Remark

By Prop. 3 ($V$, p. 107) it is enough to verify the condition (AF) for finite subsets $A_i$. We thus obtain the following result : if $(E_i)_{i \in I}$ is an algebraically free family, the same is true of $(E'_i)_{i \in I}$ if $E'_i$ is a subextension of $E_i$ for each $i \in I$; conversely, if every family $(E'_i)_{i \in I}$, where $E'_i$ is a finitely generated subextension of $E$ for each $i \in I$ is algebraically free, then $(E_i)_{i \in I}$ is algebraically free. On the other hand, for $(E_i)_{i \in I}$ to be algebraically free it is necessary and sufficient that $(E_i)_{i \in J}$ should be algebraically free for every finite subset $J$ of $I$. Speaking intuitively we may say that the algebraic independence of extensions is a property « of finite character ».