Math 210 B : Algebra , Homework 3

so ker f ⊂ imh. Hence ker f = imh so the sequence is exact at the middle term. Further, f is injective since Z[X] is a domain, so left multiplication by any element is injective. Finally, h is surjective since the constant polynomials g = a have g(0) = a for every a ∈ Z. Therefore the sequence is exact. The sequence is split over R. The requisite map q : Z→ R so that h ◦ q = 1Z. It is clear that q(a) = f(ga) +a for some ga ∈ R, since h(q(a)) = a implies that q(a) has constant term a. Since we require