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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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 Fall '11
 KeithCornell

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