# Abstract Algebra [Lecture notes] by Irena Swanson

By Irena Swanson

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**Sample text**

You may want to suggestively write the equivalence class of (a, b) in the form ab . ) Define binary operations +, · on F as (a, b) + (c, d) = (ad + bc, bd), (a, b) · (c, d) = (ac, bd). Prove that +, · are well-defined and that they make F into a field. iii) Let i : R → F be the function i(r) = (r, 1). Prove that i is an injective ring homomorphism. iv) Prove that for every f ∈ F there exists a non-zero r ∈ R such that i(r)f ∈ i(R). , P = R) ideal such that a, b ∈ R and ab ∈ P implies that a ∈ P or b ∈ P .

Nk , the kernel is n1 · · · nk , so that the obvious map ϕ above is injective. 6 (Criterion for when Zn1 ⊕ Zn2 ⊕ · · · ⊕ Znk is isomorphic to Zn1 ···nk ) Zn1 ⊕ Zn2 ⊕ · · · ⊕ Znk is isomorphic to Zn1 ···nk if and only if the ni are pairwise relatively prime. Proof. 5 and by iduction on k, the ni are pairwise relatively prime. It remains to prove the other direction. So assume that the ni are pairwise relatively prime. By the set-up just before this theorem, the obvious map ϕ : Zn1 ···nk → Zn1 ⊕ Zn2 ⊕ · · · ⊕ Znk is injective.

As m2 , . . , mk are multiples of n1 , this is the same as a1 m1 b1 mod n1 , and from the equality 1 = a1 m1 +a2 m2 + · · ·+ak mk , we deduce that a1 m1 mod n1 = 1. Thus the first component of ϕ ◦ ψ(b1 , . . , bk ) equals b1 , and similarly for the other components, so that ϕ ◦ ψ is the identity function. Then ϕ ◦ ψ ◦ ϕ = ϕ, and since ϕ is one-to-one, necessarily ψ ◦ ϕ is the identity as well. 7 Criterion for when Un1 ⊕ Un2 ⊕ · · · ⊕ Unk is isomorphic to Un1 ···nk : If every pair of the ni is relatively prime, then the isomorphism holds.