Theorem 5

*Let $K$ be a field, $L$ an extension of $K$ and $(x_i)_{i \in I}$, a family of elements of $L$. If the family $(x_i)_{i \in I}$, is a separating transcendence basis of $L$ over $K$, then the family $(dx_i)_{i \in I}$, is a basis of the vector space $\Omega_K(L)$ over $L$. The converse holds if $L$ is a finitely generated separable extension of $K$.

Put $M = K(x_i)_{i \in I}$, and denote by $\alpha$ the canonical mapping of $\Omega_K(M) \otimes_M L$ into $\Omega_K(L)$. If $(x_i)_{i \in I}$ is a separating transcendence basis of $L$ over $K$, the family $(d_{M/K} x_i)_{i \in I}$, is a basis of the vector $M$-space $\Omega_K(M)$ ($V$, p. 128, Prop. 3) and $\alpha$ is an isomorphism of vector $L$-spaces since $L$ is algebraic and separable over $M$ ($V$, p. 130, Cor. 2). Since $\alpha(d_{M/K} x_i \otimes 1) = d_{L/K} x_i$, the family $(d_{L/K} x_i)_{i \in I}$ is thus a basis of $\Omega_K(L)$ over $L$.

Conversely, suppose that $L$ is a separable finitely generated extension of $K$ and that the family $(d_{L/K} x_i)_{i \in I}$ is a basis of the vector space $\Omega_K(L)$ over $L$. By Cor. 1 of $V$, p. 134, the transcendence degree of $L$ over $K$ is equal to the dimension of $\Omega_K(L)$ over $L$, hence to the cardinal of $I$. From the exact sequence $(E_{K,M,L})$ ($V$, p. 128) we have $\Omega_M(L) = 0$; since $L$ is a finitely generated extension of $M$, Cor. 2 of $V$, p. 135 shows that $L$ is algebraic and separable over $M = K(x_i)_{i \in I}$; since the transcendence degree of $L$ over $K$ is finite and equal to the cardinal of $I$, the family $(x_i)_{i \in I}$ is a transcendence basis of $L$ over $K$ ($V$, p. 110, Cor. 1).