Definition 1

The A-module M is balanced if we have $A_M= A''_M$.

If the A-module M is balanced, then the rings $A_M$ and $A'_M$ have the same center $A_M\cap A'_M$. The module M is a balanced A-module if and only if it is a balanced $A_M$-module. The countermodule of an A-module M is faithful and balanced.

When the ring A is commutative, the bicommutant $A''_M$ of M is the center of $A'_M=$ End$_A(M)$; that M is balanced means that the center of End$_A(M)$ is reduced to the homotheties.

For any element $a$ of A, denote by $\boldsymbol{\delta }_a$ the right homothety $x\mapsto xa$ from A to A and by $\boldsymbol{\gamma }_a$ the left homothety $x\mapsto ax$ (I, §8, No. 1, p. 97). The map $a\mapsto \boldsymbol{\delta }_a$ is a ring isomorphism from $A^o$ to the commutant of the left A-module $A_s$ (cf. II, §10, No. 7, p. 349, applied to $I =\{1\}$). The map $a\mapsto \boldsymbol{\gamma }_a$ is a ring isomorphism from A to the commutant of the right A-module $A_d$ (loc. cit.). If we identify A with the commutant of $A_d$ through this map, then the countermodule of $A_d$ is identified with $A_s$; consequently, the A-module $A_d$ is balanced. Likewise, the A-module $A_s$ is balanced.

Let $n$ be an integer $\geqslant 1$. We view $A^n$ as a left $\mathbf{M}_n$(A)-module (loc. cit.). The mapping that sends $m\in \mathbf{M}_n(A)$ to the endomorphism $x\mapsto mx$ of the A-module $A^n_d$ is a ring isomorphism from $\mathbf{M}_n(A)$ to the commutant of the right A-module $A^n_d$ (loc. cit.).