Lemma 3

A right B-module P is projective, finitely generated, and generating if and only if there exist an integer $n\geqslant 0$ and an idempotent $e= (e_{ij})$ in $\mathbf{M}_n(B)$ with the following properties:

(i) The B-module P is isomorphic to $eB^n_d$.

(ii) The two-sided ideal of B generated by the elements $e_{ij}$ is equal to B.

Let P be a right B-module. Then P is projective and finitely generated if and only if it is isomorphic to a direct factor submodule of a B-module of the form $B^n_d$, where $n$ is an integer $\geqslant 0$ (II, §2, No. 2, p. 232, Corollary 1). If we identify the $k$-algebras $\mathbf{M}_n(B)$ and End(B$^n_d$), then this means that there exists an idempotent $e$ in $\mathbf{M}_n(B)$ such that P is isomorphic to $eB^n_d$.

The B-module P is generating if and only if its trace ideal $\tau (P)$ is equal to B, that is, $\tau (eB^n_d) = B$ (VIII, p. 80, Theorem 1). Let $x_1, . . . , x_n$ be the elements of $B^n_d$ corresponding to the columns of the matrix $e$, and let $x^*_i$ (for $1\leqslant i\leqslant n$) be the linear form $(b_1, . . . , b_n)\mapsto b_i$ on $eB^n_d$. The family $(x_1, . . . , x_n)$ generates the B-module $eB^n_d$, and the family $(x^*_1, . . . , x^*_n)$ generates its dual. Now, we have $\langle x^*_i, x_j\rangle =e_{ij}$, so $\tau (eB^n_d)$ is the two-sided ideal of B generated by the $e_{ij}$. This proves Lemma 3.