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A commutative ring is Artinian exactly when it has finite length as a module over itself
Statement
Assume the Axiom of Choice.
Let be a commutative ring. Then is Artinian if and only if the regular module has finite length.
Facts & Assumptions
Given: A commutative ring and the Axiom of Choice.
Proof
If has finite length, then by Composition series and length of a module it has a composition series. The forward implication of A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice therefore makes Artinian, and Left and right Artinian rings says exactly that is an Artinian ring.
Suppose now that is Artinian. Then Left and right Artinian rings says that the regular module is Artinian. Under the Axiom of Choice assumed in the Statement, Every commutative Artinian ring is Noetherian makes Noetherian, so Left and right Noetherian rings makes Noetherian. Since the same choice assumption also suffices for the dependent-choice use recorded in A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice, that theorem gives a composition series for .
By Composition series and length of a module, a module has finite length exactly when it has a composition series. So step 1.1 proves the forward implication, and step 1.2 proves the reverse implication.
Therefore a commutative ring is Artinian exactly when its regular module has finite length.
Depends on
Used by
Dependency tree · two levels
22 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., Corollary (19.15) (standard reference, not scraped)
- The Stacks Project, Section 10.53: Artinian rings (standard reference, not scraped)