# Adv Alegbra HW Solutions 105 - → R is a commutative ring...

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105 (ii) If g ( x ), f ( x ) Z [ x ] with f ( x ) ±= 0, then there exist q ( x ), r ( x ) Z [ x ] with g ( x ) = f ( x ) q ( x ) + r ( x ) , where either r ( x ) = 0o r deg ( r )< deg ( f ) . Solution. False. (iii) The gcd of 2 x 2 + 4 x + 2 and 4 x 2 + 12 x + 8in Q [ x ] is 2 x + 2. Solution. False. (iv) If R is a domain, then every unit in R [ x ] has degree 0. Solution. True. (v) If k is a f eld and p ( x ) k [ x ] is a nonconstant polynomial having no roots in k , then p ( x ) is irreducible in k [ x ] . Solution. False. (vi) For every quadratic s ( x ) C [ x ] , there are a , b C and q ( x ) C [ x ] with ( x + 1 ) 1000 = s ( x ) q ( x ) + ax + b . Solution. True. (vii) If k = F p ( x ) , where p is a prime, and if f ( x ), g ( x ) k [ x ] satisfy f ( a ) = g ( a ) for all a k , then f ( x ) = g ( x ) . Solution. True. (viii) If k is a f eld, then k [ x ] is a PID. Solution. True. (ix) Z is a Euclidean ring. Solution. True. (x) There is an integer m such that m 2 ≡− 1 mod 89. Solution. True. 3.57 Given a commutative ring R , we saw, in Exercise 3.10, that F ( R ) = { all functions
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Unformatted text preview: → R } is a commutative ring under pointwise operations. (i) If R is a commutative ring, prove that ϕ : R [ x ] → F ( R ) , given by ϕ : f ( x ) 7→ f [ , is a homomorphism. Solution. Absent. (ii) If k is an in f nite f eld, prove that ϕ is an injection. Solution. Absent. 3.58 Find the gcd of x 2 − x − 2 and x 3 − 7 x + 6 in F 5 [ x ] , and express it as a linear combination of them. Solution. The Euclidean algorithm shows that gcd ( x 2 − x − 2 , x 3 − 7 x + 6 ) = x − 2 , and x − 2 = − 1 4 ( x 3 − 7 x + 6 ) + 1 4 ( x + 1 )( x 2 − x − 2 )....
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