Lemma 3

Let k be a field and K a finite-type extension of k. There exist an extension L of K, of finite degree, and a subextension $k'$ of L over k, radicial and of finite degree over k, such that the extension L of $k'$ is separable.

Let us choose a composite extension E of K and of a perfect closure $\hat{k}$ of k, and elements $(t_1, \ldots, t_n)$ of E such that E is a separable algebraic extension of $\hat{k}(t_1, \ldots, t_n)$ (A, V, p. 130, cor.). Let I be a finite generating subset of K over k. The field $\hat{k}(t_1, \ldots, t_n)$ (resp. $\hat{k}K$) is the union of the subfields $k'(t_1, \ldots, t_n)$ (resp. $k'K$), where $k'$ runs through the set of subextensions of $\hat{k}$ of finite degree over k; hence one can find such a subextension $k'$ such that the elements $t_1, \ldots, t_n$ of E = $\hat{k}K$ belong to $k'K$ and that each element of I is algebraic and separable over $k'(t_1, \ldots, t_n)$. Then L = $k'K$ is a separable extension of $k'$, of finite degree over K, and $k'$ is a finite radicial extension of k.