Lemma 3

Let $\mathfrak{h} \in \mathscr{L}$ be of dimension s over E. Then $I(\mathfrak{h})$ belongs to $\mathscr{C}$ and we have $[E : I(\mathfrak{h})] = p^s$.

It is clear that $I(\mathfrak{h})$ is a subfield of $E$ containing $E^p$. For each $x \in E$ let $f_x$ be the E-linear form $D \mapsto D(x)$ on I). Since the intersection of the kcrncls of these linear forms is equal to 0, they generate the vector space dual to $\mathfrak{h}$ (II, p. 301, Th. 7); hence there exist $x_1, \ldots, x_s$ in $E$ such that the linear forms $f_{x_1}, \ldots, f_{x_s}$ form a basis of this dual. Let $(\Delta_1, \ldots, \Delta_s)$ be the basis of $\mathfrak{h}$ characterized by $\Delta_i(x_j) = f_{x_j}(\Delta_i) = \delta_{ij}$. Put $D_i = x_i \Delta_i$, then $(D_1, \ldots, D_s)$ is a basis of I) over E and we have $D_i(x_j) = x_i \delta_{ij}$. The derivations $D_i^p - D_i$ and $[D_i, D_j]$ for $i, j = 1, \ldots, s$ belong to I) and annihilate $x_1, \ldots, x_s$; we thus have

For i in the range 0 to s let us denote by $K_i$ the intersections of the kernels of the derivations $D_j$ for $1 \leq j \leq i$. Then $K_i$ is a subfield of E and we have Let i be between 0 and $s-1$; then $K_i$ is stable under $D_{i+1}$ because $D_{i+1}$ commutes with $D_1, \ldots, D_i$. Moreover we have $D_{i+1}^p = D_{i+1}$, $D_{i+1}(x_{i+1}) = x_{i+1} \neq 0$ and $x_{i+1} \in K_i$. Hence Lemma 1 implies that $[K_i : K_{i+1}] = p$, whence finally $[E : K] = [K_0 : K_s] = p^s$.

We now come to the proof of the theorem. Let I) $\in \mathscr{L}$ be of dimension s over E and put $K = I(\mathfrak{h})$; then $[E : K] = p^s$ by Lemma 3, whence $[\Omega_K(E) : E] = s$ by

Th. 2, c) (V, p. 103). By the universal property of the module of differentials the mapping $u \mapsto u \circ d_{E/K}$ is an isomorphism of the dual of $\Omega_K(E)$ onto $g(K)$, hence $[g(K):E] = s$. Now we have $[\mathfrak{h}:E] = s$ and $\mathfrak{h} \subset g(K)$, whence $\mathfrak{h} = g(K)$, that is $\mathfrak{h} = g(I(\mathfrak{h}))$.

Conversely, for every field $K \in \mathscr{C}$ it is clear that $g(K)$ belongs to $\mathscr{L}$ (V, p. 103, Th. 2, c)). If $x$ belongs to $I(g(K))$, we have $u(d_{E/K}x) = 0$ for each linear form $u$ on $\Omega_K(E)$, whence $d_{E/K}x = 0$ and so finally $x \in K$ by Cor. 2 of Prop. 5 (V, p. 102). Thus we have $K = I(g(K))$.