Quotient Field Properties

Abstract Algebra · Axiom Academy

LESSON Quotient Field Properties Understanding when R/I is a field or integral domain through maximal and prime ideals The quotient ring "collapses" all elements of I to zero. Each equivalence class [a] contains a and all elements that differ from a by something in I. Proof Sketch (⇒): If R/M is a field and I is an ideal with M ⊂ I ⊂ R, then I/M is an ideal of R/M. But fields have no nontrivial ideals, so I/M = 0 or I/M = R/M, giving I = M or I = R. Proof Sketch (⇐): If M is maximal and [a] ≠ [0] in R/M, then a ∉ M. The ideal (M, a) properly contains M, so by maximality (M, a) = R. Thus 1 ∈ (M, a), meaning 1 = m + ra for some m ∈ M, r ∈ R. Taking equivalence classes: [1] = [ra], so [r] is the multiplicative inverse of [a]. 3. Prime Ideals ↔ Integral Domains Recall: An ideal P is prime if P ≠ R and whenever ab ∈ P, either a ∈ P or b ∈ P. An integral domain has no zero divisors: if ab = 0, then a = 0 or b = 0. Proof Connection: In R/P, [a]·[b] = [0] means ab ∈ P. If P is prime, this forces a ∈ P or b ∈ P, so [a] = [0] or [b] = [0]. Conversely, if R/P has no zero divisors and ab ∈ P, then [a]·[b] = [0] implies [a] = [0] or [b] = [0]. J ↦ J/I (ideals of R containing I → ideals of R/I) K̄ ↦ π⁻¹(K̄) (ideals of R/I → ideals of R containing I) This powerful result lets us "lift" ideal properties from R/I back to R. If M/I is a maximal ideal of R/I, then M is a maximal ideal of R containing I. Similarly for prime ideals. 5. Applications: Constructing Fields

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