Relative Vertex-Source-Pairs of Modules of and Idempotent Morita Equivalences of Rings

Abstract

Here all rings have identities. Let R be a ring and let R-mod denote the additive category of left finitely generated R-modules. Note that if R is a noetherian ring, then R-mod is an abelian category and every R-module is a finite direct sum of indecomposable R-modules. Finite Group Modular Representation Theory concerns the study of left finitely generated $\mathcal{O}G$-modules where G is a finite group and $\mathcal{O}$ is a complete discrete valuation ring with $\mathcal{O}/J\left(\mathcal{O}\right)$ a field of prime characteristic p. Thus $\mathcal{O}G$ is a noetherian $\mathcal{O}$-algebra. The Green Theory in this area yields for each isomorphism type of finitely generated indecomposable (and hence for each isomorphism type of finitely generated simple $\mathcal{O}G$-module) a theory of vertices and sources invariants. The vertices are derived from the set of p-subgroups of G. As suggested by the above, in Basic Definition and Main Results for Rings Section, let $\mathrm{\Sigma }$ be a fixed subset of subrings of the ring R and we develop a theory of $\mathrm{\Sigma }$-vertices and sources for finitely generated R-modules. We conclude Basic Definition and Main Results for Rings Section with examples and show that our results are compatible with a ring isomorphic to R. For Idempotent Morita Equivalence and Virtual Vertex-Source Pairs of Modules of a Ring Section, let e be an idempotent of R such that $R=ReR$. Set $B=eRe$ so that B is a subring of R with identity e. Then, the functions $eR{\otimes }_R\ast :R-\mathrm{mod}\to B-\mathrm{mod}$ and $Re{\otimes }_B\ast :B-\mathrm{mod}\to R-\mathrm{mod}$ form a Morita Categorical Equivalence. We show, in this Section, that such a categorical equivalence is compatible with our vertex-source theory. In Two Applications with Idemptent Morita Equivalence Section, we show such compatibility for source algebras in Finite Group Block Theory and for naturally Morita Equivalent Algebras.

1. Basic Definition and Main Results for Rings

Our notation and terminology are standard and tend to follow [1,2,3,4]. All rings have identities and all modules over a ring are unitary and finitely generated left modules unless specifically stated. We begin with required elementary results.

Let R be a ring. Then, R-mod will denote the abelian category of left R-modules. If U and V are (left) R-modules, then $U\mid V$ signifies that U is isomorphic to a direct summand of V in R-mod.

Let S also be a ring. Then, R-mod-S will denote the abelian category of abelian groups M such that M is a left R-module and a right finitely generated unitary S-module such that $\left(rm\right)s=r\left(ms\right)$ for all $r\in R$, $m\in M$ and $s\in S$.

Assume that S is a subring of R and let X an R-module. Then, X is S-projective relative to R if $X\mid R{\otimes }_{S}U$ in R-mod for some S-module U.

If n is a positive integer and R is a ring, then $M_n\left(R\right)$ is the ring of $n\times n$-matrices over R.

Let R and S be rings and let $\alpha :R\to S$ be a ring homomorphism such that $\alpha \left(1_R\right)=1_S$. Thus $\alpha$ induces a functor $\mathcal{F}\left(\alpha \right):S-\mathrm{mod}\to R-\mathrm{mod}$ where if V is an S-module, $r\in R$ and $v\in \mathcal{F}\left(\alpha \right)V=V$, then $r·v=\alpha \left(r\right)v$.

Let $\alpha :A\to R$ be a ring isomorphism so that $\alpha \left(1_A\right)=1_R$ and let $\beta ={\alpha }^{-1}:R\to A$.

Thus $\alpha$ and $\beta$ induce inverse functor isomorphisms $\mathcal{F}\left(\alpha \right):R-\mathrm{mod}\to A-\mathrm{mod}$ and $\mathcal{F}\left(\beta \right):A-\mathrm{mod}\to R-\mathrm{mod}$.

Let S be a subring of R, let T be a subring of S and let X be a T-module.

Here, $S{\otimes }_{T}X$ is an S-module so that $\mathcal{F}\left(\alpha \right)\left(S{\otimes }_{T}X\right)$ is a $\beta \left(S\right)$-module and $\beta \left(S\right){\otimes }_{\beta \left(T\right)}\mathcal{F}\left(\alpha \right)X$ is a $\beta \left(S\right)$-module.

It is easy to check:

Lemma 1. 

The well-defined linear map $\gamma :\mathcal{F}\left(\alpha \right)\left(S{\otimes }_{T}X\right)\to \beta \left(S\right){\otimes }_{\beta \left(T\right)}\mathcal{F}\left(\alpha \right)X$ such that $s{\otimes }_{T}x\mapsto \beta \left(s\right){\otimes }_{\beta \left(T\right)}x$ for all $s\in$ and $x\in X$ is a $\beta \left(S\right)$-module isomorphism. Here ${\gamma }^{-1}:\beta \left(S\right){\otimes }_{\beta \left(T\right)}\mathcal{F}\left(\alpha \right)X\to \mathcal{F}\left(\alpha \right)\left(S{\otimes }_{T}X\right)$ is such that $\beta \left(s\right){\otimes }_{\beta \left(T\right)}x\mapsto s{\otimes }_{T}x$ for all $s\in S$ and $x\in X$.

Let S be a subring of the ring R. Clearly

Lemma 2. 

The linear map $\lambda :R{\otimes }_{S}S\to R{1}_S$ such that $r{\otimes }_{S}s\mapsto rs$ for all $r\in R$ and $s\in S$ is an R-module isomorphism with ${\lambda }^{-1}:R{1}_S\to R{\otimes }_{S}S$ such that $r{1}_S\mapsto r{1}_S{\otimes }_S{1}_S$ for all $r\in R$.

Clearly:

(1)

$R=R{1}_S\oplus R(1_R-1_S)$ in R-mod-S and if V is an S-module, then $R{\otimes }_{S}V\cong R{1}_S{\otimes }_{S}V$ in R-mod.

If R is a noetherian ring, then every (finitely generated) R-module is a direct sum of finitely many indecomposable modules [5] (1, Theorem 6.2). The next result, in this case, reduces the study of relative projectivity to indecomposable R-modules.

Lemma 3. 

Suppose that ${\displaystyle X=\underset{i=1}{\overset{n}{⨁}}=Y_i}$ in R-mod where n is a positive integer and $Y_i$ is an R-submodule of X for all $1\le i\le n$. The following two conditions are equivalent:

(a)

X is S-projective relative to R; and

(b)

$Y_i$ is S projective relative to R for all $1\le i\le n$.

Proof. 

Clearly, (a) implies (b). Assume (b). Then, $Y_i\mid R{\otimes }_S{W}_i$ in R-mod for some S-module $W_i$ for all $1\le i\le n$. Thus ${\displaystyle X\mid \underset{i=1}{\overset{n}{⨁}}(R{\otimes }_S{W}_i)\cong R{\otimes }_S(\underset{i=1}{\overset{n}{⨁}}{W}_i)}$ in R-mod and we are done. □

Let X be an R-module such that $X\mid R{\otimes }_{S}U$ in R-mod for some S-module U and let T be a subring of S.

Lemma 4. 

If $U\mid S{\otimes }_{T}W$ for some T-module W, then $X\mid R{\otimes }_{T}W\cong R{1}_T{\otimes }_{T}W$ in R-mod.

Proof. 

Here $X\mid R{\otimes }_{S}S{\otimes }_{T}W\cong R{1}_S{\otimes }_{T}W\cong R{1}_T{\otimes }_{T}W\cong R{\otimes }_{T}W$ in R-mod and we are done. □

Let X be an R-module.

Lemma 5. 

The following four conditions are equivalent:

(a)

X is S-projective relative to R;

(b)

$X\mid R{\otimes }_{S}U$ in R-mod for some S-module U;

(c)

$X\mid R{\otimes }_S{Res}_S^R\left(X\right)$ in R-mod; and

(d)

$X\mid R{\otimes }_{S}U$ in R-mod for some S-module U such that $U\mid {Res}_S^R\left(X\right)$ in S-mod.

Proof. 

Clearly, (a) and (b) are equivalent, (c) and (d) are equivalent and (c) implies (b). Here, [2] (Theorem 2.6.12) completes a proof. □

Remark 1. 

Suppose that $\mathcal{O}$ is a complete discrete valuation ring, that R is an $\mathcal{O}$-algebra, S is an $\mathcal{O}$-subalgebra of R and X is an indecomposable R-module that is S-projective. Then, X satisfies Lemma 5d with U an indecomposable S-module by the Krull–Schmidt Theorem [5] (1, Theorem 6.1).

As suggested by [1] (III, Sections 4 and 5) and [5] (Chapter 4, Section 3):

Let $\mathrm{\Sigma }$ denote a set of subrings of R.

Let $S\in \mathrm{\Sigma }$, let V be an S-module and let X be an R-module.

Our main idea is:

Definition 1. 

Suppose that

(i)

$V\mid {Res}_S^R\left(X\right)$ in S-mod;

(ii)

$X\mid R{\otimes }_{S}V$ in R-mod; and

(iii)

If $T\in \mathrm{\Sigma }$ is a subring of S and V is T-projective relative to S, then $T=S$.

In which case, S is called a Σ-vertex of X, V is called a Σ-S-source of X and $(S,V)$ is called a Σ-vertex-source pair of X.

Definition 2. 

$ITM(R,\mathrm{\Sigma },(S,V\left)\right)$ will denote a set of representatives of the isomorphism types of R-modules X such that $(S,V)$ is a Σ-vertex-source pair of X.

Lemma 6. 

Let X be an R-module and suppose that $S\in \mathrm{\Sigma }$ is a minimal element under inclusion of the set $\{T\in \mathrm{\Sigma }\mid$ X is T-projective relative to R}. Then, there is an S-module V such that $(S,V)$ is a Σ-vertex-source pair of X.

Proof. 

By Lemma 5, there is an S-module such that $V\mid {\mathrm{Res}}_S^R\left(X\right)$ in S-mod and $X\mid R{\otimes }_{S}V$ in R-mod. Assume that $V\mid S{\otimes }_{T}W$ in S-mod where T is a subring of S, $T\in \mathrm{\Sigma }$ and W is a T-module. Then, $X\mid R{\otimes }_{T}W$ in R-mod by Lemma 4. Thus $T=S$ and we are done. □

Remark 2. 

If Σ is a finite set and the R-module X is S-projective relative to R for some $S\in \mathrm{\Sigma }$, then Lemma 6 applies. A similar statement applies if R is an algebra over a field F by using a dimension over F.

Let A also be a ring and assume that $\alpha :A\to R$ is a ring isomorphism so that $\alpha \left(1_A\right)=1_R$. Set $\beta ={\alpha }^{-1}:R\to A$. Thus, $\alpha$, $\beta$ induce inverse functor isomorphisms $\mathcal{F}\left(\alpha \right):R-\mathrm{mod}\to A-\mathrm{mod}$ and $\mathcal{F}\left(\beta \right):A-\mathrm{mod}\to R-\mathrm{mod}$.

Let $\mathrm{\Delta }$ be a set of subrings of A so that $\mathrm{\Sigma }=\alpha \left(\mathrm{\Delta }\right)=\left\{\alpha \right(S)\mid S\in \mathrm{\Delta }\}$ is a set of subrings of R.

Lemma 7. 

(a)

Let X be an A-module with Δ-vertex-source pair $(Q,V)$ in A-mod. Then, the R-module $\mathcal{F}\left(\beta \right)X$ has Σ-vertex-source pair $\left(\alpha \right(Q),\mathcal{F}(\beta \left)V\right)$ in R-mod; and

(b)

Let Y be an R-module with Σ-vertex-source pair $(T,W)$ in R-mod. Then, the A-module $\mathcal{F}\left(\alpha \right)Y$ has Δ-vertex-source-pair $\left(\beta \right(T),\mathcal{F}(\alpha \left)W\right)$ in A-mod.

Proof. 

By symmetry it suffices to prove (a). Let $U\in \mathrm{\Delta }$ be a subalgebra of Q and let Y be an $\alpha \left(U\right)$-module such that $\mathcal{F}\left(\beta \right)V\mid \alpha \left(Q\right){\otimes }_{\alpha \left(U\right)}Y$ in $\alpha \left(Q\right)$-mod. Then, $V\mid \mathcal{F}\left(\alpha \right)\left(\alpha \left(Q\right){\otimes }_{\alpha \left(U\right)}Y\right)\cong Q{\otimes }_U\beta \left(Y\right)$ in Q-mod by Lemma 1. Thus, $U=Q$ and $\alpha \left(U\right)=\alpha \left(Q\right)$. □

Here, $V\mid {\mathrm{Res}}_Q^{\mathrm{\Delta }}\left(X\right)$ in Q-mod so that $\mathcal{F}\left(\beta \right)V\mid \mathcal{F}\left(\beta \right){\mathrm{Res}}_Q^{\mathrm{\Delta }}\left(X\right)\cong {\mathrm{Res}}_{\beta \left(Q\right)}^R\left(\mathcal{F}\left(\beta \right)X\right)$ in $\alpha \left(Q\right)$-mod.

Also, $X\mid A{\otimes }_{Q}V$ in A-mod so that $\mathcal{F}\left(\beta \right)X\mid \mathcal{F}\left(\beta \right)\left(A{\otimes }_{Q}V\right)\cong R{\otimes }_{\alpha \left(Q\right)}\mathcal{F}\left(\beta \right)V$ in R-mod by Lemma 1. The proof is complete.

Remark 3. 

As above, let $Q\in \mathrm{\Delta }$ and V be a Q-module. Then, α induces a bijection of $ITM(A,\mathrm{\Delta },(Q,V\left)\right)$ into $ITM(R,\alpha (\mathrm{\Delta }),\alpha (Q),\mathcal{F}(\beta \left)V\right)$.

Example 1. 

Let G be a finite group and let ${\displaystyle A=\underset{g\in G}{⨁}{A}_g}$ be a G-crossed product. Thus, $1_A=1_{A_1}$ and for each $g\in G$, there is an $x_g\in A_g\cap A^{\times }$ such that $A_g=x_g{A}_1=A_1{x}_g$ and ${\displaystyle \mathcal{G}=\bigcup _{g\in G}\left(x_g{A}_1^{\times }\right)}$ is the group of graded units of A.

For each $H\le G$, set ${\displaystyle A_H=\underset{x\in H}{⨁}{A}_x}$ and let $\mathrm{\Sigma }=\{A_H\mid H\le G\}$, a finite set of subrings of A.

Let X be an A-module so that $X\cong A{\otimes }_{A}X$ in A-mod. Thus there is a minimal subgroup Q of G such that X is $A_Q$-projective relative to A. Thus there is an $A_Q$-module V such that $(A_Q,V)$ is a Σ-vertex-source pair of X.

Example 2. 

Let k be a commutative ring such that $k/J\left(k\right)$ is a field of prime characteristic p and let G be a finite group. Let $P\in {\mathrm{Syl}}_p\left(G\right)$. By [2] (Corollary 2.6.3), every $kG$-module is $kP$-projective. Set $\mathrm{\Sigma }=\{kQ\mid Q\le P\}$. As above, every $kG$-module X has a Σ-vertex-source pair. Thus we extend the beginning of J. A. Green’s Theory of vertices and source for indecomposable modules in Finite Group Modular Representation Theory.

Example 3. 

Let $\mathcal{O}$ be a complete discrete valuation ring such that $\mathcal{O}/J\left(\mathcal{O}\right)$ is a field of prime characteristic p. Also, let G be a finite group, let b be a block of $\mathcal{O}G$, let P be a defect group of $\left(\mathcal{O}G\right)b$ and let X be an $\left(\mathcal{O}G\right)b$-module that is not necessarily indecomposable. Then, ref. [1] (III, Corollary 6.8) implies that X is $\mathcal{O}P$-projective. Set $\mathrm{\Sigma }=\left\{\right(\mathcal{O}Q)b\mid Q\le P\}$. Then, X has a Σ-vertex-source pair as an $\left(\mathcal{O}G\right)b$-module.

2. Idempotent Morita Equivalence and Virtual Vertex-Source Pairs of Modules of a Ring

Let e be an idempotent of R so that $C_R\left(e\right)=\{r\in R\mid er=re\}$ is a subring of R with identity $1_R$. Set $B=eRe$ so that B is a subring of $C_R\left(e\right)$ with identity e.

Clearly:

(1)

$Re=B\oplus (1_R-e)Re$ in $C_R\left(e\right)$-mod-$C_R\left(e\right)$.

(2)

$e\in Z\left(C_R\left(e\right)\right)$, $1_R-e\in Z\left(C_R\left(e\right)\right)$, $B=e{C}_R\left(e\right)=C_R\left(e\right)e$ and e, $(1_R-e)$ are orthogonal central idempotents of $C_R\left(e\right)$ with $C_R\left(e\right)=B\oplus (1_R-e)C_R\left(e\right)$ as ideals of $C_R\left(e\right)$.

Let $\tau :C_R\left(e\right)\to B$ denote the ring epimorphism such that $\alpha \mapsto \alpha e=e\alpha$ for all $\alpha \in C_R\left(e\right)$.

(3)

$\mathrm{Ker}\left(\tau \right)=(1_R-e)C_R\left(e\right)$ and $\tau$ induces a functor $\mathcal{F}\left(\tau \right):B-\mathrm{mod}\to C_R\left(e\right)-\mathrm{mod}$.

Let Z be a B-module so that $\mathcal{F}\left(\tau \right)Z$ is a $C_R\left(e\right)$-module and $B{\otimes }_{B}Z$ is a $C_R\left(e\right)$-module.

(4)

The linear map $\mu :B{\otimes }_{B}Z\to Z$ such that $b{\otimes }_{B}z\mapsto bz$ for all $b\in B$ and $z\in Z$ is a $C_R\left(e\right)$-module isomorphism with inverse ${\mu }^{-1}:Z\to B{\otimes }_{B}Z$ such that $z\mapsto e{\otimes }_{B}z$ for all $z\in Z$.

(5)

The linear map $\pi :B{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z\to \mathcal{F}\left(\tau \right)Z$ such that $b{\otimes }_{C_R\left(e\right)}z\mapsto bz$ for all $b\in B$ and $z\in Z$ is a well defined $C_R\left(e\right)$-module isomorphism with inverse ${\pi }^{-1}:\mathcal{F}\left(\tau \right)\to B{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z$ such that $z\mapsto e{\otimes }_{C_R\left(e\right)}z$ for all $z\in Z$.

Let T be a subring of $C_R\left(e\right)$ with identity $1_R$ so that $S=Te$ is a subring of B with identity e.

Here, by restriction, $\tau :T\to S$ is a ring epimorphism that induces a functor $\mathcal{F}\left(\tau \right):S-\mathrm{mod}\to T-\mathrm{mod}$.

Let U be an S-module so that $\mathcal{F}\left(\tau \right)U$ is a T-module. Let F be a ring and let Y be an F-mod-S-module so that $Y\mathcal{F}\left(\tau \right)$ is an F-mod-T-module.

Lemma 8. 

The linear map $\omega :Y\mathcal{F}\left(\tau \right){\otimes }_T\mathcal{F}\left(\tau \right)V\to Y{\otimes }_{S}V$ such that $y{\otimes }_{T}v\mapsto y{\otimes }_{S}v$ for all $y\in Y\mathcal{F}\left(\tau \right)=Y$ and $v\in \mathcal{F}\left(\tau \right)V=V$ is an F-module isomorphism with inverse ${\omega }^{-1}:Y{\otimes }_{S}V\to Y\mathcal{F}\left(\tau \right){\otimes }_T\mathcal{F}\left(\tau \right)V$ such that $y{\otimes }_{s}v\mapsto y{\otimes }_{T}v$ for all $y\in Y$ and $v\in V$.

Proof. 

Let $y\in Y$, $v\in V$ and $t\in T$. Then, $\left(\left(yte\right){\otimes }_{S}v\right)-\left(y{\otimes }_s\left(te\right)v\right)=0$ so that $\omega$ is a well defined F-module homomorphism. Also, if $s\in S$ so that $s=te$ for some $t\in T$, then $\left(y\left(te\right){\otimes }_{T}v\right)-\left(y{\otimes }_T\left(te\right)v\right)=(y·t{\otimes }_{T}v)-\left(y{\otimes }_T(t·v)\right)=0$. Then, ${\omega }^{-1}$ is a well defined F-module homomorphism and the proof is complete. □

Let Z be a B-module, let $\mathrm{\Sigma }$ be a set of subrings S of B such that $e\in S$. Let $S\in \mathrm{\Sigma }$ and let V be an S-module.

Theorem 1. 

The following three conditions are equivalent:

(a)

$(S,V)$ is a Σ-vertex-source-pair of Z in B-mod;

(b)

$(S,V)$ is a Σ-vertex-source-pair of $\mathcal{F}\left(\tau \right)Z$ in $C_R\left(e\right)$-mod; and

(c)

$(S,V)$, is a Σ-vertex-source-pair of $Re{\otimes }_{B}Z$ in R-mod.

Proof. 

Clearly, the requirement of Definition 1(iii) for S is satisfied in B, $C_R\left(e\right)$ and R.

Assume $Z\mid B{\otimes }_{S}V$ in B-mod so that $\mathcal{F}\left(\tau \right)Z\mid \mathcal{F}\left(\tau \right)\left(B{\otimes }_{S}V\right)\mid C_R\left(e\right){\otimes }_{S}V$ in $C_R\left(e\right)$-mod since $C_R\left(e\right)=\mathcal{F}\left(\tau \right)B\oplus C_R\left(e\right)(1_R-e)$ in $C_R\left(e\right)$-mod-S. Conversely, $\mathcal{F}\left(\tau \right)Z\mid C_R\left(e\right){\otimes }_{S}V$ in $C_R\left(e\right)$-mod implies that $Z\cong e\mathcal{F}\left(\tau \right)Z\mid e{C}_R\left(e\right){\otimes }_{S}V=B{\otimes }_{S}V$ in B-mod.

Assume $V\mid {\mathrm{Res}}_S^R\left(Z\right)$ in S-mod. Here ${\mathrm{Res}}_S^B\left(Z\right)={\mathrm{Res}}_S^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)$ in S-mod so that $V\mid {\mathrm{Res}}_S^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)$ in S-mod. Conversely, assume that $V\mid {\mathrm{Res}}_S^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)$ in S-mod. Then, $V\mid e{\mathrm{Res}}_S^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)={\mathrm{Res}}_S^R\left(Z\right)$ and (a) and (b) are equivalent.

Assume $Z\mid B{\otimes }_{S}V$ in B-mod. Then, $Re{\otimes }_{B}Z\mid Re{\otimes }_{B}B{\otimes }_{S}V\cong Re{\otimes }_{S}V$ in R-mod. Conversely, assume $Re{\otimes }_{B}Z\mid Re{\otimes }_{S}V$ in R-mod. Thus $Z\cong B{\otimes }_{B}Z\mid B{\otimes }_{S}V$ in B-mod.

Assume $V\mid {\mathrm{Res}}_S^B\left(Z\right)$ in S-mod. Since $Re=B\oplus (1_R-e)Re$ in B-mod-B, ${\mathrm{Res}}_S^R\left(Re{\otimes }_{B}Z\right)={\mathrm{Res}}_S^B\left(B{\otimes }_{B}Z\right)\oplus {\mathrm{Res}}_S^B(1_r-eRe{\otimes }_{B}Z)\cong {\mathrm{Res}}_S^B\left(Z\right)$ in S-mod and so $V\mid {\mathrm{Res}}_S^R\left(Re{\otimes }_{B}Z\right)$ in S-mod. Finally, assume that $V\mid {\mathrm{Res}}_S^R\left(Re{\otimes }_{B}Z\right)$ in S-mod. Then, since ${\mathrm{Res}}_S^R\left(Re{\otimes }_{B}Z\right)\cong {\mathrm{Res}}_S^B\left(Z\right)$, we have $V\mid {\mathrm{Res}}_S^B\left(Z\right)$ and (a) and (c) are equivalent. The proof is complete. □

For the remainder of this Section assume:

(6)

$R=ReR$

As is well known, (cf. [2] (Theorem 2.8.7)):

Proposition 1. 

The functors $Re{\otimes }_B\ast :B-mod\to R-mod$ and $eR{\otimes }_R\ast :R-mod\to B-mod$ comprise a Morita equivalence between the abelian categories B-mod and R-mod. Here, if X is an R-module, then the linear homomorphism $\delta :eR{\otimes }_{R}X\to eX$ such that if $\alpha \in R$ and $x\in X$, then $e\alpha {\otimes }_{R}c\mapsto e\alpha x$ is a well defined B-module isomorphism with ${\delta }^{-1}:eX\to eR{\otimes }_{R}X$ such that if $x\in X$, then $ex\mapsto e{\otimes }_{R}ex$. Also, if X is an R-module, then $X\cong R{\otimes }_{B}eX\cong Re{\otimes }_{B}eX$ in R-mod.

Theorem 1 implies:

Corollary 1. 

Let Σ, $S\in \mathrm{\Sigma }$ and B be as in Theorem 1. Then, the functor $Re{\otimes }_B\ast :B-mod\to R-mod$ induces a bijection of $ITM(B,\mathrm{\Sigma },(S,V\left)\right)$ onto $ITM(R,\mathrm{\Sigma },(S,V\left)\right)$.

For the remainder of this Section, assume:

(7)

$\mathrm{\Delta }$ is a set of subrings T of $C_R\left(e\right)$ such that $1_R\in T$ and $\mathrm{Ker}\left(\tau \right)\cap T=\left(0\right)$.

Thus:

(8)

If $T\in \mathrm{\Delta }$, then $\tau :T\to Te$ is a ring isomorphism with ${\tau }^{-1}:Te\to T$ and the functors $\mathcal{F}\left(\tau \right):Te-\mathrm{mod}\to T--\mathrm{mod}$ and $\mathcal{F}\left({\tau }^{-1}\right):T-\mathrm{mod}\to Te-\mathrm{mod}$ are inverse functor category isomorphisms.

Let Z be a B-module so that $\mathcal{F}\left(\tau \right)Z$ is a $C_R\left(e\right)$-module.

Lemma 9. 

(a)

The linear map $\varphi :R{\otimes }_{C_r\left(e\right)}\mathcal{F}\left(t\right)Z\to Re{\otimes }_{C_r\left(e\right)}\mathcal{F}\left(t\right)Z$ such that $r{\otimes }_{C_R\left(e\right)}z\to re{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(t\right)z$ for all $r\in R$ and $z\in Z$ is a well defined isomorphism in R-mod; and

(b)

The linear map $\sigma :Re{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z\to Re{\otimes }_{B}Z$ such that $re{\otimes }_{C_R\left(e\right)}z\mapsto re{\otimes }_{B}z$ for all $r\in R$ and $z\in Z$ is a well defined R-module isomorphism with inverse ${\sigma }^{-1}:Re{\otimes }_{B}Z\to Re{\otimes }_{C_r\left(e\right)}\mathcal{F}\left(\tau \right)Z$ such that $re{\otimes }_{B}z\mapsto re{\otimes }_{C_R\left(e\right)}z$ for all $r\in R$ and $z\in Z$.

Proof. 

For (a), note that if $z\in \mathcal{F}\left(\tau \right)Z$, then $ez=z$. Since $R=Re\oplus R(1_r-e)$ in R-mod-$C_R\left(e\right)$, (a) follows.

Let $r\in R$, $z\in Z$, and $\alpha \in C_R\left(e\right)$. Then, $\left(re\alpha {\otimes }_{B}z\right)-\left(re{\otimes }_B\left(e\alpha e\right){\otimes }_B\right)=re\left(e\alpha e\right){\otimes }_{B}z-\left(re{\otimes }_B\left(e\alpha e\right)z\right)=0$. Thus, $\sigma$ is a well defined R-module homomorphism. Also, if $b\in B$, then $\left(re\right)b{\otimes }_{C_R\left(e\right)}z)-\left(re{\otimes }_{C_R\left(e\right)}bz\right)=\left(\left(re\right)b{\otimes }_{C_R\left(e\right)}z\right)-(re{\otimes }_{C_R\left(e\right)}b·z)=0$. Thus ${\sigma }^{-1}$ is a well defined R-module homomorphism and we are done. □

Let $T\in \mathrm{\Delta }$, set $S=Te$ and let V be an S-module so that $\mathcal{F}\left(\tau \right)V$ is a T-module. The next two results follow from Lemma 8 as indicated:

Lemma 10. 

The linear map $\rho :Re{\otimes }_T\mathcal{F}\left(\tau \right)V\to Re{\otimes }_{S}V$ such that $re{\otimes }_{T}v\mapsto re{\otimes }_{S}v$ for all $r\in R$ and $v\in V$ is a well defined R-module isomorphism with inverse ${\rho }^{-1}:Re{\otimes }_{S}v\to Re{\otimes }_T\mathcal{F}\left(\tau \right)v$ such that $re{\otimes }_{S}v\mapsto re{\otimes }_{T}v$ for all $r\in R$ and $v\in V$.

Proof. 

In Lemma 8 set $Y=Re$ in R-mod-S. Since $Re\mathcal{F}\left(\tau \right)=Re$ in R-mod-T, Lemma 8 yields a proof. □

Lemma 11. 

The linear map $\lambda :B{\otimes }_T\mathcal{F}\left(\tau \right)V\to B{\otimes }_{S}V$ such that $b{\otimes }_{T}v\mapsto b{\otimes }_{S}v$ for all $b\in B=B\mathcal{F}\left(\tau \right)$ and $v\in \mathcal{F}\left(\tau \right)V=V$ is a well defined $C_R\left(e\right)$-module isomorphism with inverse ${\lambda }^{-1}:B{\otimes }_{S}V\to B{\otimes }_T\mathcal{F}\left(\tau \right)V$ such that $b{\otimes }_{S}v\mapsto b{\otimes }_{T}v$ for all $b\in B$ and $v\in V$.

Proof. 

In Lemma 8 set $Y=B$. a $C_R\left(e\right)$-mod-S module. Since $B\mathcal{F}\left(\tau \right)=B$ in $C_R\left(e\right)$-mod-T, Lemma 8 yields a proof. □

Let $U\in \mathrm{\Delta }$ be a subring of $T\in \mathrm{\Delta }$ and let X be a U-module so that $T{\otimes }_{U}X$ is a T-module. Set $S=Te$ so that $\mathcal{F}\left({\tau }^{-1}\right)X$ is a $Ue$-module.

Lemma 12. 

The linear map $\sigma :\mathcal{F}\left(t^{-1}\right)\left(T{\otimes }_{U}X\right)\to S{\otimes }_{Ue}\mathcal{F}\left(t^{-1}\right)X$ such that $t{\otimes }_{U}x\mapsto \tau \left(t\right){\otimes }_{Ue}x$ for all $t\in \mathcal{F}\left(t^{-1}\right)T=T$ and $x\in X$ is a well defined S-module isomorphism with inverse ${\sigma }^{-1}:S{\otimes }_{Ue}\mathcal{F}\left(t\right)X\to \mathcal{F}\left(t^{-1}\right)\left(T{\otimes }_{U}X\right)$ such that $s{\otimes }_{Ue}x\mapsto t^{-1}\left(s\right){\otimes }_{U}x$ for all $s\in S$ and $x\in \mathcal{F}\left(t^{-1}\right)X=X$.

Proof. 

Let $s\in S$, $t\in T$, $u\in U$ and $x\in X$. Then, $\sigma (s·\left(t{\otimes }_{U}X\right))=\sigma \left(t^{-1}\left(s\right)t{\otimes }_{U}x\right)=s\left(te\right){\otimes }_{Ue}x=s\sigma \left(t{\otimes }_{U}x\right)$. Also, $\left(\tau \left(tu\right)\right){\otimes }_{Ue}x)-\left(\tau \left(t\right){\otimes }_{Ue}ux\right)=\left(\tau \left(t\right)\left(ue\right){\otimes }_{Ue}x\right)-(\tau \left(t\right){\otimes }_{Ue}ue·x)=0$. Thus $\sigma$ is a well defined S-module homomorphism. Since $\left({\tau }^{-1}\left(sue\right){\otimes }_{U}x\right))-\left({\tau }^{-1}\left(s\right){\otimes }_{U}ux\right))=\left({\tau }^{-1}\left(s\right)u{\otimes }_{U}x\right)-\left({\tau }^{-1}\left(s\right){\otimes }_{U}ux\right)=0$, ${\sigma }^{-1}$ is a well defined S-module homomorphism and we are done. □

Let Z be a B-module so that $\mathcal{F}\left(\tau \right)Z$ is a $C_R\left(e\right)$-module and $\mathcal{F}\left(\tau \right)B\mathcal{F}\left(\tau \right)B$ is in $C_R\left(e\right)$-mod-$C_R\left(e\right)$. Set $T\in \mathrm{\Delta }$ and set $S=Te\in \mathrm{\Sigma }$, also let V be an S-module so that $\mathcal{F}\left(\tau \right)V$ is a T-module.

The main result of this section is:

Theorem 2. 

The following four conditions are equivalent:

(a)

$(S,V)$ is a Σ-vertex-source pair of Z in B-mod;

(b)

$(S,V)$ is a Σ-vertex-source pair of $\mathcal{F}\left(\tau \right)Z$ in $C_R\left(e\right)$-mod;

(c)

$(S,V)$ is a Σ-vertex-source pair of $Re{\otimes }_{B}Z$ in R-mod; and

(d)

$(T,\mathcal{F}(\tau \left)V\right)$ is a Δ-vertex-source pair of $\mathcal{F}\left(\tau \right)Z$ in $C_R\left(e\right)$-mod.

In which case, $(T,\mathcal{F}(\tau \left)V\right)$ is a Δ-vertex-source pair of $Re{\otimes }_{B}Z$ in R-mod.

Proof. 

Here, (a)–(c) are equivalent by Theorem 1. Assume (a). Let $U\in \mathrm{\Delta }$ be a subring of T such that $\mathcal{F}\left(\tau \right)V\mid T{\otimes }_{U}W$ in T-mod for some T-module W. Then, $V\mid \mathcal{F}\left({\tau }^{-1}\right)\left(T{\otimes }_{U}W\right)\cong S{\otimes }_{Ue}\mathcal{F}\left({\tau }^{-1}\right)W$ in S-mod. Thus, $Ue=Te$ and $U=T$.

Here $V\mid {\mathrm{Res}}_S^B\left(Z\right)$ in S-mod so that $\mathcal{F}\left(\tau \right)V\mid \mathcal{F}\left(\tau \right){\mathrm{Res}}_S^B\left(Z\right)={\mathrm{Res}}_T^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)$ in T-mod.

Also, $Z\mid B{\otimes }_{S}V$ in B-mod so that $\mathcal{F}\left(\tau \right)Z\mid \mathcal{F}\left(\tau \right)\left(B{\otimes }_{S}V\right)\cong \mathcal{F}\left(\tau \right)\left(B{\otimes }_T\mathcal{F}\left(\tau \right)\left(V\right)\right)$ in $C_R\left(e\right)$-mod by Lemma 11. Since $\mathcal{F}\left(\tau \right)B=B$ in $C_R\left(e\right)$-mod-$C_R\left(e\right)$, $\mathcal{F}\left(\tau \right)Z\mid C_R\left(e\right){\otimes }_T\mathcal{F}\left(\tau \right)V$ in $C_R\left(e\right)$-mod. Thus, (a) implies (d).

Assume (d). Let $Q\in \mathrm{\Sigma }$ be a subring of S such that $V\mid S{\otimes }_Q\mathcal{Z}$ in S-mod for some S-module $\mathcal{Z}$. Then, $\mathcal{F}\left(\tau \right)V\mid \mathcal{F}\left(\tau \right)\left(S{\otimes }_Q\mathcal{Z}\right)\cong T{\otimes }_{{\tau }^{-1}\left(Q\right)}\mathcal{F}\left(\tau \right)\mathcal{Z}$ in T-mod by Lemma 12. Thus, $T={\tau }^{-1}\left(Q\right)$ and $Q=S$.

Here, $\mathcal{F}\left(\tau \right)Z\mid C_R\left(e\right){\otimes }_T\mathcal{F}\left(\tau \right)V$ in $C_R\left(e\right)$-mod so that $Z\mid B{\otimes }_T\mathcal{F}\left(\tau \right)V\cong B{\otimes }_{S}V$ in B-mod by Lemma 11. Also, $\mathcal{F}\left(\tau \right)V\mid {\mathrm{Res}}_T^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)$ in T-mod so that $V\mid \mathcal{F}\left({\tau }^{-1}\right){\mathrm{Res}}_T^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)={\mathrm{Res}}_S^B\left(Z\right)$ in S-mod. Thus, (a)–(d) are equivalent.

Clearly, $T\in \mathrm{\Delta }$ and $\mathcal{F}\left(\tau \right)V$ satisfy condition (iii) of Definition 1 for $Re{\otimes }_{B}Z$ in R-mod.

Here, $Re{\otimes }_{B}Z\cong Re{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z$ in R-mod by Lemma 9(b) and $Re{\otimes }_{C_{R}e}\mathcal{F}\left(\tau \right)Z\mid Re{\otimes }_{C_R\left(e\right)}{C}_R\left(e\right){\otimes }_T\mathcal{F}\left(\tau \right)V\cong Re{\otimes }_T\mathcal{F}\left(\tau \right)V$ in R-mod so that $Re{\otimes }_{B}Z\mid Re{\otimes }_T\mathcal{F}\left(\tau \right)V$ in R-mod. Also, $\mathcal{F}\left(\tau \right)V\mid {\mathrm{Res}}_T^{C_R\left(e\right)}\left(\mathcal{F}\left(\tau \right)Z\right)$ in T-mod and ${\mathrm{Res}}_T^R\left(Re{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z\right)={\mathrm{Res}}_{C_R\left(e\right)}^R\left({\mathrm{Res}}_T^{C_R\left(e\right)}\left(Re{\otimes }_{C_R\left(e\right)}\right)\mathcal{F}\left(\tau \right)Z\right))={\mathrm{Res}}_{C_R\left(e\right)}^R\left({\mathrm{Res}}_T^{C_R\left(e\right)}\left(B{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z\right)\right)\oplus {\mathrm{Res}}^R\left(C_R\left(e\right)Re{\otimes }_T\mathcal{F}\left(\tau \right)Z\right)$ in T-mod. As $B{\otimes }_{C_R\left(e\right)}\mathcal{F}\left(\tau \right)Z\cong \mathcal{F}\left(\tau \right)Z$ in $C_R\left(e\right)$-mod by (5), $\mathcal{F}\left(\tau \right)V\mid {\mathrm{Res}}_T^R\left(Re{\otimes }_{B}Z\right)$ in T-mod and the proof is complete. □

3. Two Applications with Idemptent Morita Equivalence

3.1. Application to Source Algebras in Finite Group Block Theory

Let $\mathcal{O}$ be a complete discrete valuation ring such that $\mathcal{O}/J\left(\mathcal{O}\right)$ is a field of prime characteristic p and let G be a finite group. Also, let b be a block idempotent of $\mathcal{O}G$ so that $\left(\mathcal{O}G\right)b$ is a primitive G-algebra over $\mathcal{O}$.

Let $P_{\gamma }$ be a defect of $\left(\mathcal{O}G\right)b$ and let $j=jb=bj\in \gamma$.

Thus:

(1)

$\left(\mathcal{O}G\right)bj\left(\mathcal{O}G\right)b=\left(\mathcal{O}G\right)b=\left(\mathcal{O}G\right)j\left(\mathcal{O}G\right)$.

Set $B=j\left(\mathcal{O}G\right)j=j\left(\mathcal{O}G\right)bj$ so that B is a source algebra of $\left(\mathcal{O}G\right)b$ with identity j.

Proposition 2. 

The functor $\left(\mathcal{O}G\right)j{\otimes }_B\ast :B-mod\to \left(\mathcal{O}G\right)b-mod$ and $j\left(\mathcal{O}G\right)b{\otimes }_{\left(\mathcal{O}G\right)b}\ast :\left(\mathcal{O}G\right)b-mod\to B-mod$ comprise a Morita equivalence between the abelian categories B-mod and $\left(\mathcal{O}G\right)b$-mod. Here, if X is an $\left(\mathcal{O}G\right)b$-module, then $j\left(\mathcal{O}G\right)b{\otimes }_{\left(\mathcal{O}G\right)b}X\cong jX$ in B-mod and $X\cong \left(\mathcal{O}G\right)b{\otimes }_{B}jX$ in $\left(\mathcal{O}G\right)b$-mod.

(2)

Let $p\in P$. Then, $pj=jp$ and the $\mathcal{O}$-linear map $\tau :\mathcal{O}P\to \left(\mathcal{O}P\right)j$ such that $p\mapsto pj$ is an $\mathcal{O}$-algebra isomorphism. Thus, $\tau$ and ${\tau }^{-1}$ induce inverse functor isomorphisms $\mathcal{F}\left(\tau \right):\left(\mathcal{O}P\right)j-\mathrm{mod}\to \left(\mathcal{O}P\right)b-\mathrm{mod}$ and $\mathcal{F}\left({\tau }^{-1}\right):\left(\mathcal{O}P\right)b-\mathrm{mod}\to \left(\mathcal{O}P\right)j-\mathrm{mod}$.

From [3] (Theorem 6.4.7), we have:

(3)

Every B-module is $\left(\mathcal{O}P\right)j$-projective and $\left(\mathcal{O}P\right)j$ is an $\mathcal{O}$-submodule of B.

Set $C=C_{\left(\mathcal{O}P\right)b}\left(j\right)$ so that C is an $\mathcal{O}$-subalgebra of $\left(\mathcal{O}G\right)b$ with identity b. Thus, $B=jC=Cj$ is an $\mathcal{O}$-ideal of C where $C=B\oplus \left(\right(1-j\left)C\right)$ as ideals of C.

Extend the map $\tau$ to $\tau :C\to B$ such that $\alpha \mapsto \alpha j$ for all $\alpha \in C$ so that $\tau$ is an $\mathcal{O}$-algebra epimorphism. Thus $\tau$ induces a functor $\mathcal{F}\left(\tau \right):B-\mathrm{mod}\to C-\mathrm{mod}$.

Set $\mathrm{\Delta }=\left\{\right(\mathcal{O}Q)b\mid Q\le P\}$ and $\mathrm{\Sigma }=\tau \left(\mathrm{\Delta }\right)=\{\mathcal{O}Qj\mid Q\le P\}$. Thus, $\mathrm{\Delta }$ is a (finite) set of $\mathcal{O}$-subalgebras of C, $\mathrm{\Sigma }$ is a finite set of $\mathcal{O}$-subalgebras of B and $\tau :\mathrm{\Delta }\to \mathrm{\Sigma }$ is a bijection.

The following consequence of our analyses implies and extends [3] (Theorem 6.4.10):

Theorem 3. 

Let Z be a B-module so that $\mathcal{F}\left(\tau \right)Z$ is a C-module.

(a)

Z has a Σ-vertex-source pair in B-mod. Let $\left(\right(\mathcal{O}Q)j,V)$ where $Q\le P$ be a Σ-vertex-source pair of Z in B-mod so that $\left(\right(\mathcal{O}Q)j,V)$ is a Σ-vertex-source pair of $\left(\mathcal{O}G\right)b{\otimes }_{B}Z$ in $\left(\mathcal{O}G\right)b$-mod. Here, V is an $\left(\mathcal{O}Q\right)j$-module so that $\mathcal{F}\left(t\right)V$ is an $\left(\mathcal{O}Q\right)b$-module.

(b)

$\left(\mathcal{O}Q\right)b$ is a minimal element U of Δ such that $\mathcal{F}\left(t\right)V$ is U-projective relative to $\left(\mathcal{O}Q\right)j$;

(c)

$\left(\right(\mathcal{O}Q)b,\mathcal{F}(t\left)V\right)$ is a Δ-vertex-source pair of $\mathcal{F}\left(t\right)Z$ in C-mod;

(d)

$\left(\mathcal{O}G\right)j{\otimes }_{B}Z\cong \left(\mathcal{O}G\right)j{\otimes }_C\mathcal{F}\left(t\right)Z$ in $\left(\mathcal{O}G\right)b$-mod; and

(e)

$\left(\right(\mathcal{O}Q)b,\mathcal{F}(t\left)V\right)$ is a Δ-vertex-source pair of $\left(\mathcal{O}G\right)j{\otimes }_{B}Z$ in $\left(\mathcal{O}G\right)b$-mod.

Proof. 

Clearly, (3) yields (a). Now, Lemma 9(b) and Theorem 2 complete a proof. □

Let $Q\le P$ and let V be an $\left(\mathcal{O}Q\right)j$-module.

Corollary 2. 

The Morita equivalence of Proposition 2 induces a bijection of $ITM\left(\right(B,\mathrm{\Sigma },\left(\mathcal{O}Q\right)j),V)$ onto $ITM\left(\right(\mathcal{O}G)b,\mathrm{\Delta },(\mathcal{O}Q)b,\mathcal{F}(t\left)V\right)$.

3.2. Application to Naturally Morita Equivalent Algebras

Let k be a commutative ring, let n be a positive integer and set $S=M_n\left(R\right)$ so that S is a k-algebra. Also, let F be a k-algebra and set $A=S{\otimes }_{K}F$ so that A is a k-algebra with identity $1_c{\otimes }_k{1}_F$. Here, $S{\otimes }_{k}k{1}_F$ and $k{1}_S{\otimes }_{k}F$ are k-subalgebras of $A=S{\otimes }_{k}kF$.

The k-linear map $\alpha :F\to k{1}_S{\otimes }_{k}F$ such that $f\mapsto 1_s{\otimes }_{k}f$ for all $f\in F$ is a k-algebra isomorphism.

Let $\{E_{ij}\mid 1\le i,j\le n\}$ be the “matrix k basis” of S and set $e=E_{11}$ so that

(1)

$e^2=e$, $eSe=ke$ and $SeS=S$.

Let $W=Se={\sum }_{j=1}^{n}k{E}_{j1}$ so that W is an S-mod-k module. Thus:

(2)

$e{\otimes }_s{1}_F$ is an idempotent of $A=S{\otimes }_{k}F$, $\left(S{\otimes }_{k}F\right)\left(e{\otimes }_k{1}_f\right)\left(S{\otimes }_{k}F\right)=S{\otimes }_{k}F$ and $\left(e{\otimes }_k{1}_E\right)\left(S{\otimes }_{k}F\right)\left(e{\otimes }_k{1}_F\right)=k{1}_S{\otimes }_{k}F$.

Set $B=k{1}_S{\otimes }_{k}F$.

Proposition 3. 

The functors $\left(S{\otimes }_{k}F\right){\otimes }_B\ast :B-mod\to S{\otimes }_{k}F-mod$ and $\left(e{\otimes }_k{1}_F\right)\left(S{\otimes }_{k}F\right){\otimes }_{S{\otimes }_{k}F}\ast :S{\otimes }_{k}F-mod\to B-mod$ comprise a Morita equivalence between the abelian categories B-mod and $S{\otimes }_{k}F$-mod. Here, if X is an $S{\otimes }_{e}F$-mod module, then $\left(e{\otimes }_k{1}_F\right)\left(S{\otimes }_{k}F\right){\otimes }_{S{\otimes }_{k}P}X\cong \left(e{\otimes }_k{1}_F\right)X$ in B-mod and $X\cong \left(S{\otimes }_{k}F\right){\otimes }_B\left(e{\otimes }_k{1}_F\right)X$ in $S{\otimes }_{k}F$-mod.

Clearly:

Lemma 13. 

(a)

The k-linear map $\tau :F\to B=ke{\otimes }_{k}F$ such that $f\mapsto e{\otimes }_{k}f$ for all $f\in F$ is a k-algebra isomorphism with ${\tau }^{-1}:B=ke{\otimes }_{k}F\to F$ such that $\alpha e{\otimes }_{k}f\mapsto \alpha f$ for all $\alpha \in k$ and $f\in F$; and

(b)

τ and ${\tau }^{-1}$ induce inverse functor isomorphisms $\mathcal{F}\left(\tau \right):B-mod\to F-mod$ and $\mathcal{F}\left({\tau }^{-1}\right):F-mod\to B-mod$.

Let $\mathrm{\Delta }$ be a set of k-subalgebras T of F with $1_F\in T$ and set $\mathrm{\Sigma }=\{ke{\otimes }_{k}U\mid U\in \mathrm{\Delta }\}$ so that $\mathrm{\Sigma }$ is a set of k-subalgebras W of B such that $e{\otimes }_k{1}_k\in W$ and $\tau :\mathrm{\Delta }\to \mathrm{\Sigma }$ is a bijection.

Let Z be an F-module so that $\mathcal{F}\left({\tau }^{-1}\right)Z$ is a B-module.

Theorem 4. 

Let $T\in \mathrm{\Delta }$ and let V be a T-module. Then, the following three conditions are equivalent:

(a)

$(T,V)$ is a Δ-vertex-source pair of Z in F-mod;

(b)

$(\tau \left(T\right),\mathcal{F}\left({\tau }^{-1}\right)V)$ is a Σ-vertex-source pair of $\mathcal{F}\left({\tau }^{-1}\right)Z$ in B-mod; and

(c)

$(\tau \left(T\right),\mathcal{F}\left({\tau }^{-1}\right)V)$ is a Σ-vertex-source pair of $\left(S{\otimes }_{k}F\right){\otimes }_B\mathcal{F}\left({\tau }^{-1}\right)Z$ in $A=S{\otimes }_{k}F$-mod.

Proof. 

Clearly, Lemma 7 implies that (a) and (b) are equivalent and Theorem 1 implies the equivalence of (b) and (c). The proof is complete. □

Corollary 3. 

Let $T\in \mathrm{\Delta }$ and let V be a T-module. Then:

(a)

$\alpha :F\to S{\otimes }_{k}F$ induces a bijection from $ITM(F,\mathrm{\Delta },(T,V\left)\right)$ onto $ITM(B,\mathrm{\Sigma },\tau \left(T\right),\mathcal{F}\left({\tau }^{-1}\right)V)$; and

(b)

the Morita equivalence of Proposition 3 induces a bijection from $ITM(B,\mathrm{\Sigma },\mathcal{F}\left({\tau }^{-1}\right)V)$ onto $ITM(S{\otimes }_{k}F,\mathrm{\Sigma },(\tau \left(T\right),\mathcal{F}\left({\tau }^{-1}\right),V))$.

Proposition 4. 

The results above prove and extend [6] (Proposition 2.6 (ii)).