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Suppose that also f (x) = q (x)g(x) + r (x) with q (x), r (x) ∈ F [x] and deg r (x) < deg g(x). We have r(x) − r (x) = [f (x) − q(x)g(x)] − [f (x) − q (x)g(x)] = [q (x) − q(x)]g(x). If q(x) = q (x), then we get the contradiction deg g(x) > max{deg r(x), deg r (x)} ≥ deg[r(x) − r (x)] = deg[(q (x) − q(x))g(x)] = deg[q (x) − q(x)] + deg g(x) ≥ deg g(x), since deg[q (x) − q(x)] ≥ 0. Therefore, q (x) = q(x) and the earlier equation gives r (x) = r(x) as well. This completes the proof. 4 Zeros of a polynomial Let R be a commutative ring and let f (x) = ni=0 ai xi be a polynomial over R.

Define ϕ : S → (S +I)/I by ϕ(s) = s+I. Then ϕ is a homomorphism (it is simply the restriction to S of the canonical epimorphism R → R/I). For s ∈ S we have s ∈ ker ϕ ⇐⇒ ϕ(s) = I ⇐⇒ s + I = I ⇐⇒ s ∈ S ∩ I, so ker ϕ = S ∩ I. Let x ∈ (S + I)/I. Then x = (s + a) + I = s + I for some s ∈ S and a ∈ I, and we have ϕ(s) = s + I = x, so ϕ is surjective. 1), S/(S ∩ I) = S/ ker ϕ ∼ = im ϕ = (S + I)/I, and the proof is complete. 4 Third Isomorphism Theorem Let R be a ring and let I and J be ideals of R with J ⊇ I.

Let I be an ideal of R. An element r of R is in the kernel of π if and only if I = π(r) = r + I, which occurs if and only if r ∈ I. Therefore, I = ker π. 6 Homomorphism is injective iff kernel is trivial Here is a reminder of a useful criterion from group theory for checking injectivity of a homomorphism. Let ϕ : R → R be a homomorphism of rings. 1 Theorem. ϕ is injective if and only if ker ϕ = {0}. Proof. Assume that ϕ is injective. Let r ∈ ker ϕ. Then ϕ(r) = 0. 3. So ϕ(r) = ϕ(0) and injectivity of ϕ gives r = 0.

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Abstract Algebra II by Randall R. Holmes


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