Theorem 1

Let M be a left A-module. The following properties are equivalent:

(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.