TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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For a finite module, support is the set of primes containing the annihilator

Statement

If $M$ is a finitely generated left $R$ -module, then ${Supp}_{R} (M) = {p : {Ann}_{R} (M) \subseteq p} .$

Facts & Assumptions

Given: A commutative ring $R$ and a finitely generated left $R$ -module $M$ .

[L1]

A prime ideal lies in ${Supp}_{R} (M)$ exactly when some element of $M$ has annihilator inside it (A prime lies in the support exactly when some element has annihilator inside it).

[L2]

If $m_{1}, \dots, m_{r}$ generate $M$ , then ${Supp}_{R} (M) = ⋃_{i} {Supp}_{R} (R / {Ann}_{R} (m_{i}))$ (A finite module has the union of its generator-cyclic supports).

[L3]

The annihilator of $M$ is ${Ann}_{R} (M) = {r \in R : r m = 0 for every m \in M}$ (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct

1.1L1L3

If $p \in {Supp}_{R} (M)$ , [L1] gives $m \in M$ with ${Ann}_{R} (m) \subseteq p$ . Since every element of ${Ann}_{R} (M)$ kills every element of $M$ , one has ${Ann}_{R} (M) \subseteq {Ann}_{R} (m) \subseteq p$ .

1.2L2L3choose

Choose generators $m_{1}, \dots, m_{r}$ of $M$ . If ${Ann}_{R} (M) \subseteq p$ and no ${Ann}_{R} (m_{i})$ is contained in $p$ , choose $t_{i} \in {Ann}_{R} (m_{i}) ∖ p$ for every $i$ . Then $t = t_{1} \dots t_{r} \notin p$ , but $t$ annihilates every generator and hence all of $M$ , so $t \in {Ann}_{R} (M) \subseteq p$ , a contradiction. Thus ${Ann}_{R} (m_{i}) \subseteq p$ for some $i$ , and [L2] gives $p \in {Supp}_{R} (M)$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the support-annihilator formula.

Depends on

Used by

Dependency tree · two levels

11 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (13.27) (standard reference, not scraped)
The Stacks Project, Lemma 10.40.5 (standard reference, not scraped)