Remark

(5) Let $(a_i)_{1 \leq i \leq n}$ be a basis of an A-module E and let $(b_i)_{1 \leq i \leq n}$ be a family of elements of E given by the relations

where $\lambda_{ii}$ is invertible in A; then $(b_i)_{1 \leq i \leq n}$ is a basis of E. It suffices to argue by induction on $n$, the proposition being obvious for $n = 1$. If E' is the submodule of E generated by the family $(a_i)_{1 \leq i \leq n-1}$, it follows from the induction hypothesis that $(b_i)_{1 \leq i \leq n-1}$ is a basis of E'; on the other hand, it follows from (46) that if $\mu b_n \in E'$ with $\mu \in A$, then also $\mu \lambda_{nn} a_n \in E'$, whence $\mu = 0$ since $\lambda_{nn}$ is invertible. The family $(b_i)_{1 \leq i \leq n}$ is thus free and, as

it is seen that $(b_i)_{1 \leq i \leq n}$ is a generating system of E, which completes the proof. This result is easily generalized to a family $(a_i)_{i \in I}$ whose indexing set I is well ordered.