1.12, 1.14, 1.16 JOHN DOYLE Exercise 1.3: Let | · | be a norm on a field k . a) If x ∈ 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 = ⇒ | x n | = | 1 | = ⇒ | x | n = | 1 | = ⇒ | x | n = 1 = ⇒ | x | is a positive real n th root of 1. Therefore | x | = 1. b) ( (i) = ⇒ (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 + ··· + 1), the n-fold sum of 1. Then either 2 = 0 (hence k has characteristic 2), or there exists m ∈ 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 ∈ F p [ x ] for some m ∈ N , it follows that k is algebraic over F p . ( (ii) = ⇒ (i): ) Suppose k has characteristic p and is algebraic over F p , and let 0 6 = α ∈ k . Since k is algebraic, there exists a polynomial f ∈ F p [ x ] such that f ( α ) = 0. Let n := deg f . Then α is contained a subfield of k that is a degree n extension of F p , i.e., α ∈ F p n . Therefore α satisfies x p n = x ; since α is nonzero, it has an inverse, so we conclude that α p n- 1 = 1. Thus α 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....
View Full Document

This note was uploaded on 10/26/2011 for the course MATH 8410 taught by Professor Staff during the Fall '11 term at University of Georgia Athens.

Page1 / 4

8410BestofJohnDoyle - SOLUTIONS TO EXERCISES 1.3, 1.12,...

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