Adv Alegbra HW Solutions 107

Adv Alegbra HW - x = x − a 1 ·· x − a n ∈ R x where R is a commutative ring Show that f x has no repeated roots(that is all the a i are

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107 Solution. If F = Frac ( R ) , use the division algorithm in F [ x ] to compute the quotient and remainder for dividing g ( x ) by f ( x ) . If either quotient or remainder in R [ x ] were not unique, then they would not be unique in F [ x ] . 3.65 Let k be a f eld, and let f ( x ), g ( x ) k [ x ] be relatively prime. If h ( x ) k [ x ] , prove that f ( x ) | h ( x ) and g ( x ) | h ( x ) imply f ( x ) g ( x ) | h ( x ) . Solution. The proof is the same as for Exercise 1.58, mutatis mutandis. 3.66 If k is a f eld in which 1 + 1 ±= 0, prove that 1 x 2 / k ( x ) , where k ( x ) is the f eld of rational functions. Solution. Suppose, on the contrary, that 1 x 2 = f ( x )/ g ( x ) , where f ( x ), g ( x ) k [ x ] ; we may assume that f ( x )/ g ( x ) is in lowest terms; that is, f ( x ) and g ( x ) are relatively prime. Cross multiply and square, obtaining f ( x ) 2 = g ( x ) 2 ( 1 x 2 ) = g ( x ) 2 ( 1 x )( 1 + x ). Since 1 + 1 ±= 0, the polynomials 1 x and 1 + x are relatively prime and irreducible, Euclid s lemma gives 1 x | f ( x ) and 1 + x | f ( x ) ; that is, f ( x ) = ( 1 x 2 ) h ( x ) for some h ( x ) k [ x ] . After substituting and canceling, h ( x ) 2 ( 1 x 2 ) = g ( x ) 2 . Repeat the argument to obtain 1 x | g ( x ) , and this contradicts f ( x ) and g ( x ) being relatively prime. Therefore, 1 x 2 / k ( x ) . 3.67 Let f
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Unformatted text preview: ( x ) = ( x − a 1 ) ··· ( x − a n ) ∈ R [ x ] , where R is a commutative ring. Show that f ( x ) has no repeated roots (that is, all the a i are distinct) if and only if the gcd ( f , f ) = 1, where f is the derivative of f . Solution. Exercise 3.37 says that x − a is a common divisor of f ( x ) and f if and only if ( x − a ) 2 | f ( x ) ; that is, if and only if f ( x ) has repeated roots. 3.68 Let ∂ be the degree function of a Euclidean ring R . If m , n ∈ N and m ≥ 1, prove that ∂ is also a degree function on R , where ∂ ( x ) = m ∂( x ) + n for all x ∈ R . Conclude that a Euclidean ring may have no elements of degree 0 or degree 1. Solution. If x , y ∈ R are nonzero, then ∂ ( x ) = m ∂( x ) + n ≤ m ∂( xy ) + n = ∂ ( xy )....
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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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