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The support of any module is closed under specialisation

Statement

Assume the Axiom of Choice.

Let R be a commutative ring, let M be an R-module, and let p,q∈Spec⁡(R) with p⊆q. If p∈Supp⁡R(M), then q∈Supp⁡R(M). Equivalently, the support of any module is closed under specialisation.

Facts & Assumptions

Given: A commutative ring R, an R-module M, prime ideals p⊆q, and the Axiom of Choice.

[L1]

A prime ideal lies in the support exactly when some module element has annihilator contained in that prime (A prime lies in the support exactly when some element has annihilator inside it).

[L2]

In a prime spectrum, specialisation is reverse inclusion (Specialisation in a prime spectrum is reverse inclusion).

Proof

technique · direct
1.1

Since p∈Supp⁡R(M), fact [L1] gives an element m∈M with Ann⁡R(m)⊆p. Because p⊆q, the same annihilator satisfies Ann⁡R(m)⊆q.

L1givenchoose
2.1

Applying [L1] again, step 1.1 implies q∈Supp⁡R(M). The equivalent specialisation language follows from [L2].

L1L2step 1.1
3.1

Therefore the support of any module is closed under specialisation.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

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