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Exercise 1880Algebra › Rings and fieldsL1Added 3 Nov 2026

Prime and maximal ideals

#prime ideal#maximal ideal#quotient

Statement

Let RR be a commutative ring with unit. An ideal II is:
- prime if I≠RI\neq R and ab∈I⇒a∈Iab\in I\Rightarrow a\in I or b∈Ib\in I;
- maximal if I≠RI\neq R and the only ideals containing II are II and RR.

Show that:
1. II is prime iff R/IR/I is an integral domain;
2. II is maximal iff R/IR/I is a field;
3. maximal ⇒\Rightarrow prime.

Hint

Use the correspondence between ideals of RR containing II and ideals of R/IR/I.

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Solution

1. If II prime: aˉbˉ=0\bar a\bar b=0 in R/IR/I iff ab∈Iab\in I, iff a∈Ia\in I or b∈Ib\in I, iff aˉ=0\bar a=0 or bˉ=0\bar b=0. So R/IR/I is an integral domain. Conversely, if R/IR/I integral domain, II prime.

2. If II maximal: any nonzero ideal of R/IR/I lifts to an ideal containing II, so equals R/IR/I. So R/IR/I is a field (every nonzero element invertible). Conversely, if R/IR/I is a field, every ideal containing II is II or RR, so II maximal.

3. A field is an integral domain, so maximal implies prime.

Conclusion. The quotient ring structure reflects the ideal's type: integral domain for prime, field for maximal.

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