Math 210B Notes

Proposition 0.1. Let R be a commutative ring. TFAE: 1. Every ideal in R is finitely generated. 2. Every chain of ideals I1 ⊆ I2 ⊆ · · · is stable. That is, there exists n such that In = In+1 = · · · . 3. Every nonempty set of ideals has a maximal element wrt inclusion. Proof. 1. (1) =⇒ (2). I = ⋃ k Ik ideal in R. I is generated by x1, x2, . . . xm. There exists n such that x1, . . . , xm ∈ In. In = I, this implies In = In+1 = In+2 = · · · = I. 2. (2) =⇒ (3). Let A be nonempty set of ideals. SUppose A has no maximal element. PIck I ∈ A. Find chain in A I1 ( I2 ( I3 ( · · · a contradiction. 3. (3) =⇒ (1). I ⊆ R ideal. Suppose I is not finitely generated. Construct x1, x2, · · · ∈ I. x1 ∈ I, x2 ∈ I \ 〈x1〉, x3 ∈ I \ 〈x1, x2〉. This yields 〈x1〉 ( 〈x1, x2〉 ( 〈x1, x2, x3〉 ( · · · Then A = {〈x1, x2, . . . , xi : i ≥ 1} has no maximal element, a contradiction.