By Randall R. Holmes
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Extra resources for Abstract Algebra II
Proof. 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.
Its inverse map A → A is given by S → ϕ−1 (S ). (b) For S, T ∈ A, ϕ(S) ⊆ ϕ(T ) if and only if S ⊆ T , and in this case |ϕ(T ) : ϕ(S)| = |T : S|. (c) For S, T ∈ A, ϕ(S) ϕ(T )/ϕ(S) ∼ = T /S. ϕ(T ) if and only if S T , and in this case Proof. 3, if S is a subring of R, then ϕ(S) is a subring of R so the map is well defined. By this same section, if S is a subring of R , then ϕ−1 (S ) is a subring of R, and this latter subring contains ker ϕ since ϕ(k) = 0 ∈ S for all k ∈ ker ϕ. Therefore, the indicated inverse map is also well defined.
We conclude that f (x) is irreducible as claimed. Other irreducibility criteria are given in Section 10. 6 F[x] is PID if F is field Let F be a field. 1 Theorem. The polynomial ring F [x] is a PID. Proof. We first check that F [x] is an integral domain. Any two monomials axi and bxj (a, b ∈ F ) commute since axi bxj = abxi+j = baxi+j = bxj ai . That F [x] is commutative now follows from the distributive law and the fact that polynomials are sums of monomials. The constant polynomial 1 is an identity in F [x].
Abstract Algebra II by Randall R. Holmes