Lemma 2

Suppose that B is a semisimple ring that is a finitely generated module over its center Z. Let M and N be $(B,A)$-bimodules. Suppose that they have finite length (which is, in particular, the case when they are right A-modules of finite length). If M and N are isomorphic as $(Z,A)$-bimodules, then they are isomorphic as $(B,A)$-bimodules.

A) First consider the case when B is the endomorphism ring of a vector space S of finite dimension $d$ over a commutative field L. We then have Z = L; we view S as a $(B$, Z)-bimodule. The ring B is simple, S is a simple B-module, and Z is the commutant of S; every B-module is isotypical of type S (VIII, p. 122, Proposition 2 a)). Let $(V, \alpha )$ (resp. $(W, \beta )$) be a description of the B-module M (resp. N). The set V (resp. W) is endowed with a $(Z$, A)-bimodule structure such that $\alpha$ (resp. $\beta$ ) is an isomorphism of $(B$, A)-bimodules (VIII, p. 64, Remark 2). As $(Z$, A)-bimodules, M is isomorphic to $V^d$ and N to $W^d$, and there exists an isomorphism from the set of $(Z$, A)-sub-bimodules of V, ordered by inclusion, to that of the $(B$, A)-sub-bimodules of M (loc. cit.). Hence V is a $(Z$, A)-bimodule of finite length, and so is W. Since the $(Z,A)$-bimodules $V^d$ and $W^d$ are isomorphic, the $(Z$, A)-bimodules V and W are isomorphic by Theorem 2, d) of VIII, p. 37 applied to the ring $Z\otimes_{\mathbf{Z}}A^o$. Finally, the $(B$, A)-bimodules M and N are isomorphic.

B) Now consider the case when B is a simple ring that is finitely generated as a Z-module. Then Z is a field, and B is a central simple algebra of finite degree over the field Z. By Theorem 1 of VIII, p. 252, there exists an extension $Z'$ of Z of finite degree over Z such that the $Z'$-algebra $B'= B_{(Z')}$ is isomorphic to the endomorphism algebra of a finite-dimensional $Z'$-vector space. Set $M'= M_{(Z')}$ and $N'= N_{(Z')}$. Then $M'$ and $N'$ are $(B'$, A)-bimodules of finite length; viewed as $(Z'$, A)-bimodules, $M'$ and $N'$ are isomorphic. By the case treated in $A), M'$ and $N'$ are isomorphic as $(B'$, A)-bimodules and a fortiori as $(B$, A)-bimodules. Set $r= [Z': Z]$. The $(B$, A)-bimodule $M'= Z'\otimes_ZM$ is isomorphic to $M^r$, and, likewise, the $(B$, A)-bimodule $N'$ is isomorphic to $N^r$. Since M and N are $(B$, A)-bimodules of finite length, it follows from Theorem 2, d) of VIII, p. 37 that the $(B$, A)-bimodules M and N are isomorphic.

C) Finally, consider the general case, when B is a semisimple ring that is finitely generated as a Z-module. Let $\mathscr{S}$ be the set of classes of simple B-modules; it is finite (VIII, p. 136, Proposition 1). For any $\lambda \in \mathscr{S}$, denote by $M_{\lambda}$ (resp. $N_{\lambda}$) the isotypical component of type $\lambda$ of the B-module M (resp. N); it is a $(B$, A)-sub-bimodule of M (resp. N) (Remark, VIII, p. 67). For $\lambda \in \mathscr{S}$, denote the annihilator of the B-module $\lambda$ by $\mathfrak{b}_{\lambda}$, and set $B_{\lambda}= B/\mathfrak{b}_{\lambda}$; let $Z_{\lambda}$ be the center of $B_{\lambda}$. For $\lambda \in \mathscr{S}$, the $(B_{\lambda}$, A)-bimodules $M_{\lambda}$ and $N_{\lambda}$ have finite length. We can then identify B with the product of the simple rings $B_{\lambda}$ and Z with the product of the $Z_{\lambda}$ (VIII, p. 141, Proposition 8). Moreover, we can identify M with $\prod_{\lambda\in\mathscr{S}}M_{\lambda}$ and N with $\prod_{\lambda\in\mathscr{S}}N_{\lambda}$. By assumption, M and N are isomorphic as $(Z$, A)-bimodules; it follows that for $\lambda \in \mathscr{S}, M_{\lambda}$ and $N_{\lambda}$ are isomorphic $(Z_{\lambda}$, A)-bimodules. By the case treated in B), the $(B_{\lambda}$, A)-bimodules $M_{\lambda}$ and $N_{\lambda}$ are isomorphic, and so the $(B$, A)-bimodules M and N are isomorphic.