Proposition 3

Let A be a simple ring. An A-module M is free if and only if its length is divisible by the length of A. If that is the case, then all bases of M have the same cardinal, denoted by dim$_A(M)$ (II, §7, No. 3, p. 294, Remark 2) and determined by the relation

(1) long$_A(M) =$ long(A) $\cdot$ dim$_A(M)$.

Suppose that M is free, and let $(e_i)_{i\in I}$ be a basis of M. The A-module M is the direct sum of the A-modules $Ae_i$, which are each isomorphic to $A_s$. Set $r=$ long$_A(A_s)$; this is an integer greater than or equal to 1 (VIII, p. 119, Proposition 1). We have long$_A(M) =r$ Card(I) by formula (13) of VIII, p. 73.

Conversely, suppose that the cardinal long$_A(M)$ is divisible by $r$. Let $\mathfrak{a}$ be the cardinal such that long$_A(M) =r\mathfrak{a}$. Then the A-module M has the same length as $A^{(\mathfrak{a})}_s$, hence is isomorphic to it by Proposition 2. This proves that M is free.