Adv Alegbra HW Solutions 119

Adv Alegbra HW Solutions 119 - 119 (xiii) For every...

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Unformatted text preview: 119 (xiii) For every positive integer n , there exists a field with exactly 10n elements. Solution. False. (xiv) For every positive integer n , there exists a field with exactly 9n elements. Solution. True. (xv) There exists a field E of characteristic 2 such that x 4 + x + 1 is a product of linear factors in E [x ]. Solution. True. 3.93 For every commutative ring R , prove that R [x ]/(x ) ∼ R . = Solution. Define ϕ : R [x ] → R by f (x ) → constant term of f (x ). It is easy to check that ϕ is a surjective homomorphism with ker ϕ = (x ). 3.94 (Chinese Remainder Theorem in k [x ]) (i) Prove that if k is a field and f (x ), f (x ) ∈ k [x ] are relatively prime, then given b(x ), b (x ) ∈ k [x ], there exists c(x ) ∈ k [x ] with c − b ∈ ( f ) and c − b ∈ ( f ); moreover, if d (x ) is another common solution, then c − d ∈ ( f f ). Solution. Every solution of c − b ∈ ( f ) has the form c = b + g f , where g (x ) ∈ k [x ]. Hence, we must find g such that g f − (b − b ) ∈ ( f ). Since ( f , f ) = 1, there are polynomials s , t ∈ k [x ] with s f + t f = 1. Define g = s (b − b). Then g f = s f (b − b ) = (1 − t f )(b − b ) = (b − b ) − t (b − b ) f ∈ ( f ). If d is another common solution, then both f and f divide c − d . By Exercise 3.94, f f | c − d , and so c − d ∈ ( f f ). (ii) Prove that if k is a field and f (x ), g (x ) ∈ k [x ] are relatively prime, then k [x ]/( f (x )g (x )) ∼ k [x ]/( f (x )) × k [x ]/(g (x )). = Solution. We adapt the proof of Theorem 2.131. ...
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