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Serre free-summand theorem for modules
PDF p. 1: Serre splitting theorem
Statement
Let be a commutative unital ring whose maximal spectrum , with its subspace topology, is noetherian of finite dimension . Let be a finitely presented -module. Suppose that for every maximal ideal , the localization has a free direct summand of rank greater than . Then for some -module . The dimension of a topological space is the supremum of the lengths of strict chains of irreducible closed subsets.
External input
This is Theorem 15.129.5 of the cited Stacks Project section. It is imported as an established result; its underlying dimension-avoidance argument is not reconstructed in this item.
Facts & Assumptions
Given: A ring , maximal spectrum , integer , and module satisfying the hypotheses in the statement.
Goal: Obtain a free rank-one direct summand of .
The cited external theorem asserts exactly this conclusion under the displayed hypotheses.
Proof
The given noetherianity, finite dimension, finite presentation, and local free-summand bounds are precisely the hypotheses of the imported theorem.
Applying that theorem yields an -module and an isomorphism . This is an application of the external input, not a local reconstruction of its proof.
Depends on
Used by · 1
Dependency tree · 1 level
Arrows run from prerequisites to the items that use them. External source nodes link directly to the cited text.
External dependencies
If the maximal spectrum of a ring is noetherian of finite dimension e and a finitely presented module has, at every maximal ideal, a free localized direct summand of rank greater than e, then the module is isomorphic to the ring plus a complementary module.
Hypotheses: Noetherian finite-dimensional maximal spectrum; finite presentation; the stated localized free-summand rank bound at every maximal ideal.
Reading limits: The theorem statement and its short displayed proof were read. The proof invokes Proposition 15.129.4 and other lemmas; their proofs were not reconstructed here. The original Serre paper was not read.
Sources
- The Stacks Project, More on Algebra, Section 15.129: Splitting off a free module · Theorem 15.129.5 (tag 0GVC), statement and displayed proof