CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30

How statement and proof provenance work

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.

Stalk and fibre are different

Statement refuted

False claim: for every scheme $X$ , every $O_{X}$ -module $F$ and every point $x \in X$ , the canonical surjection from the stalk to the fibre, $F_{x} ⟶ F (x) = F_{x} / m_{x} F_{x},$ is an isomorphism, so that the stalk of $F$ at $x$ is already a vector space over the residue field $κ (x)$ (Fibre of a module sheaf at a point).

The structure sheaf of the affine line at its origin refutes this. Let $k$ be a field, $A = k [t]$ , $X = Spec A$ and $p = (t) \in X$ , the origin of the affine line (The underlying space of an affine spectrum). For $F = O_{X}$ the stalk at $p$ is the local ring $F_{p} = O_{X, p} ≅ A_{p} = k [t]_{(t)},$ with nonzero maximal ideal $p A_{p} = (t) A_{p}$ , while the fibre is $F (p) = F_{p} / p F_{p} ≅ k [t]_{(t)} / (t) k [t]_{(t)} ≅ k,$ a one-dimensional vector space over $κ (p) = k$ . The reduction $k [t]_{(t)} \to k$ is surjective and its kernel $(t) k [t]_{(t)}$ contains the nonzero element $t$ ; it is therefore not injective, and it is not an isomorphism of $O_{X, p}$ -modules or of rings. So the stalk and the fibre of a module sheaf must be distinguished.

Facts & Assumptions

Given: The Axiom of Choice, a field $k$ , the polynomial ring $A = k [t]$ , the scheme $X = Spec A$ , its point $p = (t)$ , the structure sheaf $O_{X}$ , and the $O_{X}$ -modules $F = O_{X}$ with stalk $F_{p}$ and fibre $F (p)$ .

[L1]

$F (p) = F_{p} \otimes_{O_{X, p}} κ (p) ≅ F_{p} / p F_{p}$ and the fibre is a vector space over $κ (p) = O_{X, p} / p O_{X, p}$ (Fibre of a module sheaf at a point, The residue field at a point of an affine scheme).

[L2]

For $p \in Spec A$ the stalk of the structure sheaf is $O_{Spec A, p} ≅ A_{p}$ (The stalk of the affine structure sheaf at a prime is A_p); for any $A$ -module $M$ the stalk of the associated sheaf is $(\tilde{M})_{p} ≅ M_{p}$ (The stalk of an associated sheaf is the localisation).

[L3]

In $X = Spec A$ the point $p = (t)$ is the prime generated by $t \in k [t]$ , and the local ring $A_{p}$ has maximal ideal $p A_{p}$ (The residue field at a point of an affine scheme); $p$ is prime since $k [t] / (t) = k$ is a field.

[L5]

Localisation commutes with quotients: $S^{- 1} (A / I) ≅ S^{- 1} A / S^{- 1} I$ , so $A_{p} / I A_{p} ≅ (A / I)_{p}$ (Localisation commutes with quotient modules and arbitrary direct sums).

[L6]

The refuted claim: the canonical map $F_{x} \to F (x)$ is an isomorphism for every $O_{X}$ -module $F$ and every point $x$ .

Proof technique: direct; compute stalk, maximal ideal and fibre for $O_{Spec k [t]}$ at $p = (t)$ , and show the reduction map has nonzero kernel.

Proof

1.1L2L3algebra

With $A = k [t]$ and $p = (t)$ , the stalk of $F = O_{X}$ at $p$ is $F_{p} ≅ A_{p} = k [t]_{(t)}$ by [L2], and by [L3] its maximal ideal is $p A_{p} = (t) A_{p}$ , which contains the nonzero element $t$ : if $t / 1 = 0$ , some $s \notin (t)$ would satisfy $s t = 0$ in the domain $k [t]$ , a contradiction.

1.2L1L5

The fibre is $F (p) ≅ F_{p} / p F_{p} ≅ A_{p} / (t) A_{p}$ by [L1], and by [L5] applied to the principal ideal $(t) \subseteq A$ this is $≅ (A / (t))_{p} = k_{p} = k$ , the residue field $κ (p) = A_{p} / p A_{p} ≅ k$ being one-dimensional over itself; so the fibre is a one-dimensional $k$ -vector space.

2.1L1step 1.1step 1.2

The canonical map of [L1] is the quotient map $A_{p} \to A_{p} / (t) A_{p} ≅ k$ , which is surjective, and it is not injective: the element $t \in A_{p}$ is nonzero and is mapped to zero; equivalently its kernel is the nonzero ideal $(t) A_{p}$ .

2.2step 1.1step 1.2

The stalk is not isomorphic to the fibre even as a ring: if $t$ were a unit of $A_{p}$ , there would be $a \in A$ and $s \in A ∖ (t)$ with $a t / s = 1$ , hence $u (a t - s) = 0$ in $A$ for some $u \in A ∖ (t)$ , which is impossible in the polynomial ring $k [t]$ because the right-hand side of $a t = s$ is not divisible by $t$ ; so $t$ is a nonunit, $A_{p}$ is a local ring that is not a field, while the fibre is the field $k$ , and the two are not isomorphic.

3.1L1L2step 2.1step 2.2∎

By steps 2.1 and 2.2 the canonical surjection $F_{p} \to F (p)$ of [L1] is surjective but not injective, hence not an isomorphism, and the stalk $k [t]_{(t)}$ and the fibre $k$ are different objects; the Axiom of Choice is inherited through [L2] only, no new choice being made in this computation.

Depends on

Used by

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Sources

The Stacks Project, Schemes, §§26.5, 26.7, 26.24 (standard reference, not scraped)
Ravi Vakil, The Rising Sea, 29 August 2022, Chapters 6, 14, 17 (standard reference, not scraped)