Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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Associated primes of a cyclic quotient are colon primes

Statement

Let R be a commutative ring and let I⊴R be an ideal. Then

Ass⁡R(R/I)={p∈Spec⁡(R):p=(I:r) for some r∉I}.

Facts & Assumptions

Given: A commutative ring R and an ideal I⊴R.

[L1]

A prime ideal belongs to Ass⁡R(M) exactly when it is the annihilator of some element of M (Associated primes of a module).

[L2]

The quotient module R/I consists of the cosets r+I (Quotient module M/N with scalar multiplication on additive cosets).

[L3]

The annihilator of an element x is the set of scalars that kill x (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct
1.1

For any r∈R, the annihilator of r+I in R/I is Ann⁡R(r+I)={a∈R:a(r+I)=0+I}={a∈R:ar∈I}=(I:r).

L2L3given
2.1

If p∈Ass⁡R(R/I), then [L1] gives a nonzero class r+I with p=Ann⁡R(r+I). Since r+I≠0, one has r∉I, and step 1.1 yields p=(I:r).

L1step 1.1
2.2

Conversely, if p=(I:r) for some r∉I and if p is prime, then step 1.1 gives p=Ann⁡R(r+I) with r+I≠0, so p∈Ass⁡R(R/I) by [L1].

L1step 1.1
3.1

Steps 2.1 and 2.2 prove the stated description of Ass⁡R(R/I).

step 2.1step 2.2∎

Depends on

Used by

Dependency tree · two levels

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Sources