8410Kate - SOLUTIONS TO PROBLEMS 3.5, 3.9.5, 3.11, 3.13.5...

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Unformatted text preview: SOLUTIONS TO PROBLEMS 3.5, 3.9.5, 3.11, 3.13.5 KATE THOMPSON 3.5 Let p be an odd prime number, and let u Z be a quadratic nonresidue modulo p . Then [ Q p : Q 2 p ] = 4, and a set of coset representatives is 1 ,p,u,pu . Solutions: First, we intend to use Hensels Lemma for this proof; given that, we feel obligated to state which version we will use. Lemma. (Hensels Lemma) Let ( K,v ) be a Henselian (e.g. complete) valued field with valuation ring R and residue field k . Let f R [ t ] . If there exists an R such that v ( f ( )) > 2 v ( f ( )) , then there exists a R with f ( ) = 0 and v ( - ) > v ( f ( )) . Let P ( X ) = X 2- a for a a p-adic integer. It is clear that there is a root in Z p only if v ( a ) = v ( x 2 ) = 2 v ( x ) is even. Then, by dividing a by an appropriate power of p 2 m , we can reduce to the case v ( a ) = 0, or a Z p . Now we claim that if a Z p is a p-adic unit and if there exists such that 2 a (mod p Z p ), then a is a square of an element in Z p . To see this, apply Hensels lemma to X 2- a . Since p 6 = 2 and a Z p , we are guaranteed to have 2 6 0 (mod p ) so were fine. Next, we claim that x Q p is a square if and only if x = p 2 n y 2 for n Z , and y Z p a p- adic unit. One direction is obvious. The second direction is also fairly obviousjust refer to the previous case, and also note that p is itself NOT a square in Q p (since if it had a square root, then that square root would have valuation 1 2 ). We can conclude then that Q p : Q 2 p = 4, and is given by { 1 ,p,u,pu } for u a quadratic non-residue in Z p . Referring to the text below this Exercise in the course notes, we can moreover conclude that Q p / ( Q 2 p ) = Z / 2 Z Z / 2 Z 3.9.5 A very special and important quadratic form is q H ( x 1 ,x 2 ) = x 1 x 2 , the so-called hyperbolic plane . a) Let K be any field of characteristic different from 2. Give an explicit change of variables that diagonalizes q H . Solutions: Short answer: let P = 1- 1 2- 1 2- 1 Why does this definitely work? Since the characteristic of our field is not 2, we can represent the quadratic form x 1 x 2 as the following symmetric matrix: M = 2- 1 2- 1 Thus, using matrix theory, we claim that P works. To confirm, we compute: 1 2- 1- 1 2- 1 2- 1 2- 1 1- 1 2- 1 2- 1 = 2- 2 2- 1 2- 2- 2- 1 1- 1 2- 1 2- 1 = 2- 1- 2- 1 Last, note that since the characteristic of a field is NOT 2, 2- 1 exists. 1 b) Show by brute force that q H cannot be diagonalized over F 2 . See part (c) for a less forceful method. , c) Show that q H cannot be diagonalized over any field of characteristic 2....
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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.

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8410Kate - SOLUTIONS TO PROBLEMS 3.5, 3.9.5, 3.11, 3.13.5...

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