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Unformatted text preview: Math 302 Modern Algebra Spring 2009 Homework Solutions 4.2 DIVISIBILITY IN F [ x ] 4. (a) Prove: For all f ( x ) , g ( x ) ∈ F [ x ], if f ( x )  g ( x ) and g ( x )  f ( x ), then there exists c ∈ F , c 6 = 0 F such that f ( x ) = cg ( x ). This has symbolic form ° ∀ f ( x ) ∈ F [ x ] ¢ ° ∀ g ( x ) ∈ F [ x ] ¢ £ f ( x )  g ( x ) ∧ g ( x )  f ( x ) → ( ∃ c 6 = 0 F ) [ f ( x ) = cg ( x )] § . Proof: Suppose f ( x ) and g ( x ) are arbitrary nonzero polynomials in F [ x ]. (We need to assume both f ( x ) and g ( x ) are nonzero since each is a divisor of the other.) Suppose f ( x )  g ( x ) and g ( x )  f ( x ) . We need to show that there exists c ∈ F , c 6 = 0 F such that f ( x ) = cg ( x ). Since g ( x )  f ( x ), then there exists a ( x ) ∈ F [ x ] such that f ( x ) = a ( x ) g ( x ) . Since f ( x )  g ( x ), then there exists b ( x ) ∈ F [ x ] such that g ( x ) = b ( x ) f ( x ) . Combining these two equalities, we obtain f ( x ) = a ( x )[ b ( x ) f ( x )] = [ a ( x ) b ( x )] f ( x ) . Since F is a field, then F [ x ] is an integral domain by Corollary 4.3. Then the Cancellation Law holds in F [ x ] by Theorem 3.10, so this equation implies a ( x ) b ( x ) = 1 F . This implies a ( x ) is a unit in F [ x ] and hence is a nonzero constant polynomial. Therefore a ( x ) = c for some c ∈ F , c 6 = 0 F . Since f ( x ) = a ( x ) g ( x ), then this implies f ( x ) = cg ( x ) for some c ∈ F , c 6 = 0 F , as required. (b) Prove: If f ( x ) and g ( x ) in part (a) are monic, then f ( x ) = g ( x ). Proof: Suppose f ( x ) and g ( x ) in part (a) are monic. We need to show that f ( x ) = g ( x ). From part (a), we know that f ( x ) = cg ( x ) for some nonzero c ∈ F . Since g ( x ) is monic, then the leading coeﬃcient of cg ( x ) is c 1 F = c . Since f ( x...
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This note was uploaded on 10/20/2009 for the course MATH 302 taught by Professor Edwards during the Spring '03 term at CSU Fullerton.
 Spring '03
 Edwards
 Algebra

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