CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.
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The componentwise formula can fail to define a tensor action

Statement refuted

For every unital associative $k$ -algebra $A$ and every pair of left $A$ -modules $M, N$ , the formula $a \cdot (m \otimes n) = (a m) \otimes (a n)$ defines a left $A$ -module action on $M \otimes_{k} N$ .

Facts & Assumptions

Given: The algebra $A = k = C$ , the left modules $M = N = C$ with their usual scalar actions, and the componentwise formula in the Statement.

[F1]

Representations of unital associative algebras: a representation of a unital associative $k$ -algebra is a $k$ -linear map to ${End}_{k} (V)$ preserving multiplication and the unit.

[F2]

Associative algebra representations are unital modules: the representation $ρ (a)$ acts by $a \cdot v = ρ (a) (v)$ and gives a unital left module.

[F3]

The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ : multiplication $λ_{N} : k \otimes_{k} N \to N$ , $c \otimes n \mapsto c n$ , is an isomorphism with inverse $n \mapsto 1 \otimes n$ .

[F4]

The module category of a bialgebra and its tensor action: the bialgebra tensor action uses extra coproduct data, $h \cdot (v \otimes w) = \sum h_{(1)} v \otimes h_{(2)} w$ .

Counterexample

1.1givenF1F2construct

The identity map $ρ : C \to {End}_{C} (C)$ , $a \mapsto a id$ , is a unital representation. Thus $M$ and $N$ are unital left $A$ -modules by [F1]–[F2].

2.1givenF3step 1.1algebra

Put $u = 1 \otimes 1 \in C \otimes_{C} C$ . By [F3], multiplication sends $u$ to $1$ , so $u \neq 0$ . The proposed formula sends $u$ to $a^{2} u$ for $a \in C$ .

3.1givenF4step 2.1algebra∎

If this formula were an $A$ -module action, it would be additive in $a$ . But it gives $2 \cdot u = 4 u$ from the formula for $a = 2$ , whereas additivity and $2 = 1 + 1$ require $2 \cdot u = 1 \cdot u + 1 \cdot u = 2 u$ . Since $u \neq 0$ and $4 u \neq 2 u$ over $C$ , the formula fails the module axiom. The coproduct-based formula in [F4] is a distinct construction requiring additional bialgebra data.

The example refutes this componentwise formula; it does not rule out other actions on a tensor product. A coproduct, when supplied as part of a bialgebra, gives the distinct action in [F4].

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Sources

Pavel Etingof, Tensor Categories, MIT 18.769 complete course notes, §§1.21.2–1.21.3, printed pp. 44–45 (standard reference, not scraped)
Kalle Kytölä, Introduction to Hopf Algebras and Representations, §3.5, printed pp. 41–42 (standard reference, not scraped)