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The radicals in a minimal primary decomposition are intrinsic
Statement
Let be a Noetherian commutative ring, let be a finitely generated left -module, and let . Let be a minimal primary decomposition in which each is -primary for a prime ideal . Then the set of component radicals is uniquely determined by and equals
Facts & Assumptions
Given: A Noetherian commutative ring , a finitely generated left -module , a submodule , and a minimal primary decomposition with each -primary for a prime ideal .
In the Noetherian finite-module setting, if a minimal primary decomposition has each component -primary for a prime ideal , then its radicals are exactly the associated primes of the quotient (The radicals in a minimal primary decomposition are exactly the associated primes of the quotient).
Proof
For the given minimal primary decomposition, [L1] gives
The right-hand side depends only on the quotient , not on the chosen decomposition. Hence the set of component radicals is intrinsic.
Depends on
Used by
Dependency tree · two levels
5 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., Theorem (18.20) (standard reference, not scraped)
- J. S. Milne, A Primer of Commutative Algebra, v4.03, Theorem 19.10 (standard reference, not scraped)