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LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-27
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Primes of a quotient lie over the kernel

Statement

Let R be a commutative ring, let I⊴R be an ideal, and let π:R→R/I be the quotient map. If q∈Spec⁡(R/I), then π−1(q) is a prime ideal of R containing I. If p∈Spec⁡(R) contains I, then p/I is a prime ideal of R/I. Both assignments preserve strict inclusion.

Facts & Assumptions

Given: A commutative ring R, an ideal I⊴R, and the quotient map π:R→R/I.

[L1]

Ideals of R/I correspond to ideals of R containing I (Correspondence theorem: ideals of R/I correspond to ideals of R containing I).

[L2]

A prime ideal is a proper ideal that absorbs factors of a product (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct
1.1L2givenalgebra

Let q∈Spec⁡(R/I). Because 0+I∈q, the contraction π−1(q) contains I. If ab∈π−1(q), then (a+I)(b+I)=ab+I∈q, so [L2] gives a∈π−1(q) or b∈π−1(q). Also 1∉π−1(q) because q is proper. Hence π−1(q) is prime.

1.2L1L2givenalgebra

Let p∈Spec⁡(R) with I⊆p. By [L1], p/I is an ideal of R/I. If (a+I)(b+I)=ab+I∈p/I, then ab∈p, so [L2] gives a+I∈p/I or b+I∈p/I. Properness is inherited from 1∉p. Inclusion preservation is immediate from [L1].

2.1step 1.1step 1.2∎

Therefore primes of the quotient and primes of R above I correspond by extension and contraction, with strict inclusions preserved.

Depends on

Used by

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Sources