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The projective Nakayama pairing and the symmetric-algebra specialization

Statement

Let $A ≃ A - m o d$ be a finite $k$ -linear abelian category with module model, and let $N^{r} = A^{*} \otimes_{A} -$ be its Nakayama functor (Left and right Nakayama functors by finite kernel calculus, Nakayama kernels give well-defined adjoint functors). For every finite-dimensional projective left $A$ -module $P$ (Projective modules and the lifting property) and every finite-dimensional left $A$ -module $X$ there is a natural isomorphism $D {Hom}_{A} (P, X) ≅ {Hom}_{A} (X, A^{*} \otimes_{A} P) = {Hom}_{A} (X, N^{r} (P)),$ where $D = {Hom}_{k} (-, k)$ is the $k$ -dual (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Linear map between vector spaces over the same field). If moreover the module model is supplied with an isomorphism $A^{*} ≅ A$ of $A$ -bimodules (the symmetric-algebra condition), then $N^{r} ≅ 1$ and $N^{l} ≅ 1$ ; this is a conditional specialization, and no claim is made that every finite-dimensional $k$ -algebra is symmetric or self-injective. No commutativity of $A$ and no choice are used.

Facts & Assumptions

Given: A finite-dimensional unital $k$ -algebra $A$ , a finite $k$ -linear abelian category with module model $A ≃ A - m o d$ , a finite-dimensional projective left $A$ -module $P$ (Projective modules and the lifting property), a finite-dimensional left $A$ -module $X$ , and the Nakayama functors $N^{r} = A^{*} \otimes_{A} -$ , $N^{l} ≅ {Hom}_{A} (A^{*}, -)$ of the model (Left and right Nakayama functors by finite kernel calculus, Nakayama kernels give well-defined adjoint functors).

[F1]

The algebra $A$ is an $(A, A)$ -bimodule by left and right multiplication, and $A^{*} = {Hom}_{k} (A, k)$ is the $(A, A)$ -bimodule with $(a \cdot λ) (b) = λ (b a)$ and $(λ \cdot a) (b) = λ (a b)$ (Unital left and right modules over a ring; unqualified module means left module, $(S, R)$ -bimodules and commuting left and right scalar actions, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[F2]

For a left $A$ -module $P$ , the space ${Hom}_{A} (P, A)$ is a right $A$ -module under $(f \cdot a) (p) = f (p) a$ , since $(f \cdot a) (b p) = b f (p) a$ . Its $k$ -dual is a left $A$ -module under $(a μ) (f) = μ (f \cdot a)$ . Left multiplication on the values of $f$ need not preserve $A$ -linearity when $A$ is noncommutative (Unital left and right modules over a ring; unqualified module means left module, Linear map between vector spaces over the same field).

[F3]

A finite-dimensional left $A$ -module has a finite $k$ -basis, and that finite set generates it as an $A$ -module (through $k$ -linear combinations and the unit), so it is finitely generated (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).

[F4]

For a unital ring $B$ and a left $B$ -module $P$ that is finitely generated and projective, the evaluation map ${Hom}_{B} (P, B) \otimes_{B} Y \to {Hom}_{B} (P, Y)$ , $φ \otimes y \mapsto (p \mapsto φ (p) y)$ , is an isomorphism for every left $B$ -module $Y$ , natural in $Y$ ; with $B = A$ and $Y = X$ it identifies ${Hom}_{A} (P, X)$ with ${Hom}_{A} (P, A) \otimes_{A} X$ (The dual-basis isomorphism for a finitely generated projective bimodule, Projective modules and the lifting property).

[F5]

For unital rings $A, B$ , a $(B, A)$ -bimodule $M$ , a left $A$ -module $X$ and a left $B$ -module $Y$ , currying ${Hom}_{B} (M \otimes_{A} X, Y) \to {Hom}_{A} (X, {Hom}_{B} (M, Y))$ , $F \mapsto (x \mapsto (m \mapsto F (m \otimes x)))$ , is a bijection natural in $X$ and $Y$ ; with $B = k$ , $Y = k$ it gives ${Hom}_{k} (M \otimes_{A} X, k) ≅ {Hom}_{A} (X, {Hom}_{k} (M, k))$ for every right $A$ -module $M$ (Tensor-Hom adjunction for bimodules over arbitrary unital rings).

[F6]

For a finite-dimensional $(A, B)$ -bimodule $Z$ the $k$ -dual $Z^{*} = {Hom}_{k} (Z, k)$ is a $(B, A)$ -bimodule under $(b \cdot λ) (z) = λ (z b)$ and $(λ \cdot a) (z) = λ (a z)$ ; $(-)^{*}$ is a contravariant equivalence carrying isomorphisms to isomorphisms, and the evaluation ${ev}_{Z} : Z \to Z^{* *}$ is a natural isomorphism (Finite module duality is exact with commuting bimodule actions, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , Natural isomorphism).

[F7]

The balanced tensor product is unital and functorial in the module argument: $A \otimes_{A} X ≅ X$ naturally in $X$ , and a homomorphism of right $A$ -modules induces a natural transformation between the functors $- \otimes_{A} -$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ , Module homomorphisms induce tensor-product homomorphisms functorially).

Proof

technique · direct

1.1givenF1F2

Define $γ : A^{*} \otimes_{A} P \to {Hom}_{A} (P, A)^{*}$ by $γ (λ \otimes p) (f) = λ (f (p))$ for $λ \in A^{*}$ , $p \in P$ and $f \in {Hom}_{A} (P, A)$ . It is well defined: $γ (λ \cdot a \otimes p) (f) = (λ \cdot a) (f (p)) = λ (a f (p))$ equals $γ (λ \otimes a p) (f) = λ (f (a p))$ by [F1] and [F2], so the $A$ -balanced relation $λ \cdot a \otimes p = λ \otimes a p$ is respected. It is left $A$ -linear with respect to the left action on $A^{*} \otimes_{A} P$ and the left action $(a \cdot μ) (f) = μ (f \cdot a)$ on ${Hom}_{A} (P, A)^{*}$ : $γ (a \cdot (λ \otimes p)) (f) = (a \cdot λ) (f (p)) = λ (f (p) a)$ and $(a \cdot γ (λ \otimes p)) (f) = γ (λ \otimes p) (f \cdot a) = λ ((f \cdot a) (p)) = λ (f (p) a)$ , using [F1] and [F2].

2.1step 1.1F5F6algebra

The map $γ$ is an isomorphism: its transpose $γ^{*} : {Hom}_{A} (P, A)^{* *} \to (A^{*} \otimes_{A} P)^{*}$ is a bijection, because under the double-duality isomorphism $ev : {Hom}_{A} (P, A) \to {Hom}_{A} (P, A)^{* *}$ of [F6] and the bijection $(A^{*} \otimes_{A} P)^{*} \to {Hom}_{A} (P, A)$ , $θ \mapsto (p \mapsto {ev}_{A}^{- 1} (λ \mapsto θ (λ \otimes p)))$ , given by currying [F5] with $M = A^{*}$ followed by [F6], the composite corresponds to the identity: $γ^{*} (ev (f)) (λ \otimes p) = ev (f) (γ (λ \otimes p)) = λ (f (p))$ , and $p \mapsto {ev}_{A}^{- 1} (λ \mapsto λ (f (p))) = f (p)$ , so the composite sends $f$ to $f$ . Since both comparison maps are bijections, $γ^{*}$ is bijective; all spaces here are finite-dimensional, so double duality [F6] reflects the isomorphism of $γ^{*}$ , and $γ$ is bijective and hence an isomorphism of left $A$ -modules.

3.1step 2.1F3F4F5F6

Since a finite-dimensional module is finitely generated by [F3], the evaluation map of [F4] with $B = A$ and $Y = X$ gives a natural isomorphism ${Hom}_{A} (P, X) ≅ {Hom}_{A} (P, A) \otimes_{A} X$ ; dualizing it by [F6] and currying by [F5] with $M = {Hom}_{A} (P, A)$ and $Y = k$ gives a natural isomorphism $D {Hom}_{A} (P, X) = {Hom}_{k} ({Hom}_{A} (P, X), k) ≅ {Hom}_{A} (X, {Hom}_{A} (P, A)^{*})$ , and composing with ${Hom}_{A} (X, γ^{- 1})$ from step 2.1 gives the natural isomorphism $D {Hom}_{A} (P, X) ≅ {Hom}_{A} (X, A^{*} \otimes_{A} P) = {Hom}_{A} (X, N^{r} (P))$ . All three isomorphisms are natural in $X$ (and in $P$ , since the evaluation formula of [F4] and the formula of $γ$ are natural in $P$ ), so the composite is a natural isomorphism.

4.1step 3.1F1F7∎

Assume now that the model carries an isomorphism $α : A^{*} \to A$ of $A$ -bimodules. Then $N^{r} = A^{*} \otimes_{A} - ≅ A \otimes_{A} - ≅ 1$ , the first natural isomorphism induced by $α$ through functoriality in the first variable and the second the unit isomorphism of [F7]; and $N^{l} ≅ {Hom}_{A} (A^{*}, -) ≅ {Hom}_{A} (A, -) ≅ 1$ , where precomposition with $α$ and with $α^{- 1}$ are mutually inverse natural bijections ${Hom}_{A} (A, -) \to {Hom}_{A} (A^{*}, -)$ and $f \mapsto f (1_{A})$ with inverse $x \mapsto (a \mapsto a x)$ is the natural isomorphism ${Hom}_{A} (A, -) ≅ 1$ supplied by the module axioms of [F1]. This is a conditional specialization: only the supplied bimodule isomorphism is used, and no claim is made that an arbitrary finite-dimensional $k$ -algebra is symmetric or self-injective. Only the finite-dimensional data and finitely many operations enter, so no commutativity of $A$ and no choice are used.

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