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Exact finite tensor functors have projective right-module kernels

Statement

Throughout, a bimodule over k-algebras means a k-vector space with k-bilinear commuting actions and agreeing scalar actions: (c1B)m=m(c1A)=cm for c∈k in a (B,A)-bimodule. This compatibility is an additional requirement beyond the ring-bimodule definition (S,R)-bimodules and commuting left and right scalar actions.

Let A,B be finite-dimensional unital algebras over a field k and let M be a finite-dimensional (B,A)-bimodule with associated tensor functor TM=M⊗A−:A-mod→B-mod. Then the following are equivalent: (1) TM is exact on finite-dimensional left A-modules, equivalently M is flat as a right A-module; (2) M is a projective right A-module. Since M is finite-dimensional, (2) is also equivalent to M being a direct summand of a finite free right A-module and to M being finitely generated and projective. No choice is used.

Facts & Assumptions

Given: The agreeing scalar convention above, a field k, finite-dimensional unital k-algebras A,B, and a finite-dimensional (B,A)-bimodule M, with TM=M⊗A−:A-mod→B-mod.

[F1]

A right A-module N is flat exactly when N⊗A− is exact on left A-modules; every projective right A-module is flat, without choice (Left and right flat modules over an arbitrary ring, Projective left and right modules are flat over an arbitrary ring).

[F2]

The choice-free direction 1⇒4 of the projective-module characterizations produces, from the canonical free cover, an identification of a projective module with a direct summand of a free module; for a finitely generated module the cover may be taken over a finite generating set, so the free module is finite; conversely a direct summand of a free module whose basis is finite is projective, since lifts of the finitely many basis elements can be chosen (Equivalent characterizations of projective modules, Projective modules and the lifting property, Every module is a quotient of a free module, Generated submodule, cyclic and finitely generated modules, module basis and free module).

[F3]

For a (B,A)-bimodule N and finite-dimensional left A-modules X, duality and the tensor-hom adjunction give a natural isomorphism (N⊗AX)∗≅Hom⁡Aop(N,X∗) of left Bop-modules: a functional ω corresponds to u↦(x↦ω(u⊗x)), and the balancing relation for ω is exactly Aop-linearity of that map, because u⋅a denotes the right A-action on N and λ⋅a the right A-action on X∗ (Finite left exact functors are Hom functors with dual bimodule kernels, Finite module duality is exact with commuting bimodule actions, Universal property of the tensor product for balanced maps into abelian groups, Tensor-Hom adjunction for bimodules over arbitrary unital rings).

[F4]

Duality X↦X∗ is an exact contravariant equivalence between finite-dimensional left A-modules and finite-dimensional left Aop-modules, so a functor on one side is exact exactly when the corresponding functor on the other side is (Finite module duality is exact with commuting bimodule actions).

[F5]

The category A-mod is abelian; a finite-dimensional left A-module is finitely generated, and a finite-dimensional right A-module has a finite free cover An↠N built from a finite k-basis (Modules over a ring form an abelian category, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis, Every module is a quotient of a free module, Universal property of a direct sum of modules, The direct sum of an indexed family of modules).

[F6]

If N is a direct summand of the free right module An with inclusion h:N→An and retraction p:An↠N, then N is projective: given a surjection e:E↠N′ and f:N→N′, lift fp:An→N′ by choosing preimages under e of the images of its finitely many basis vectors; precomposing the resulting lift with h lifts f (Equivalent characterizations of projective modules, Projective modules and the lifting property, The splitting lemma for short exact sequences of modules, Split monomorphism, split epimorphism, retraction, and section).

Proof

technique · direct
1.1F1F2

(2)⇒(1). If M is a projective right A-module, then M is flat by [F1], that is, M⊗A− is exact on left A-modules; restricting to finite-dimensional left modules, TM is exact on A-mod. Equivalently, M is a direct summand of a free right module by [F2], tensoring with a free module is a direct sum of copies of the identity, and a direct summand of an exact functor is exact.

1.2F3F4

((1), transport of exactness.) Suppose TM is exact on finite-dimensional left A-modules. By [F3] there are natural isomorphisms (M⊗AX)∗≅Hom⁡Aop(M,X∗) for finite-dimensional left A-modules X; since X↦X∗ is an exact contravariant equivalence between A-mod and Aop-mod by [F4], exactness of M⊗A− on the finite left modules is equivalent to exactness of Hom⁡Aop(M,−) on the finite right A-modules.

2.1F5F6step 1.2chooseconstruct

((1)⇒(2), the cover splits.) Keep the hypothesis of step 1.2. Since M is finite-dimensional it is finitely generated as a right A-module, so a finite k-basis of M induces a surjection p:An↠M of right A-modules by [F5]. Exactness of Hom⁡Aop(M,−) at this surjection gives that Hom⁡Aop(M,An)→Hom⁡Aop(M,M) is surjective, so the identity of M lifts to h:M→An with ph=1M; hence M is a direct summand of the finite free right A-module An, and is projective by [F6].

3.1F1F2F5step 1.1step 2.1

The finite-generation clause: a finite-dimensional module is finitely generated by [F5]; conversely a finitely generated projective right module is a direct summand of a finite free module by [F2], giving the stated equivalences. Steps 1.1 and 2.1 prove (2)⇒(1) and (1)⇒(2), so the exactness of TM, the flatness of M, the projectivity of M, and the finite-summand and finite-generation conditions are all equivalent.

4.1step 1.1step 1.2step 2.1step 3.1given∎

All covers, bases and summands used above are finite (they come from finite k-bases of finite-dimensional modules), and no commutativity of A or B was used, so no choice is used.

Depends on

Used by

Dependency tree · two levels

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Sources