CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6.1-sol)

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A Deligne kernel need not be one external tensor factor

Statement refuted

Under the identification $A^{o p} ⊠ B ≃ (B, A) - b i m o d$ of The opposite Deligne product is the category of finite bimodules, not every object is one external tensor factor $\overset{ˉ}{a} ⊠ b$ . Witness: for the upper triangular $k$ -algebra $A_{0}$ with basis $e_{1}, e_{2}, u$ , unit $1 = e_{1} + e_{2}$ , $e_{1}^{2} = e_{1}$ , $e_{2}^{2} = e_{2}$ , $e_{1} u = u = u e_{2}$ and all other basis products zero (Algebras over a commutative ring, central structure maps, and algebra homomorphisms, Vector space over a field, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis), the regular bimodule $A_{0}$ has dimension $3$ , and it is not isomorphic to $b \otimes_{k} a^{*}$ for any finite-dimensional left $A_{0}$ -modules $a, b$ : if it were, one of $\dim_{k} b, \dim_{k} a$ would be $1$ , and a one-dimensional left or right $A_{0}$ -module has $u$ acting as zero (Simple module: a nonzero module with no proper nonzero submodule), forcing the left (if $\dim_{k} b = 1$ ) or right (if $\dim_{k} a = 1$ ) multiplication by $u$ on $b \otimes_{k} a^{*}$ to vanish; on the regular bimodule left multiplication by $u$ sends $e_{2}$ to $u \neq 0$ and right multiplication by $u$ sends $e_{1}$ to $u \neq 0$ ( $(S, R)$ -bimodules and commuting left and right scalar actions, Unital left and right modules over a ring; unqualified module means left module, Linear map between vector spaces over the same field), a contradiction in either case.

Facts & Assumptions

Given: A field $k$ and the $k$ -algebra $A_{0}$ with $k$ -basis $e_{1}, e_{2}, u$ , unit $1 = e_{1} + e_{2}$ , $e_{1}^{2} = e_{1}$ , $e_{2}^{2} = e_{2}$ , $e_{1} u = u = u e_{2}$ and all remaining products of basis elements zero; the regular bimodule $_{A_{0}} (A_{0})_{A_{0}}$ ; and finite-dimensional left $A_{0}$ -modules $a, b$ .

[F1]

A left $A_{0}$ -module is an abelian group with a scalar action satisfying $r (m + n) = r m + r n$ , $(r + s) m = r m + s m$ , $(r s) m = r (s m)$ and $1 m = m$ , and dually on the right (Unital left and right modules over a ring; unqualified module means left module); on a one-dimensional module the action is a $k$ -linear map into scalars, so all products and sums of actions are computed by the corresponding relations in $A_{0}$ (Linear map between vector spaces over the same field, Vector space over a field); a one-dimensional module has no nonzero proper submodule, hence is simple (Simple module: a nonzero module with no proper nonzero submodule).

[F2]

For finite-dimensional $k$ -vector spaces the dimension is the cardinality of a basis, and the products $x_{i} \otimes λ_{j}$ of bases $(x_{i})$ of $b$ and $(λ_{j})$ of $a^{*}$ form a basis of $b \otimes_{k} a^{*}$ by the universal property of the tensor product; hence $\dim_{k} (b \otimes_{k} a^{*}) = \dim_{k} b \cdot \dim_{k} a^{*}$ , and $\dim_{k} a^{*} = \dim_{k} a$ for $a^{*} = {Hom}_{k} (a, k)$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Generated submodule, cyclic and finitely generated modules, module basis and free module, Universal property of the tensor product for balanced maps into abelian groups, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[F3]

An isomorphism of $(A_{0}, A_{0})$ -bimodules is a bijection that is linear over $k$ and intertwines both actions, so it preserves $k$ -dimensions and the vanishing of the two multiplications ( $(S, R)$ -bimodules and commuting left and right scalar actions); the external objects $\overset{ˉ}{a} ⊠ b$ in the identified category correspond to the bimodules $b \otimes_{k} a^{*}$ (The opposite Deligne product is the category of finite bimodules).

Counterexample

1.1givenF1F2

The specified algebra is the upper triangular $2 \times 2$ matrix algebra under $e_{1} \mapsto E_{11}$ , $e_{2} \mapsto E_{22}$ , $u \mapsto E_{12}$ , so the products are associative and define a unital algebra. In $A_{0}$ the relations $e_{1} + e_{2} = 1$ , $e_{1} e_{2} = e_{2} e_{1} = 0$ and $e_{1} u = u = u e_{2}$ hold with $e_{1}, e_{2}, u$ a $k$ -basis of the regular bimodule, so left multiplication by $u$ sends $e_{2}$ to $u e_{2} = u \neq 0$ and right multiplication by $u$ sends $e_{1}$ to $e_{1} u = u \neq 0$ , while $A_{0}$ has $k$ -dimension $3$ [F1, F2].

2.1step 1.1F1

On a one-dimensional left or right module, the action of $u$ is multiplication by a scalar $c \in k$ . Since $u^{2} = 0$ , the module law gives $c^{2} = 0$ , hence $c = 0$ because $k$ is a field. Thus $u$ acts as zero on every one-dimensional module on either side.

3.1step 2.1F2F3∎

Suppose the regular bimodule $A_{0}$ were isomorphic to $b \otimes_{k} a^{*}$ . By [F3] the two sides have the same $k$ -dimension and the same vanishing pattern of the two multiplications, and by [F2] $3 = \dim_{k} A_{0} = \dim_{k} b \cdot \dim_{k} a^{*} = \dim_{k} b \cdot \dim_{k} a$ , so one of the two factors is one-dimensional. If $\dim_{k} b = 1$ , then left multiplication by $u$ is zero on $b$ by step 2.1, hence zero on $b \otimes_{k} a^{*}$ because $u \cdot (x \otimes λ) = (u x) \otimes λ = 0$ , contradicting step 1.1, where left multiplication by $u$ sends $e_{2}$ to $u \neq 0$ . If $\dim_{k} a = 1$ , then $u$ acts as zero on $a$ , so $(λ \cdot u) (x) = λ (u x) = 0$ for every $λ \in a^{*}$ , and right multiplication by $u$ is zero on $b \otimes_{k} a^{*}$ because $(y \otimes λ) \cdot u = y \otimes (λ \cdot u) = 0$ , contradicting step 1.1, where right multiplication by $u$ sends $e_{1}$ to $u \neq 0$ . Both alternatives contradict the assumed bimodule isomorphism, so the regular bimodule $A_{0}$ is not isomorphic to any external tensor factor $b \otimes_{k} a^{*}$ .

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