Adv Alegbra HW Solutions 29

Adv Alegbra HW Solutions 29 - p k u k p k a k p = max {k a...

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29 A similar argument shows that if p e | p r i for i 1, then p e | p r m i . By the fundamental theorem of arithmetic, the total number of factors p occurring on each side must be the same. Therefore, the total number of p s dividing ap r must equal the total number of p s dividing p r m . Since p - m , the highest power of p dividing p r m is p r , and so the highest power of p dividing ap r is p r ; that is, p - a = ³ p r m p r ´ = ³ n p r ´ , as desired. 1.73 (i) For all rationals a and b , prove that k ab k p =k a k p k b k p and k a + b k p max {k a k p , k b k p } . Solution. If a = p e p e 1 1 ··· p e n n and b = p f p f 1 1 ··· p f n n , then ab = p e + f p e 1 + f 1 1 ··· p e n + f n n . Hence k ab k p = p e f = p e p f =k a k p k b k p . Assume e f , so that f ≤− e and k a k p = max {k a k p , k b k p } . a + b = p e p e 1 1 ··· p e n n + p f p f 1 1 ··· p f n n = p e ³ p e 1 1 ··· p e n n + p f e p f 1 1 ··· p f n n ´ . If u = p e 1 1 ··· p e n n + p f e p f 1 1 ··· p f n n , then either u = 0or k u k p = p 0 = 1. In the f rst case, k a + b k p = 0, and the result is true. Otherwise, k a + b k p = p e k u k p =k a k
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Unformatted text preview: p k u k p k a k p = max {k a k p , k b k p } . (ii) For all rationals a , b , prove p ( a , b ) 0 and p ( a , b ) = 0 if and only if a = b . Solution. p ( a , b ) 0 because k c k p 0 for all c . If a = b , then p ( a , b ) = k a b k p = k k p = 0; conversely, if p ( a , b ) = 0, then a b = 0 because 0 is the only element c with k c k p = 0. (iii) For all rationals a , b , prove that p ( a , b ) = p ( b , a ) . Solution. p ( a , b ) = p ( b , a ) because k c k p = k 1 k p k c k p = k c k p ....
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