Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-27
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Prime ideals of a quotient ring are exactly the prime ideals containing the ideal

Statement

Let R be a commutative ring, let I⊴R be an ideal, and let π:R→R/I be the quotient map. Then contraction along π induces an inclusion-preserving bijection Spec⁡(R/I)→V(I), sending q to π−1(q). Its inverse sends a prime ideal p⊇I to p/I.

Facts & Assumptions

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

[L1]

Primes of R/I correspond to primes of R containing I, and strict inclusions are preserved (Primes of a quotient lie over the kernel).

[L2]

Every quotient map induces a spectrum map by contraction (A ring map induces a contraction map on prime spectra).

Proof

technique · direct
1.1L1L2

By [L2], contraction along π gives a map Spec⁡(R/I)→Spec⁡(R). The quotient-prime correspondence [L1] shows that its values are precisely prime ideals containing I, so the map lands in V(I).

1.2L1

The same correspondence [L1] provides the inverse assignment p↦p/I on V(I), and it also shows that extension and contraction undo one another and preserve inclusion.

2.1step 1.1step 1.2∎

Therefore contraction along π identifies Spec⁡(R/I) with V(I).

Depends on

Used by

…and 1 more result.

Dependency tree · two levels

7 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