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Adv Alegbra HW Solutions 96

# Adv Alegbra HW Solutions 96 - 96(iii Denote a formal power...

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96 (iii) Denote a formal power series σ = ( s 0 , s 1 , s 2 , . . . , s n , . . . ) by σ = s 0 + s 1 x + s 2 x 2 + · · · . Prove that if σ = 1 + x + x 2 + · · · , then σ = 1 /( 1 x ) is in R [[ x ]] . Solution. We have 1 + x + x 2 + · · · = 1 + x ( 1 + x + x 2 + · · · ). Hence, if σ = 1 + x + x 2 + · · · , then σ = 1 + x σ. Solving for σ gives σ = 1 /( 1 x ) . 3.40 If σ = ( s 0 , s 1 , s 2 , . . . , s n , . . . ) is a nonzero formal power series, de fi ne ord (σ) = m , where m is the smallest natural number for which s m = 0. (i) Prove that if R is a domain, then R [[ x ]] is a domain. Solution. If σ = ( s 0 , s 1 , . . . ) and τ = ( t 0 , t 1 , . . . ) are nonzero power series, then each has an order ( σ = 0 if and only if it has an order); let ord (σ) = p and ord (τ) = q . Write στ = ( c 0 , c 1 , . . .). For any n 0, we have c n = i + j = n s i t j . In particular, if n < p + q , then i < p and s i = 0 or j < q and t j = 0; it follows that c n = 0 because each summand s i t j = 0. The same analysis shows that c p + q = s p t q , for all the other terms are 0. Since R is a domain, s p = 0 and t q = 0 imply s p t q = 0. Therefore, ord (στ) = ord (σ)
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