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Adv Alegbra HW Solutions 97

Adv Alegbra HW Solutions 97 - 97 induction on n 0 Dene b0 =...

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97 induction on n 0. De fi ne b 0 = a 1 0 . If v exists, then the equa- tion u v = 1 would imply that i + j = n a i b j = 0 for all n > 0. Assuming that b 0 , . . . , b n 1 have been de fi ned, then we have 0 = a 0 b n + i + j = n j < n a i b j , and this can be solved for b n because a 0 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 = a n x n + a n + 1 x n + 1 + a n + 2 x n + 2 + · · · = x n ( a n + a n + 1 x + a n + 2 x 2 + · · · ). As a n = 0, we have a n + a n + 1 x + a n + 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 ho- momorphism, 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
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