Theorem 2

— *Let K be a field of characteristic 0, L an extension of K and $(x_i)_{i \in I}$, a family of elements of L. a) For $(x_i)_{i \in I}$, to be algebraically free over K it is necessary and sufficient that the differentials $dx_i$ (for $i \in I$) in $\Omega_K(L)$ should be linearly free over L. b) For L to be algebraic over $K(x_i)_{i \in I}$ it is necessary and sufficient that the differentials $dx_i$, for $i \in I$ generate the vector space $\Omega_K(L)$ over L. c) For $(x_i)_{i \in I}$ to be a transcendence basis of L over K, it is necessary and sufficient that the family $(dx_i)_{i \in I}$ should be a basis of $\Omega_K(L)$ over L. For every $i \in I$ let $E_i$ be the subfield $K(x_j)_{j \in I, j \neq i}$ of L. For the family $(x_i)_{i \in I}$, to be algebraically free over K it is necessary and sufficient ($V$, p. 108, Prop. 5) that $x_i$ should be transcendental over $E_i$ for each $i \in I$. By Cor. 2 of Th. 1 this means that for each $i \in I$ the differential $dx_i$ is not a linear combination with coefficients in L of the differentials $dx_j$ with $j \neq i$ in I; this latter condition just means that the family $(dx_i)_{i \in I}$ is free over L, whence a). Assertion b) follows at once from Cor. 2 of Th. 1 and c) is a consequence of a) and b).