CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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Every surjective endomorphism of a Noetherian module is injective

Statement

Every surjective endomorphism of a Noetherian module is injective, hence an automorphism. See Finite generation, ACC, and maximal-condition characterizations of Noetherian modules.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left $R$ -module $M$ , the following are equivalent: every submodule is finitely generated; every ascending chain of submodules stabilizes; and every nonempty family of submodules has a maximal member. The implication from ACC to the maximal condition uses dependent choice; the other displayed implications are choice-free. (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules).

[L2]

For a left $R$ -module $M$ , define ${End}_{R} (M) : = {Hom}_{R} (M, M) .$ Addition is pointwise and multiplication is composition, $(f g) (m) : = f (g (m))$ . The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring ${End}_{R} (M)$ under addition and composition).

Proof

technique · direct

1.1

For a surjective endomorphism $f$ , the ascending chain $\ker f \subseteq \ker f^{2} \subseteq \dots$ stabilizes.

L1L2givenalgebra

2.1

Choose $n$ with $\ker f^{n} = \ker f^{n + 1}$ . For $x \in \ker f$ , surjectivity of $f^{n}$ gives $y$ with $f^{n} (y) = x$ . Then $f^{n + 1} (y) = 0$ , so $y \in \ker f^{n + 1} = \ker f^{n}$ and $x = f^{n} (y) = 0$ .

step 1.1givenalgebra

3.1

The zero module is admitted: its only endomorphism is the identity, which is injective, so the conclusion holds there and the argument of step 2.1 is not vacuous by accident. This proves the stated claim.

step 2.1givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 15 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Keith Conrad, Noetherian Modules, Sections 1-2 (standard reference, not scraped)