Remark 2

Let K be a field, $F_1, \ldots, F_m$ polynomials in $K[X_1, \ldots, X_n]$ and L an extension of K generated by the elements $x_1, \ldots, x_n$. Suppose that the polynomials $F_1, \ldots, F_m$ generate the ideal of $K[X_1, \ldots, X_n]$ consisting of all polynomials F such that $F(x_1, \ldots, x_n) = 0$. From Prop. 1 (V, p. 127) and the universal property of the module of differentials (III, p. 569) we easily deduce the following result: the vector space $\Omega_K(L)$ over L is generated by $dx_1, \ldots, dx_n$; we have the relations

finally if $u_1, \ldots, u_n$ are elements of L such that $\sum_{i=1}^n u_i \cdot dx_i = 0$, there exist elements $v_1, \ldots, v_m$ of L such that $u_i = \sum_{j=1}^m D_i F_j(x_1, \ldots, x_n) \ v_j$ for $1 \leq i \leq n$. Let us denote by r the rank of the matrix $(D_i F_j(x_1, \ldots, x_n))$ with n rows and m columns; let s be the transcendence degree of $L$ over $K$. Then we have $[ \Omega_K(L) : L ] = n - r$. Therefore the extension $L$ of $K$ is separable if and only if $r + s = n$ (Cor. 1). and it is algebraic and separable if and only if $r = n$ (Cor. 2).