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8410Kate

# 8410Kate - SOLUTIONS TO PROBLEMS 3.5 3.9.5 3.11 3.13.5 KATE...

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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 Hensel’s Lemma for this proof; given that, we feel obligated to state which version we will use. Lemma. (Hensel’s 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 α R such that v ( f ( α )) > 2 v ( f ( α )) , then there exists a β R with f ( β ) = 0 and v ( α - β ) > v ( f ( α )) . 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 α such that α 2 a (mod p Z p ), then a is a square of an element in Z × p . To see this, apply Hensel’s lemma to X 2 - a . Since p = 2 and a Z × p , we are guaranteed to have 2 α 0 (mod p ) so we’re 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 obvious–just 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 = 0 2 - 1 2 - 1 0 Thus, using matrix theory, we claim that P works. To confirm, we compute: 1 2 - 1 - 1 2 - 1 0 2 - 1 2 - 1 0 1 - 1 2 - 1 2 - 1 = 2 - 2 2 - 1 2 - 2 - 2 - 1 1 - 1 2 - 1 2 - 1 = 2 - 1 0 0 - 2 - 1 Last, note that since the characteristic of a field is NOT 2, 2 - 1 exists. 1

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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. Solutions: Suppose that it can be diagonalized. This means that for some a, b, c, d F , ad - bc = 0, we have a c b d 0 1 0 0 a b c d = e 0 0 f Working this out, we see that we have: a c b d 0 1 0 0 a b c d = 0 a 0 b a b c d = ac ad bc bd This means that ad = bc = 0; however, that would contradict our assumption that ad - bc = 0.
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8410Kate - SOLUTIONS TO PROBLEMS 3.5 3.9.5 3.11 3.13.5 KATE...

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