Adv Alegbra HW Solutions 109

Adv Alegbra HW Solutions 109 - 109 Solution. There are q ,...

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109 Solution. There are q , r R with b i = qb i + 1 + r , where either r = 0or ∂( r )<∂( b i + 1 ) .I f r = 0, then b i = qb i + 1 ; canceling, 1 = qb , and this contradicts the hypothesis that b is not a unit. Therefore, r ±= 0. But r = b i qb i + 1 = b i ( 1 qb ), so that ∂( b i ) ∂( b i ( 1 qb )) = ∂( r ) (1 qb is nonzero because b is not a unit). Thus, ∂( b i ) ∂( r )<∂( b i + 1 ). 3.72 If k is a f eld, prove that the ring of formal power series k [[ x ]] is a PID. Solution. Let I be an ideal in k [[ x ]] ; we may assume that I ±={ 0 } because { 0 }= ( 0 ) . Choose τ I of smallest order, say, τ = a m x m + a m + 1 x m + 1 + ···= x m v , where v = a m + a m + 1 x +··· ; note that v is a unit in k [[ x ]] , by Exercise 3.40. We claim that I = ( x m ) .Now x m = v 1 τ I , so that ( x m ) I . For the reverse inclusion, if σ k [[ x ]] is nonzero, then σ = b n x n + b n + 1 x n + 1 +···= x n w , where n
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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