ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (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.

Flat dual-number lifts of the affine line and their automorphisms

Example

Let $k$ be any field, let $R = k [ϵ] / (ϵ^{2})$ be the dual-number ring, and let $k [x]$ be the polynomial algebra in one indeterminate. A flat marked lift of $k [x]$ is a commutative unital $R$ -algebra $D$ which is flat as an $R$ -module, together with a specified $k$ -algebra isomorphism $D / ϵ D ≅ k [x]$ . A marked isomorphism is an $R$ -algebra isomorphism inducing the identity of this specified quotient.

There is exactly one marked isomorphism class of such lifts, represented by $R [x]$ with its coefficient-reduction marking. Its marked automorphisms are exactly $ϕ_{p} (x) = x + ϵ p (x), p (x) \in k [x],$ with $ϕ_{p} ϕ_{q} = ϕ_{p + q}$ , $ϕ_{p}^{- 1} = ϕ_{- p}$ , and identity $ϕ_{0}$ . In particular uniqueness of the marked isomorphism class does not imply uniqueness of marked isomorphisms or triviality of automorphisms.

Example references

Flat and faithfully flat modules and ring homomorphisms

Facts & Assumptions

Given: a field $k$ , $R = k \oplus k ϵ$ with $ϵ^{2} = 0$ , and the polynomial algebra $k [x]$ .

[F1]

The explicitly defined ring $R = k \oplus k ϵ$ has multiplication $(a + b ϵ) (c + d ϵ) = a c + (a d + b c) ϵ$ and coefficient-reduction kernel $(ϵ)$ , which squares to zero.

[F2]

Flat and faithfully flat modules and ring homomorphisms defines flatness by tensor exactness. A free module is flat because tensor with a direct sum of copies of $R$ is a direct sum of the original exact sequence.

Verification

Given: [F1], [F2] and the marked-lift convention of the Example.

1.1F1F2givenalgebra

The algebra $R [x]$ is free over $R$ on $1, x, x^{2}, \dots$ , so it is flat, and its quotient modulo $ϵ$ is exactly $k [x]$ with the stated marking. Its kernel $ϵ k [x]$ has square zero. If $D$ is any flat marked lift, tensor $0 \to (ϵ) \to R \to k \to 0$ with $D$ . Exactness gives an injective multiplication map $(ϵ) \otimes_{R} D \to D$ with image $ϵ D$ . Since $(ϵ) ≅ k$ as $R$ -modules, the induced map $D / ϵ D \to ϵ D$ , $\overset{ˉ}{d} \mapsto ϵ d$ , is an isomorphism. This is the exact kernel computation needed below.

2.1F1F2step 1.1choosealgebra

Choose $y \in D$ mapping to the marked variable $x$ . The polynomial map $u : R [t] \to D$ , $t \mapsto y$ , is a marked isomorphism. Indeed any $d \in D$ differs from a polynomial in $y$ with coefficients in $k$ by $ϵ e$ ; replacing $e$ by a polynomial with the same reduction changes $ϵ e$ by zero. Thus $u$ is surjective. If $u (f) = 0$ , reduction gives $\overset{ˉ}{f} (x) = 0$ in $k [x]$ , so $f = ϵ g$ for some $g \in R [t]$ . Step 1.1 then implies $\overline{g (y)} = 0$ in $D / ϵ D$ , hence $\overset{ˉ}{g} (x) = 0$ and $g = ϵ h$ . Therefore $f = ϵ^{2} h = 0$ , proving injectivity. Renaming $t$ as $x$ gives the representative $R [x]$ . Neither existence nor uniqueness here invokes an obstruction theorem.

2.2F1step 1.1algebraconstruct

A marked endomorphism of $R [x]$ is determined by its value on $x$ , and that value must reduce to $x$ . The kernel of $R [x] \to k [x]$ is precisely $ϵ k [x]$ , so the value is uniquely $x + ϵ p (x)$ for $p \in k [x]$ . Conversely each such value determines a unital $R$ -algebra homomorphism by polynomial substitution. For any polynomial $q \in k [x]$ , binomial expansion gives $ϵ q (x + ϵ p (x)) = ϵ q (x)$ ; every other term contains $ϵ^{2}$ . Consequently composing $ϕ_{p}$ and $ϕ_{q}$ sends $x$ to $x + ϵ (p (x) + q (x))$ . The inverse is $ϕ_{- p}$ . This proves the complete classification and group law in every characteristic.

3.1step 2.1step 2.2algebra∎

Taking $p = 1$ gives a nonidentity automorphism since $ϵ \neq 0$ in $R$ . Any two marked isomorphisms from a chosen lift to $R [x]$ differ by exactly one of the automorphisms in step 2.2, by composing one with the inverse of the other. Thus the set of marked isomorphism classes is a singleton while the automorphism group is the nonzero additive group of $k [x]$ . This verifies the distinction in the Example explicitly.

Flat dual-number lifts of the affine line and their automorphisms

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources