CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Normalization of the node is two-to-one over the node

Statement refuted

False claim: the normalization morphism of a classical variety is injective.

Facts & Assumptions

Assume the Axiom of Choice.

Given: AC, an algebraically closed field $k$ of characteristic not two, the nodal curve $X = V (y^{2} - x^{2} (x + 1)) \subseteq A^{2}$ , and its normalization $ν ⁣ : A^{1} \to X$ .

[F1]

The normalization of the node is $ν ⁣ : A^{1} \to X$ , $t \mapsto (t^{2} - 1, t (t^{2} - 1))$ , a finite birational morphism, and the fibre of $ν$ over the node $(0, 0)$ consists exactly of the two distinct points $t = 1$ and $t = - 1$ (Normalizing the nodal plane curve, The normalization of an irreducible affine variety, Normalization is finite, surjective and birational).

[F2]

The fibre $ν^{- 1} (x)$ is the set-theoretic preimage of the point $x$ under $ν$ (Images and fibres of a regular map).

[F7]

AC is inherited through the classical localization, normalization, or finite-morphism suppliers cited above (The Axiom of Choice).

Counterexample

1.1F1F2givenF7

By [F1] the fibre of the normalization over the node has the two distinct elements $1$ and $- 1$ , so by [F2] the set-theoretic preimage $ν^{- 1} ((0, 0)) = {1, - 1}$ has two elements.

2.1F1step 1.1∎

The map $ν$ therefore sends two distinct points of $A^{1}$ to the same point of $X$ , so it is not injective; the claim is refuted. In dimension one a normalization need not be injective; in contrast, when the target is normal the normalization is an isomorphism and hence injective.

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Algebraic Geometry (2025 version), Ch. 8 Example 8.6(b): the node has two branches (standard reference, not scraped)