Adv Alegbra HW Solutions 97

Adv Alegbra HW Solutions 97 - 97 − induction on n ≥ 0...

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Unformatted text preview: 97 − induction on n ≥ 0. Define b0 = a0 1 . If v exists, then the equation u v = 1 would imply that i + j =n ai b j = 0 for all n > 0. Assuming that b0 , . . . , bn −1 have been defined, then we have 0 = a0 bn + ai b j , i+ j = n j<n and this can be solved for bn because a0 is invertible. (iii) Prove that if σ ∈ k [[x ]] and ord(σ ) = n , then σ = x n u, where u is a unit in k [[x ]]. Solution. Since ord(σ ) = n , we have σ = an x n + an +1 x n +1 + an +2 x n +2 + · · · = x n (an + an +1 x + an +2 x 2 + · · · ). As an = 0, we have an + an +1 x + an +2 x 2 + · · · a unit, by part (ii). 3.41 True or false with reasons. (i) If R and S are commutative rings and f : R → S is a ring homomorphism, then f is also a homomorphism from the additive group of R to the additive group of S . Solution. True. (ii) If R and S are commutative rings and if f is a homomorphism from the additive group of R to the additive group of S with f (1) = 1, then f is a ring homomorphism. Solution. False. (iii) If R and S are isomorphic commutative rings, then any ring homomorphism f : R → S is an isomorphism. Solution. False. (iv) If f : R → S is a ring homomorphism, where S is a nonzero ring, then ker f is a proper ideal in R . Solution. True. (v) If I and J are ideals in a commutative ring R , then I ∩ J and I ∪ J are also ideals in R . Solution. False. (vi) If ϕ : R → S is a ring homomorphism and if I is an ideal in R , then ϕ( I ) is an ideal in S . Solution. False. ...
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