97induction onn≥0. Defineb0=a−10. Ifvexists, then the equa-tionuv=1 would imply that∑i+j=naibj=0 for alln>0.Assuming thatb0, . . . ,bn−1have been defined, then we have0=a0bn+i+j=nj<naibj,and this can be solved forbnbecausea0is invertible.(iii)Prove that ifσ∈k[[x]]and ord(σ)=n, thenσ=xnu,whereuis a unit ink[[x]].Solution.Since ord(σ)=n, we haveσ=anxn+an+1xn+1+an+2xn+2+ · · ·=xn(an+an+1x+an+2x2+ · · ·).Asan=0, we havean+an+1x+an+2x2+· · ·a unit, by part (ii).3.41True or false with reasons.(i)IfRandSare commutative rings andf:R→Sis a ring ho-momorphism, thenfis also a homomorphism from the additivegroup ofRto the additive group ofS.Solution.True.(ii)IfRandSare commutative rings and iffis a homomorphismfrom the additive group ofRto the additive group of
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