Proposition 1

Let $k_0$ be a field, A a $k_0$-algebra which is a separated and complete local ring, K a sub-$k_0$-extension of $\kappa_A$ which possesses a separating transcendence basis $(\xi_{\lambda})_{\lambda \in \Lambda}$ over $k_0$ (A, V, p. 130, déf. 1). For every $\lambda \in \Lambda$, let $x_{\lambda}$ be a representative of $\xi_{\lambda}$ in A. There exists a unique subfield L of A, containing $k_0$ and the elements $x_{\lambda}$, and such that the canonical homomorphism $\pi$ from A onto $\kappa_A$ induces an isomorphism of L onto K.

Let $\varphi$ be the $k_0$-homomorphism of the polynomial ring $k_0[(X_{\lambda})_{\lambda \in \Lambda}]$ into A which maps $X_{\lambda}$ onto $x_{\lambda}$ for every $\lambda \in \Lambda$. Let $u$ be a nonzero element of $k_0[(X_{\lambda})_{\lambda \in \Lambda}]$; one has $\pi(\varphi(u)) \neq 0$, because the family $(\xi_{\lambda})_{\lambda \in \Lambda}$ is algebraically free over $k_0$ in $\kappa_A$; consequently, $\varphi(u)$ is invertible in the local ring A. It follows that $\varphi$ extends to a homomorphism $\psi$ of the field $k_1 = k_0((X_{\lambda})_{\lambda \in \Lambda})$ into A. Then A is a $k_1$-algebra, $\kappa_A$ is an extension of $k_1$ and K a sub-extension of $\kappa_A$ which is algebraic and separable over $k_1$. It remains to prove that there exists a unique subfield L of A containing $\psi(k_1)$ and such that $\pi(L) = K$.

a) Existence of L: Let S be the set of subfields L of A, containing $\psi(k_1)$ and such that $\pi(L) \subset K$; it is inductive for the inclusion relation. Let L be a maximal element of S; we consider K as an extension (algebraic and separable, by A, V, p. 40, prop. 9) of L. Let $\xi \in K$ and let $P \in L[X]$ be its minimal polynomial over L. Since $\xi$ is a simple root of P, Hensel's lemma (III, § 4, no. 5, cor. 1 of th. 2) ensures the existence of an element x of A such that $\pi(x) = \xi$ and $P(x) = 0$. The subring $L[X]$ of A belongs to S; by the maximal character of L, we therefore have $x \in L$, whence $\xi \in \pi(L)$. Finally one has $\pi(L) = K$.

b) Uniqueness of L: Let L and L’ be two subfields of A containing $\psi(k_1)$ and such that $\pi(L) = \pi(L') = K$. Let $\xi \in K$, and let $x \in L$ and $x' \in L'$ be the elements such that $\pi(x) = \pi(x') = \xi$. If $P \in k_1[X]$ is the minimal polynomial of $\xi$ over $k_1$, then $\xi$ is a simple root of P, and one has $P(x) = P(x') = 0$. By Hensel's lemma (loc. cit.) one has $x = x'$. Therefore $L = L'$.