Adv Alegbra HW Solutions 106

# Adv Alegbra HW Solutions 106 - 106 3.59 Let k be a ﬁeld...

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Unformatted text preview: 106 3.59 Let k be a ﬁeld, let f (x ) ∈ k [x ] be nonzero, and let a1 , a2 , . . . , at be some distinct roots of f (x ) in k . Prove that f (x ) = (x − a1 )(x − a2 ) · · · (x − at )g (x ) for some g (x ) ∈ k [x ]. Solution. Absent. 3.60 If R is a domain and f (x ) ∈ R [x ] has degree n , show that f (x ) has at most n roots in R . Solution. Let F = Frac( R ); if f (x ) had more than n roots in R , then it would have more than n roots in F (which contains R ), and this is a contradiction. 3.61 Let R be an arbitrary commutative ring. If f (x ) ∈ R [x ] and if a ∈ R is a root of f (x ), that is, f (a ) = 0, prove that there is a factorization f (x ) = (x − a )g (x ) in R [x ]. Solution. Absent. 3.62 (i) Show that the following pseudocode implements the Euclidean algorithm ﬁnding gcd f (x ) and g (x ) in k [x ], where k is a ﬁeld. Input: g , f Output: d d := g ; s := f WHILE s = 0 DO rem := remainder(d , s ) d := s s := rem END WHILE a := leading coefﬁcient of d d := a −1 d Solution. Absent. (ii) Find ( f , g ), where f (x ) = x 2 + 1, g (x ) = x 3 + x + 1 ∈ I3 [x ]. Solution. Absent. 3.63 Prove the converse of Euclid’s lemma. Let k be a ﬁeld and let f (x ) ∈ k [x ] be a polynomial of degree ≥ 1; if, whenever f (x ) divides a product of two polynomials, it necessarily divides one of the factors, then f (x ) is irreducible. Solution. The proof is the same as for the corresponding result for integers. 3.64 Let f (x ), g (x ) ∈ R [x ], where R is a domain. If the leading coefﬁcient of f (x ) is a unit in R , then the division algorithm gives a quotient q (x ) and a remainder r (x ) after dividing g (x ) by f (x ). Prove that q (x ) and r (x ) are uniquely determined by g (x ) and f (x ). ...
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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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