ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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A left-projective tensor bimodule need not be right-flat

Example

Let $k$ be a field, let $B = k$ with its trivial grading and let $A = k [ε] / (ε^{2})$ with $ε$ in degree $0$ . Let $π : A ↠ k$ be the augmentation with $π (ε) = 0$ , and let $M = k$ be the graded $(B, A)$ -bimodule concentrated in degree $0$ whose left $B$ -action is ordinary multiplication and whose right $A$ -action is $m \cdot a : = π (a) m$ .

Then $M$ is finite projective as a left $B$ -module, but $M \otimes_{A} -$ is not exact: it destroys the monomorphism $(ε) ↪ A$ in the exact sequence $0 \to (ε) \to A \to k \to 0$ of graded left $A$ -modules, because the induced map $k \otimes_{A} (ε) \to k \otimes_{A} A$ is the zero map while its source is a copy of $k$ .

Facts & Assumptions

Given: A field $k$ , the graded $k$ -algebras $B = k$ and $A = k [ε] / (ε^{2})$ concentrated in degree $0$ , the augmentation $π : A \to k$ , and the graded $(B, A)$ -bimodule $M = k$ with $b \cdot m = b m$ and $m \cdot a = π (a) m$ .

[L1]

Graded algebras, graded modules, degree-zero maps and graded submodules are defined in Associative graded algebras, bimodules, and internal shifts; all modules here are concentrated in degree $0$ , so all module maps are degree-zero.

[L2]

The tensor product carries the total-degree grading and the outer actions, in particular $(m \otimes a) c = m \otimes (a c)$ and $b (m \otimes a) = (b m) \otimes a$ (Graded balanced tensor product and homogeneous Hom).

[L3]

${GrMod}_{0} (A)$ is abelian with degreewise exactness, so a sequence concentrated in degree $0$ is exact exactly when the underlying sequence of $A$ -modules is (Graded modules with degree-zero maps form an abelian category).

[L4]

If $M$ is flat as a right $A$ -module then $M \otimes_{A} -$ is exact, and if $M$ is finite graded projective as a left $B$ -module then $M \otimes_{A} -$ preserves finite graded projectives (Bimodule tensor exactness and preservation of finite projectives have separate hypotheses).

Verification

1.1

The two actions on $M$ commute, since $(b \cdot m) \cdot a = π (a) b m = b \cdot (m \cdot a)$ , and are additive and unital, so $M$ is a $(B, A)$ -bimodule; both preserve degree $0$ , so $M$ is graded.

1.2

$M = k = B$ is the free left $B$ -module of rank one, hence a finite direct sum of shifts $B {0}$ and therefore a finite graded projective left $B$ -module.

L1L4

1.3

In $A = k [ε] / (ε^{2})$ the ideal $(ε) = k ε$ has $ε^{2} = 0$ , so $a ε = π (a) ε$ for every $a \in A$ ; hence $(ε)$ is isomorphic to $k = A / (ε)$ as a left $A$ -module by $1 \mapsto ε$ , and the sequence $0 \to (ε) \to A \overset{π}{\to} k \to 0$ is exact with all maps degree-zero and $A$ -linear.

L1L3

2.1

Tensoring the sequence of step 1.3 with $M$ : the unit isomorphisms identify $M \otimes_{A} A ≅ M = k$ and $M \otimes_{A} k ≅ k$ , and by step 1.3 also $M \otimes_{A} (ε) ≅ M \otimes_{A} k ≅ k$ . The induced map $M \otimes_{A} (ε) \to M \otimes_{A} A$ sends $1 \otimes ε$ to $1 \otimes ε = (1 \cdot ε) \otimes 1 = 0$ , by the balancing relation and the right action $1 \cdot ε = π (ε) = 0$ on $M$ . So the induced map is zero while its source is $k \neq 0$ , and it is not injective.

step 1.2step 1.3L2

3.1

By step 2.1 the functor $M \otimes_{A} -$ fails to preserve the monomorphism $(ε) \to A$ , so it is not exact and $M$ is not flat as a right $A$ -module; by step 1.2 $M$ is nevertheless finite graded projective over $B$ . Hence finite left $B$ -projectivity of a bimodule does not imply right $A$ -flatness, the hypothesis that the exactness clause of [L4] requires.

step 1.2step 2.1L4

4.1

The example therefore exhibits a bimodule that is finite projective on the tensoring-out side but whose tensor functor is not exact. ∎

step 3.1

Depends on

Used by

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Sources

Stacks Project, Algebra, §10.12, tag 00CV (standard reference, not scraped)
Charles A. Weibel, An Introduction to Homological Algebra, ch. 3, §3.2, printed pp. 68-69 (standard reference, not scraped)