TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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$R / P$ is an integral domain if and only if $P$ is a prime ideal

Statement

$R / P$ is an integral domain if and only if $P$ is a prime ideal.

Here $R$ is commutative and $P ⊴ R$ is an ideal.

Facts & Assumptions

Given: A commutative ring $R$ and a two-sided ideal $P ⊴ R$ .

[L1]

A prime ideal is proper and satisfies $a b \in P \Rightarrow a \in P$ or $b \in P$ (Prime ideals and maximal ideals in a commutative ring).

[L2]

Products in $R / P$ are $(a + P) (b + P) = a b + P$ (For a two-sided ideal $I$ , the additive cosets form a ring $R / I$ with identity $1 + I$ ).

[L3]

An integral domain is a nonzero commutative ring without zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

[L4]

The canonical projection $R \to R / P$ has kernel $P$ ; equivalently, $a + P = 0 + P$ exactly when $a \in P$ (The canonical projection $R \to R / I$ is a surjective ring homomorphism with kernel $I$ ).

Proof

technique · direct

1.1

If $P$ is prime, then $P$ is proper, so $1 + P \neq 0 + P$ by [L4]; and $(a + P) (b + P) = 0 + P$ gives $a b \in P$ , hence $a + P = 0 + P$ or $b + P = 0 + P$ , while commutativity of $R$ makes the quotient commutative.

L1L2L3L4givenalgebra

2.1

If $R / P$ is a domain and $a b \in P$ , then $(a + P) (b + P) = 0 + P$ , so [L3] and [L4] give $a \in P$ or $b \in P$ ; its nonzero identity gives $1 \notin P$ , hence $P ⊊ R$ .

step 1.1L1L2L3L4givenalgebra

3.1

These implications prove the equivalence.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ernst, An Inquiry-Based Approach to Abstract Algebra, Maximal and Prime Ideals (standard reference, not scraped)

R/PR/P is an integral domain if and only if PP is a prime ideal