a) The mapping $(u_i)\mapsto \prod u_i$ from $\prod$ End$_{\mathbf{Z}}(M_i)$ to End$_{\mathbf{Z}}(M)$ (II, §1, No. 5, p. 200) restricts to ring isomorphisms from $\prod(A_i)_{M_i},\prod(A_i)'_{M_i}$, and $\prod(A_i)''_{M_i}$ to $A_M,A'_M$, and $A''_M$, respectively.

b) The mapping $(u_i)\mapsto \bigoplus u_i$ from $\prod$ End$_{\mathbf{Z}}(M_i)$ to End$_{\mathbf{Z}}(N)$ (II, §1, No. 6, p. 203) restricts to ring isomorphisms from $\prod(A_i)_{M_i},\prod(A_i)'_{M_i}$, and $\prod(A_i)''_{M_i}$ to $A_N,A'_N$, and $A''_N$, respectively.

The ring isomorphisms defined in Proposition 1 are called canonical. The product $\prod(A_i)_{M_i}$ is often identified with $A_M$, and $\prod(A_i)'_{M_i}$ with $A'_M$, etc., through these isomorphisms.

The mapping $\varphi : (u_i)\mapsto \prod u_i$ from $\prod$ End$_{\mathbf{Z}}(M_i)$ to End$_{\mathbf{Z}}(M)$ is an injective ring homomorphism. By the definition of the A-module structure on M, we have $\varphi (\prod(A_i)_{M_i}) = A_M$. Let $u\in A'_M$. For every $i\in I$, denote by $h_i$ the element of A with all components equal to 1 except that of index $i$, which is equal to 0 . If $x$ is an element of M with component of index $i$ equal to 0, then we have $x=h_ix$, and therefore pr$_i(u(x)) =$ pr$_i(u(h_ix)) =$ pr$_i(h_iu(x)) = 0$. Consequently, there exists a unique group homomorphism $u_i: M_i\mapsto M_i$ such that pr$_i(u(y)) =u_i$(pr$_i(y)$) for every $y\in M$. We have $u=\prod u_i$. Since the mapping $u$ is A-linear, the mapping $u_i$ is $A_i$-linear for every $i\in I$. This proves that the image of $\prod(A_i)'_{M_i}$ under $\varphi$ contains $A'_M$; the reverse inclusion is obvious. By applying this to the countermodule of M, we deduce that $\varphi$ induces an isomorphism from $A''_M$ to $\prod_i(A_i)''_{M_i}$. This proves assertion a).

The proof of b) is the same as that of a) mutatis mutandis.

For $i\in I$, we denote the projection homomorphism from $M^{(I)}$ to M by $\pi_i: (x_j)_{j\in I}\mapsto x_i$ and the corresponding canonical injection (I, §4, No. 9, p. 47) by $\iota_i: M\rightarrow M^{(I)}$.

Let $u$ be an element of End$_A(M^{(I)})$. For all $i, j\in I$, the composition $u_{i,j}=\pi_j\circ u\circ \iota_i$ belongs to the commutant $A'_M$ of M. For every element $b$ of $A''_M$ and every $(x_i)\in M^{(I)}$, we have the relations

The homothety $b_{M^{(I)}}$ therefore belongs to the bicommutant of the A-module $M^{(I)}$.

Conversely, let $b$ be an element of the bicommutant of $M^{(I)}$. For $i, j\in I$, set $b_{i,j}=\pi_j\circ b\circ \iota_i$. Take $i, j\in I$ with $i\not=j$. Since $\iota_j\circ \pi_j$ belongs to the commutant of the A-module $M^{(I)}$, we have

Likewise, we have

Moreover, $b_{i,i}$ belongs to $A''_M$. It follows that $b$ coincides with a homothety of the $A''_M$-module $M^{(I)}$.

(i) The A-module M is generating.

(ii) For every A-module N, there exist a set I and a surjective A-linear mapping from $M^{(I)}$ to N.

(iii) There exist an integer $n\geqslant 0$ and a surjective A-linear mapping from $M^n$ to $A_s$.

(iv) There exists an integer $n\geqslant 0$ such that $A_s$ is isomorphic to a direct factor submodule of $M^n$.

(v) The trace ideal $\tau (M)$ is equal to A.

(vi) There exist an integer $n\geqslant 0$, elements $x_1, . . . , x_n$ of M, and elements $x^*_1, . . . , x^*_n$ of $M^*$ satisfying $\sum^n_{i=1}\langle x_i, x^*_i\rangle = 1$.

(i) $\Rightarrow$ (ii): Suppose that M is generating, and let N be a left A-module. There exists a family $(u_i)_{i\in I}$ of A-linear mappings from M to N such that we have $N =\sum_{i\in I}u_i(M)$. The mapping $(x_i)\mapsto \sum_{i\in I}u_i(x_i)$ from $M^{(I)}$ to N is A-linear and surjective.

(ii) $\Rightarrow$ (iii): Assume that property (ii) holds, and take $N = A_s$. We then have a set I and a surjective linear mapping $u: M^{(I)}\rightarrow A_s$. Since $A_s$ is generated by the element 1, there exists a finite subset J of I such that $u(M^{(J)}) = A_s$, whence (iii).

(iii) $\Rightarrow$ (iv): This follows from Proposition 21 of II, §1, No. 12, p. 218.

(iv) $\Rightarrow$ (v): Let $n\geqslant 1$ be an integer such that $A_s$ is isomorphic to a direct factor submodule of M. We have $A =\tau (A_s)\subset \tau (M^n) =\tau (M)$ and therefore $\tau (M) = A$.

(v) $\Rightarrow$ (vi): This is clear.

(vi) $\Rightarrow$ (i): Let $n$ be an integer $\geqslant 0,x_1, . . . , x_n$ elements of M, and $x^*_1, . . . , x^*_n$ elements of $M^*$ satisfying $\sum^n_{i=1}\langle x_i, x^*_i\rangle = 1$. Let N be a left A-module and $y$ an element of N. The mappings $u_i:x\mapsto \langle x, x^*_i\rangle y$ from M to N are A-linear, and we have $y=\sum^n_{i=1}u_i(x_i)$. This proves that M is a generating A-module.

Let $a\in A$ satisfy $aM = 0$. Using the implication (i) $\Rightarrow$ (iv) of Theorem 1, we obtain $aA_s= 0$, and therefore $a= 0$. The corollary follows (II, §1, No. 12, p. 219).

(i) $\mathfrak{a}M = M$.

(ii) Ann(M) $+\mathfrak{a}= A$.

(iii) There exists an element $a$ of $\mathfrak{a}$ such that $am=m$ for every $m\in M$.

(i) $\Rightarrow$ (ii): Let $(x_1, . . . , x_n)$ be a generating family of the A-module M. If $\mathfrak{a}M = M$, then each $x_i$ can be written as $\sum^n_{j=1}c_{ij}x_j$, where the $c_{ij}$ belong to $\mathfrak{a}$. Denote the matrix $(c_{ij})$ by C and the column matrix with entries $x_1, . . . , x_n$ by X. We have ($I_n-C$)X = 0. Let $d$ be the determinant and V the cofactor matrix of the matrix $I_n-C$. By formula (26) of III, §8, No. 6, p. 532, we have $dX=^tV$($I_n-C$)X = 0, so that $d\in$ Ann(M). On the other hand, since the $c_{ij}$ belong to $\mathfrak{a}$, we have $d\equiv$ 1(mod $\mathfrak{a}$) (III, §8, No. 5, p. 529, (18)), and therefore $1\in$ Ann(M) $+\mathfrak{a}$.

(ii) $\Rightarrow$ (iii): Assuming that (ii) holds, there exist $a\in \mathfrak{a}$ and $b\in$ Ann(M) such that $a+b= 1$. We then have $am=m$ for every $m\in M$.

(iii) $\Rightarrow$ (i): This is clear.

Let P be a finitely generated projective A-module. We have $\tau (P)P = P$ (VIII, p. 80).

If the A-module P is faithful, then the ideal $\tau (P)$ is equal to A (Lemma 1) and the A-module P is generating (Theorem 1); this leads to the first assertion. The second follows by Lemma 2 below.

Let $(x_i)_{i\in I}$ be a generating family of the A-module M. There exists a family $(x^*_i)_{i\in I}$ of linear forms on the A-module M such that for every $x\in M$, the family $(\langle x, x^*_i\rangle )_{i\in I}$ has finite support and $x=\sum_{i\in I}\langle x, x^*_i\rangle x_i$ (II, §2, No. 6, p. 238, Proposition 12). For every $i\in I$, the mapping $x\mapsto \langle x, x^*_i\rangle_M$ is a linear form on the $A_M$-module M, and we have $x=\sum_{i\in I}\langle x, x^*_i\rangle_Mx_i$ for every $x\in M$. By loc. cit., M is a projective $A_M$-module.

Let A be a ring, and let M be a generating A-module; by definition, there exist an integer $n\geqslant 0$, elements $x_1, . . . , x_n$ of M, and elements $x^*_1, . . . , x^*_n$ of the dual $M^*$ of M satisfying $\sum^n_{i=1}\langle x_i, x^*_i\rangle = 1$. Recall (II, §4, No. 2, p. 271) that we define a group homomorphism $\theta : M^*\otimes_AM\rightarrow$ End$_A(M)$ by the formula $\theta (x^*\otimes y)(x) =\langle x, x^*\rangle y$. If $u$ is an element of the bicommutant of M, then it commutes with End$_A(M)$; therefore, for every $y\in M$, we have

Consequently, $u$ belongs to $A_M$, and M is balanced.

This is clear if the module is zero, and a nonzero free module is generating (VIII, p. 81, Example 2).

Let M be the right A-module $A^n_d$. It is balanced by Corollary 1. Consequently, the centers of $A_M, A'_M$, and $A''_M$ coincide. Corollary 2 then follows from the fact that $A_M$ is identified with A and $A'_M$ with $\mathbf{M}_n(A)$.

Indeed, a finitely generated projective module M is a generating $A_M$-module by Proposition 3 of VIII, p. 82. The corollary therefore follows from the remark above.

Indeed, a finitely generated module M over a principal ideal domain A is a generating $A_M$-module (VIII, p. 81, Example 4).

(i) There exists a polynomial P in K[X] such that $v= P(u)$.

(ii) The endomorphism $v$ commutes with every endomorphism of V that commutes with $u$.

Take A to be the ring K[X] and M to be the K[X]-module deduced from V and $u$ (VII, §5, No. 1, p. 28). Assertion (i) means that $v\in K[X]_M$ and (ii) that $v\in K[X]''_M$. Corollary 5 is therefore a particular case of Corollary 4.

This follows from Example 3 of VIII, p. 81 and the remark.

Consider the A-module $M =\prod_{i\in I}S_i$. Since I is finite, we also have $M =\bigoplus_{i\in I}S_i$, and the image of $S_i$ in M is the isotypical component of M of type $S_i$. Consequently, the endomorphisms of the A-module M are the mappings $(s_i)\mapsto (s_id_i)$, where $(d_i)_{i\in I}$ runs through $\prod_{i\in I}D_i$ (VIII, p. 66, Proposition 5). Since I is finite and for every $i\in I$, the right $D_i$-vector space $S_i$ is finite-dimensional, the countermodule of M is finitely generated. By Proposition 4, the A-module M is balanced. Now, the bicommutant of the A-module M consists of elements of End$_{\mathbf{Z}}(M)$ of the form $\prod_{i\in I}u_i$, where $(u_i)\in \prod_{i\in I}$ End$_{D_i}(S_i)$ (VIII, p. 78, Proposition 1) because End$_{D_i}(S_i)$ is the bicommutant of $S_i$. The corollary follows from this.

This corollary applies, in particular, when A is an algebra over a commutative field K and each $S_i$ is a simple A-module that is finite-dimensional as a K-vector space: indeed, $D_i$ then contains the homotheties $\alpha_{S_i}$, where $\alpha$ runs through K, and $S_i$ is finite-dimensional over $D_i$ because it is so over K.

Indeed, the field of endomorphisms of the A-module S consists of the homotheties $\alpha_S$ with $\alpha \in K$ (VIII, p. 47, Theorem 1). We then apply Corollary 1 to the simple A-module S.

b) For every $\lambda \in \mathscr{S}_M$, the B-module $V_{\lambda}$ is simple, and its commutant is equal to $(D_{\lambda})_{V_{\lambda}}$.

c) The mapping $\lambda \mapsto$ cl(V$_{\lambda}$) is a bijection from the support of the A-module M to the support of its countermodule.

d) For every $\lambda \in \mathscr{S}_M$, the B-submodule $M_{\lambda}$ is the isotypical component of type $V_{\lambda}$ of the B-module M, and the multiplicity of $V_{\lambda}$ in M is equal to dim$_{D_{\lambda}}(S_{\lambda})$.

e) For $s\in S$, denote the mapping $\varphi \mapsto \varphi (s)$ from $V_{\lambda}=$ Hom$_A(S_{\lambda},M)$ to M by $\widetilde{s}$. It is B-linear. The resulting mapping $s\mapsto \widetilde{s}$ from $S_{\lambda}$ to Hom$_B(V_{\lambda},M)$ is an isomorphism of $(A,D_{\lambda})$-bimodules.

Let $\lambda \in \mathscr{S}_M$. Denote the ring End$_{D_{\lambda}}(V_{\lambda})$ by $E_{\lambda}$; since $V_{\lambda}$ is a nonzero $D_{\lambda}$-vector space, it is a simple $E_{\lambda}$-module, (VIII, p. 45, Example 3), and its commutant is equal to $(D_{\lambda})_{V_{\lambda}}$ (VIII, p. 82, Corollary 1 of Theorem 2). Since $E_{\lambda}$ is the ring of homotheties of the B-module $V_{\lambda}$ (VIII, p. 71, Corollary of Proposition 9), this proves b).

The canonical description $\alpha_M$ of M defines an isomorphism $\alpha_{\lambda}$ from $V_{\lambda}\otimes_{D^o_{\lambda}}S_{\lambda}$ to $M_{\lambda}$. Since $V_{\lambda}$ is a simple B-module, the B-module $V_{\lambda}\otimes_{D^o_{\lambda}}S_{\lambda}$ is isotypical of type $V_{\lambda}$ (VIII, p. 61, Proposition 1); the same therefore holds for the B-module $M_{\lambda}$, which proves a).

By Remark 1 above, there exist elements $e_{\lambda}$ of B, for $\lambda$ running through $\mathscr{S}_M$, such that $(e_{\lambda})_{V_{\lambda}}= 1_{V_{\lambda}}$ and $(e_{\lambda})_{V_\mu}$ = 0 for $\mu\in \mathscr{S}_M$ with $\mu\not=\lambda$. The simple B-modules $V_{\lambda}$ are therefore pairwise nonisomorphic, which proves c) and the first assertion of d). The B-module $M_{\lambda}$ is isomorphic to $V_{\lambda}\otimes_{D^o_{\lambda}}S_{\lambda}$, so dim$_{D_{\lambda}}(S_{\lambda})$ is the multiplicity of $V_{\lambda}$ in M (II, §3, No. 7, p. 255, Corollary 1).

The mapping $\sum_{\lambda\in\mathscr{S}_M}\alpha_{\lambda}$ from $\bigoplus_{\lambda\in\mathscr{S}_M}V_{\lambda}\otimes_{D^o_{\lambda}}S_{\lambda}$ to M provides a description (VIII, p. 69, Definition 5) of the semisimple B-module M. By VIII, p. 70, Proposition 8, b), for every $\lambda \in \mathscr{S}_M$, the mapping from $S_{\lambda}$ to Hom$_B(V_{\lambda},M)$ described in e) is bijective and $D_{\lambda}$-linear. Since it is obviously A-linear, this proves e).

b) For every $\lambda \in \mathscr{S}_M,M_{\lambda}$ is a simple $(A,B^o)$-sub-bimodule of M.

c) For every $(A,B^o)$-sub-bimodule N of M, there exists a unique subset Λ of $\mathscr{S}_M$ such that N is equal to $\oplus_{\lambda\in\Lambda}M_{\lambda}$.

Let $\lambda$ be in $\mathscr{S}_M$. The left A-module $S_{\lambda}$ and the right B-module $V_{\lambda}$ are simple, and $D_{\lambda}$ is the opposite ring of the field of endomorphisms of $S_{\lambda}$. By Corollary 2 of VIII, p. 63, the $(A,B^o$)-bimodule $S_{\lambda}\otimes_{D_{\lambda}}V_{\lambda}$ is simple, and the same holds for $M_{\lambda}$, which is isomorphic to it. This proves b), and a) follows.

If $\lambda$ and $\mu$ are distinct in $\mathscr{S}_M$, then $M_{\lambda}$ and $M_\mu$ are not isomorphic as A-modules, nor a fortiori as $(A,B^o$)-bimodules. Assertion c) follows by Corollary 2 of VIII, p. 68.

b) Endow the $S_{\lambda}$, for $\lambda$ running through $\mathscr{S}_M$, with the C-module structure defined by a). Then the canonical mapping $\alpha_M$ from $\bigoplus_{\lambda\in\mathscr{S}_M}S_{\lambda}\otimes_{D_{\lambda}}V_{\lambda}$ to M is an isomorphism of $(C,B^o)$-bimodules.

c) The mapping $c\mapsto (c_{\lambda})_{\lambda\in\mathscr{S}_M}$ is an isomorphism from C to $\prod_{\lambda\in\mathscr{S}_M}$ End$_{D_{\lambda}}(S_{\lambda})$.

Assertions a) and c) follow from VIII, p. 71, Proposition 9 because the canonical mapping $\alpha_M$ from $\bigoplus_{\lambda\in\mathscr{S}}(S_{\lambda}\otimes_{D_{\lambda}}W_{\lambda})$ to M provides a description of the B-module M (Remark 2). Moreover, $\alpha_M$ is $(C,B^o$)-linear, which proves b).

b) The A-module M is isotypical and nonzero if and only if Z is a field.

c) Let Λ be a subset of $\mathscr{S}_M$. Denote by $e_{\Lambda}$ the unique element of Z such that $(e_{\Lambda})_{S_{\lambda}}= 1_{S_{\lambda}}$ for $\lambda \in \Lambda$ and $(e_{\Lambda})_{S_{\lambda}}= 0$ for $\lambda \in \mathscr{S}_M-\Lambda$. We have $(e_{\Lambda})_{M_{\lambda}}= 1_{M_{\lambda}}$ for $\lambda \in \Lambda$ and $(e_{\Lambda})_{M_{\lambda}}= 0$ for $\lambda \in \mathscr{S}_M-\Lambda$.

d) If the support $\mathscr{S}_M$ of M is finite, then the mapping $\Lambda \mapsto e_{\Lambda}Z$ is a bijection from the set of subsets of $\mathscr{S}_M$ to the set of ideals of Z, and the mapping $\mathfrak{a}\mapsto \mathfrak{a}M$ is a bijection from the set of ideals of Z to the set of $(A,B^o)$-sub-bimodules of M. These bijections are isomorphisms of ordered sets. The reverse bijection sends an $(A,B^o)$-sub-bimodule N of M to the ideal consisting of the elements $z$ of Z that send N into M.

For $\lambda \in \mathscr{S}_M, Z_{\lambda}$ is the common center of the commutant $D_{\lambda}$ and the bicommutant $C_{\lambda}$ of the A-module $S_{\lambda}$. By Proposition 7, c) above, the mapping $c\mapsto (c_{\lambda})_{\lambda\in\mathscr{S}_M}$ is an isomorphism from C to $\prod_{\lambda\in\mathscr{S}_M}C_{\lambda}$. By restriction to the centers, we obtain the isomorphism $z\mapsto (z_{S_{\lambda}})_{\lambda\in\mathscr{S}_M}$ from Z to $\prod_{\lambda\in\mathscr{S}_M}Z_{\lambda}$, whence a).

The ring $\prod_{\lambda\in\mathscr{S}_M}Z_{\lambda}$ is a field if and only if the set $\mathscr{S}_M$ has a single element, whence b).

Assertion c) follows from Proposition 7, a). Suppose that $\mathscr{S}_M$ is finite. It follows from a) and Proposition 8 of I, §8, No. 10, p. 109, that the mapping $\Lambda \mapsto e_{\Lambda}Z$ is an isomorphism of ordered sets from $\mathfrak{P}(\mathscr{S}_M)$ to the set of ideals of Z. Let Λ be a subset of $\mathscr{S}_M$. By c), we have the relation $e_{\Lambda}$ZM $=e_{\Lambda}M =$ $\bigoplus_{\lambda\in\Lambda}M_{\lambda}$; because of Proposition 6, c) of VIII, p. 86, it remains to describe the inverse bijection. But $z\in Z$ sends M into $\bigoplus_{\lambda\in\Lambda}M_{\lambda}$ if and only if $z=e_{\Lambda}z$, that is, $z\in e_{\Lambda}Z$.

Since M is a finite-dimensional vector space over K that is the direct sum of the family of nonzero submodules $(M_{\lambda})_{\lambda\in\mathscr{S}_M}$, the set $\mathscr{S}_M$ is finite, and each of the spaces $S_{\lambda}$, for $\lambda \in \mathscr{S}_M$, is finite-dimensional over K. Since the field K is algebraically closed, we have $D_{\lambda}= Z_{\lambda}= K$ (VIII, p. 47, Theorem 1), and the mapping $z\mapsto (z_{S_{\lambda}})_{\lambda\in\mathscr{S}_M}$ is an isomorphism from Z to $K^{\mathscr{S}_M}$ (Proposition 8, a)). The corollary then follows from part c) of Proposition 8.

(D) For every finite subset F of M, there exists an element $a$ of A such that $c$ coincides with $a_M$ on F.

First, suppose that $c$ satisfies condition (D). Let $u$ be an element of $A'_M$. Let $x$ be an element of M, and apply condition (D) to the subset $F =\{x, u(x)\}$. There exists an element $a$ of A such that $c(x) =ax$ and $c(u(x)) =au(x)$, so that $u(c(x)) =u(ax) =au(x) =c(u(x))$. Since $x$ is arbitrary, we have $cu=uc$; as this holds for all $u$, we have $c\in A''_M$.

For the converse, we use the following lemma.

Let N be an A-submodule of M. By Corollary 2 of VIII, p. 56, there exists a projector $p$ of the A-module M with image N; it belongs to $A'_M$. Since we have the relation $pb=bp$ for every $b\in B$, we obtain that N is a B-submodule of M.

Let us conclude the proof of Theorem 3. Suppose that $c$ belongs to $A''_M$, and let F = $\{x_1, . . . , x_n\}$ be a finite subset of M. Denote the element $(x_1, . . . , x_n)$ of $M^n$ by $x$. The A-module $M^n$ is semisimple and, by Proposition 2 of VIII, p. 79, its bicommutant coincides with the homotheties of the $A''_M$-module $M^n$. By Lemma 3, the A-submodule $Ax$ of $M^n$ is an $A''_M$-submodule of $M^n$. Let $a\in A$ be such that $(cx_1, . . . , cx_n)$ is equal to $ax$. Then $c$ coincides with $a_M$ on $\{x_1, . . . , x_n\}$, which implies condition (D).