ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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.

The dual-numbers tensor functor is right exact but not left exact

Example

Let $k$ be a field, let $A = k [ε] / (ε^{2})$ be the algebra of dual numbers (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ) and let $S = A / (ε)$ be its simple module (Simple module: a nonzero module with no proper nonzero submodule). Then the tensor functor $T_{S} = S \otimes_{A} - : A - m o d \to A - m o d$ is right exact but not left exact. Explicitly, applied to the non-split short exact sequence $0 \to (ε) \to A \to S \to 0$ of finite-dimensional $A$ -modules it yields, under the identifications $S \otimes_{A} (ε) ≅ S$ , $S \otimes_{A} A ≅ S$ and $S \otimes_{A} S ≅ S$ , the sequence $S \overset{0}{\to} S \overset{≅}{\to} S \to 0$ ; the comparison map $S \otimes_{A} (ε) \to S \otimes_{A} A$ is zero and therefore is not injective, so $T_{S}$ is not left exact. Consistently, $T_{S}$ has a right adjoint ${Hom}_{A} (S, -)$ but no left adjoint, and its kernel $S$ is not a projective right $A$ -module. No choice is used.

Facts & Assumptions

Given: A field $k$ , the algebra $A = k [ε] / (ε^{2})$ of dual numbers, and its simple module $S = A / (ε)$ .

[L1]

The algebra $A = k [ε] / (ε^{2})$ is a commutative unital $k$ -algebra in which the class $ε$ of the indeterminate satisfies $ε^{2} = 0$ and every element has the form $a + b ε$ with $a, b \in k$ ; hence $(ε) = k ε$ and $S = A / (ε)$ is a one-dimensional $k$ -vector space with $ε S = 0$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

[L2]

For a finite-dimensional $(B, A)$ -bimodule $M$ with agreeing $k$ -scalar actions, the functor $T_{M} = M \otimes_{A} -$ is a $k$ -linear right exact functor $A - m o d \to B - m o d$ , and it is left adjoint to ${Hom}_{B} (M, -)$ ; in particular $S$ , a module over the commutative algebra $A$ , is an $(A, A)$ -bimodule and $T_{S}$ is right exact with right adjoint ${Hom}_{A} (S, -)$ (Finite Eilenberg–Watts for right exact linear functors, Tensor-Hom adjunction for bimodules over arbitrary unital rings).

[L3]

The unit isomorphism $S \otimes_{A} A ≅ S$ sends $s \otimes a$ to $s a$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ), and a right exact functor carries an exact sequence $X \to Y \to Z \to 0$ to an exact sequence; the kernel $T_{S} (A)$ equals $S \otimes_{A} A$ (Exact sequences and short exact sequences of modules, Module homomorphism and isomorphism, kernel, image and cokernel, Finite left exact functors are Hom functors with dual bimodule kernels).

[L4]

If a short exact sequence $0 \to X^{'} \to X \to X^{''} \to 0$ splits, then $X ≅ X^{'} \oplus X^{''}$ with the given maps (The splitting lemma for short exact sequences of modules). If $T_{S}$ were left exact it would have a left adjoint; if $S$ were a projective right $A$ -module then $T_{S}$ would be exact (Finite one-sided exactness is equivalent to the existence of the corresponding adjoint, Exact finite tensor functors have projective right-module kernels, Projective modules and the lifting property, Left and right flat modules over an arbitrary ring).

Verification

technique · direct

1.1L1L2algebra

By [L1] every element of $A$ is $a + b ε$ with $ε^{2} = 0$ , so $(ε) = k ε$ , the quotient $S = A / (ε)$ is one-dimensional over $k$ with $ε S = 0$ , and $A$ acts on $S$ through $k$ ; the $A$ -submodules of $S$ are therefore exactly its $k$ -subspaces, so $S \neq 0$ is simple. The quotient map $π : A \to S$ has kernel $(ε)$ , and the map $φ : A \to (ε)$ , $φ (a) = a ε$ , has image $(ε)$ and kernel $(ε)$ , since $(a + b ε) ε = a ε$ vanishes only for $a = 0$ ; hence $φ$ induces an $A$ -module isomorphism $S = A / (ε) ≅ (ε)$ and $0 \to (ε) \to A \overset{π}{\to} S \to 0$ is a short exact sequence.

2.1L1L4step 1.1algebra

The sequence $0 \to (ε) \to A \to S \to 0$ does not split. If it split, then by [L4] there would be an $A$ -module isomorphism $A ≅ (ε) \oplus S$ , and $(ε) ≅ S$ by step 1.1, so $A ≅ S \oplus S$ ; by step 1.1 the element $ε$ acts as zero on each copy of $S$ , hence as zero on $S \oplus S$ and therefore, through the isomorphism, as zero on $A$ . But $ε \cdot 1_{A} = ε \neq 0$ in $A$ . Contradiction, so the sequence does not split.

2.2L2L3step 1.1

The functor $T_{S} = S \otimes_{A} -$ is right exact by [L2], so applying it to $0 \to (ε) \to A \to S \to 0$ gives the exact sequence $S \otimes_{A} (ε) \to S \otimes_{A} A \to S \otimes_{A} S \to 0$ by [L3].

2.3L2L3step 1.1

The three outer identifications of the statement hold: $S \otimes_{A} A ≅ S$ by the unit isomorphism of [L3]; $S \otimes_{A} (ε) ≅ S \otimes_{A} S$ because $(ε) ≅ S$ by step 1.1; and $S \otimes_{A} S ≅ S$ , because applying the right exact functor $- \otimes_{A} S$ to $A \to S \to 0$ identifies $S \otimes_{A} S$ with the cokernel of $(ε) \otimes_{A} S \to A \otimes_{A} S ≅ S$ , whose image is $ε S = 0$ .

3.1L2L3step 2.3algebra

Under these identifications the first map $S \otimes_{A} (ε) \to S \otimes_{A} A$ is zero: it is induced by the inclusion $(ε) ↣ A$ , and the generator $1_{S} \otimes ε$ maps to $1_{S} \otimes ε$ , which corresponds under $S \otimes_{A} A ≅ S$ to $1_{S} ε = 0$ because $ε$ annihilates $S$ ; since $S \otimes_{A} (ε) ≅ S$ is generated as an $A$ -module by $1_{S} \otimes ε$ , the map is zero. The second map $S \otimes_{A} A \to S \otimes_{A} S$ is induced by the quotient $A ↠ S$ and corresponds under the identifications to the identity of $S$ , hence is an isomorphism. So the image sequence is $S \overset{0}{\to} S \overset{≅}{\to} S \to 0$ .

4.1L1step 1.1step 2.3step 3.1

The sequence $S \overset{0}{\to} S \overset{≅}{\to} S \to 0$ is exact, as the second map is an isomorphism and its kernel is zero, so right exactness of $T_{S}$ is exhibited directly. But $T_{S}$ is not left exact: the extended sequence $0 \to S \otimes_{A} (ε) \to S \otimes_{A} A \to S \otimes_{A} S \to 0$ fails to be exact at $S \otimes_{A} (ε)$ , because the map $S \otimes_{A} (ε) \to S \otimes_{A} A$ is zero while $S \otimes_{A} (ε) ≅ S \neq 0$ by steps 1.1 and 2.3.

5.1L3L4step 4.1

Consistently, $T_{S}$ has the right adjoint ${Hom}_{A} (S, -)$ by [L2] but no left adjoint, since a functor with a left adjoint is left exact by [L4] and $T_{S}$ is not left exact by step 4.1; and its kernel $T_{S} (A) = S \otimes_{A} A ≅ S$ is not a projective right $A$ -module, since by [L4] a projective kernel would make $T_{S}$ exact, while $T_{S}$ is not left exact by step 4.1.

6.1step 1.1step 2.2step 2.3step 3.1step 4.1step 5.1∎

All modules and sequences above are finite-dimensional and the computations use only the finitely many structure maps of $A$ and $S$ , so no choice is used.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

98 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Etingof, Gelaki, Nikshych, Ostrik, Tensor Categories, §1.8 (Definitions 1.8.1–1.8.6), printed pp.9–11 (standard reference, not scraped)
Fuchs, Schaumann, Schweigert, Eilenberg–Watts calculus for finite categories and a bimodule Radford S^4 theorem, arXiv:1612.04561v3, §2.1 (Lemma 2.1) (standard reference, not scraped)