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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 $P\mathrel{\trianglelefteq}R$ is an ideal.

Facts & Assumptions

Given: A commutative ring $R$ and a two-sided ideal $P\mathrel{\trianglelefteq}R$ .

[L1]

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

[L2]

Products in $R/P$ are $(a+P)(b+P)=ab+P$ (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 $1 \ne 0$ 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 $1+P\ne0+P$ by [L4]; and $(a+P)(b+P)=0+P$ gives $ab\in P$ , 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 $ab\in P$ , 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 $P\subsetneq R$ .

step 1.1L1L2L3L4givenalgebra

3.1

These implications prove the equivalence.

step 2.1∎

Depends on

Used by

Every maximal ideal of a commutative ring is prime Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

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