Adv Alegbra HW Solutions 95

Adv Alegbra HW Solutions 95 - 95 Solution. Assume that x a...

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95 Solution. Assume that x a | f ( x ) .I f ( x a ) 2 | f ( x ) , then f ( x ) = ( x a ) 2 h ( x ) , and so f 0 ( x ) = 2 ( x a ) h ( x ) + ( x a ) 2 h 0 ( x ) . Hence, x a | f 0 . (The assumption ( x a ) | f ( x ) is not needed in this direction.) Conversely, assume that x a | f ( x ) and x a | f 0 ; hence f ( x ) = ( x a ) g ( x ) and f 0 ( x ) = ( x a ) k ( x ) .Now ( x a ) k ( x ) = f 0 ( x ) = (( x a ) g ) 0 = g + ( x a ) g 0 , and so x a | g ; that is, g ( x ) = ( x a )`( x ) . Hence, f ( x ) = ( x a ) g ( x ) = ( x a ) 2 `( x ) . 3.38 (i) If f ( x ) = ax 2 p + bx p + c F p [ x ] , prove that f 0 ( x ) = 0. Solution. If f ( x ) = ax 2 p + bx p + c F p [ x ] , then f ( x ) = 2 pax 2 p 1 + pbx p 1 = 0 . (ii) State and prove a necessary and suf f cient condition that a polyno- mial f ( x ) F p [ x ] have f 0 ( x ) = 0. Solution. The condition is that there should be a polynomial g ( x ) = b n x n with f ( x ) = g ( x p ) ; that is, f ( x ) = a n x np and b p n = a n for all n . As in part (i), if f ( x ) has this form, then f 0 = 0. Conversely, if f ( x ) = a n x n has a n ±= 0 for some n not divisible by p , then f 0 has a nonzero coef f cient na n and so f 0 ±= 0. 3.39 If R is a commutative ring, de f ne R [[ x ]] to be the set of all formal power series over R . (i) Show that the formulas de f ning addition and multiplication on R [ x ] make sense for R [[ x ]] , and prove that R [[
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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