We have ${L'}^p \subset K \subset K'$ and $L' = K'(L)$, whence ${L'}^p = {K'}^p({L'}^p) \subset K'$. In other words, L' is a p-radical extension of height $\leq 1$ of K'. The other assertions of the corollary follow at once from Prop. 1 and from V, p. 14, Prop. 5.

Suppose first that S is p-free and let $T \neq S$ be a subset of S. Denote by A the set of families with finite support $a = (\alpha_s)_{s \in S}$ of integers between 0 and $p-1$; let $A'$ be the subset of A consisting of the families $a = (\alpha_s)_{s \in S}$ such that $a_s = 0$ for all s in $S - T$. We also put $u_a = \prod_{s \in S} s^{\alpha_s}$ for $a \in A$. Then (V, p. 98, Prop. 1) the family $(u_a)_{a \in A'}$ is a basis of $K(S)$ over K and the subfamily $(u_a)_{a \in A'}$ is a basis of $K(T)$ over K. Since $A' \neq A$, we have $K(T) \neq K(S)$.

Suppose now that S is not p-free. Then there exists an integer $n \geq 1$ and a sequence of elements $x_1, ..., x_n$ of S such that $(x_1, ..., x_{n-1})$ is p-free but $(x_1, ..., x_n)$ is not. We have $[K(x_1, ..., x_{n-1}):K] = p^{n-1}$ and $[K(x_1, ..., x_n):K] < p^n$. Since $[K(x_1, ..., x_n):K]$ is a multiple of $[K(x_1, ..., x_{n-1}):K]$, we thus have

whence $x_n \in K(x_1, ..., x_{n-1})$. This shows that $K(S - \{x_n\}) = K(S)$.

Let K be a field. A family $(x_i)_{i \in I}$, is said to be a (absolute) p-basis of K if it is a p-basis of K over $K^p$. For every integer $n \geq 1$ we denote by $\Lambda(n)$ the subset of $N^{(I)}$ consisting of all $a = (\alpha_i)_{i \in I}$, such that $\alpha_i < p^n$ for all $i \in I$.

For $n = 1$ the assertion reduces to Prop. 1 (V, p. 98). The set $\Lambda(n)$ consists of elements of $N^{(I)}$ of the form $a = \beta + p^{n-1}\gamma$ with $\beta \in \Lambda(n-1)$ and $\gamma \in \Lambda(1)$. Such a decomposition is unique and we have $x^\alpha = x^\beta (x^\gamma)^{p^{n-1}}$. Moreover, the family $(x_i^{p^{n-1}})_{i \in I}$ is clearly a p-basis of $K^{p^{n-1}}$ over $K^{p^n}$, hence the family $(x^\gamma)^{p^{n-1}}$ is a basis of $K^{p^{n-1}}$ over $K^{p^n}$. We thus obtain the conclusion by induction, using II, p. 222, Prop. 25.

By $V$, p. 24, Prop. 1, $\{1,\ x,\ \ldots,\ x^{p-1}\}$ is a basis of $L$ over $K$. For every element $u = a_0 + a_1x + \ldots + a_{p-1}x^{p-1}$ of $L$ (with $a_0,\ \ldots,\ a_{p-1}$ in $K$) we put $D(u) = A(a_0) + x\Delta(a_1) + \ldots + x^{p-1}\Delta(a_{p-1})$. It is clear that $D$ extends $A$ and satisfies (1); so it is enough to establish the relation

when $u$ is of the form $ax^i$ and $v$ of the form $bx^j$ with $a,\ b$ in $K,\ 0 \leq i < p$ and $0 \leq j < p$.

When $i + j < p$ we have $uv = x^{i+j}ab$, whence $D(uv) = x^{i+j}\Delta(ab)$. Otherwise we have $0 \leq i + j - p < p$ and so $uv = x^{i+j-p}(abx^p)$, with $abx^p \in K$, but since $A(x^p) = 0$, we have $A(abx^p) = x^pA(ab)$, whence again $D(uv) = x^{i+j}\Delta(ab)$. So we have in all cases

By Zorn's lemma (E, III, p. 20) there exists a maximal extension of $A$ to a derivation $D_0 : L_0 \to V$, where $L_0$ is a subfield of $L$ containing $K$. Let $x \in L$; we have $x^p \in L_0$ and $A(x^p) = 0$, whence $D_0(x^p) = 0$. By Prop. 5,

D_0 extends to a derivation defined on L_0(x); by the maximal character of D_0 we thus have L_0(x) = L, whence x ∈ L_0. Since x was arbitrary we conclude that L_0 = L.

Let x be an element of L not belonging to E. Then {1, x, ..., x^{p^{l-1}}} is a basis of E(x) over E. Let A be the E-linear mapping of E(x) into L such that Δ(x^i) = ix' for 0 ≤ i < p. We verify at once that A is an E-derivation of E(x) into L, and in particular a K-derivation. By Cor. 1 to Prop. 5 there exists a K-derivation D of L into L extending A. By the universal property of Ω_K(L), there exists a linear form u on this vector L-space such that D = u ∘ d_{L/K}. We have D(x) = x and D|E = 0; therefore we have u(d_{L/K}x) ≠ 0 and u|U = 0, whence d_{L/K}x ∉ U.

The last assertion is the particular case E = K; then we have U = 0.

a) For (x_i)_{i ∈ I} to be p-free over K it is necessary and sufficient that the family (dx_i)_{i ∈ I} should be free in the vector L-space Ω_K(L).

b) For (x_i)_{i ∈ I} to generate L over K it is necessary and sufficient that the family (dx_i)_{i ∈ I} should generate the vector L-space Ω_K(L).

c) For (x_i)_{i ∈ I} to be a p-basis of L over K it is necessary and sufficient that the family (dx_i)_{i ∈ I} should be a basis of the vector L-space Ω_K(L).

To begin with let us remark that the differential dx of an element of the field K((x_i)_{i ∈ I}) is a linear combination with coefficients in L of the differentials dx_i, i ∈ I. For the family (x_i)_{i ∈ I} to be p-free it is necessary and sufficient that x_i ∉ K(x_j)_{j ∈ I - {i}} for all i ∈ I (V, p. 99, Prop. 2). By Cor. 2 of Prop. 5 this means that dx_i is not a linear combination of the dx_j for j ≠ i in I. Assertion a) follows from this. Now b) follows at once from Cor. 2 to Prop. 5 and c) follows from a) and b).

The following corollary makes Cor. 1 of Prop. 5 more specific:

By Cor. 1 of Prop. 5 there exists a derivation D_0 of L into V extending A. The derivations of L in V extending A are precisely the mappings of the form D = D_0 + u ∘ d_{L/K} where u is an L-linear mapping of Ω_K(L) into V. We have D(x_i) = u_i if and only if the linear mapping u satisfies the conditions u(dx_i) = u_i − D_0(x_i). Since the family dx_i is a basis of Ω_K(L), this determines u uniquely and the corollary follows.

Let L be a p-radical extension of height $\leq 1$ of K and $(x_i)_{i \in I}$ a p-basis of L over K. By the Cor. to Th. 1 there exists for each $i \in I$ a K-derivation $D_i$ of L into L characterized by $D_i(x_j) = \delta_{ij}$ (Kronecker symbol); $D_i$ is sometimes called the partial derivation with respect to $x_i$. When $I$ is finite, the family $(D_i)_{i \in I}$ is a basis of the vector space over L consisting of the K-derivations of L into L.

Th. 1 allows the study of p-bases to be reduced to the study of bases of a vector space. For example we have the following result:

Assertions a) and b) are immediate (II, p. 292). If $[L : K]$ is finite, there exists by a) a finite p-basis $(x_1, ..., x_n)$ of $L$ over $K$; then the monomials $x_1^{\alpha_1} ... x_n^{\alpha_n}$ with $0 \leq \alpha_i < p$ for $1 \leq i \leq n$ form a basis of $L$ over $K$ and the differentials $dx_1, ..., dx_n$ form a basis of $\Omega_K(L)$ over $L$. We thus have $[L : K] = p^n$ and $[\Omega_K(L) : L] = n$. Conversely if $\Omega_K(L)$ is of finite dimension over $L$, there exists a finite p-basis of L over K (V, p. 102, Th. 1) and $[L : K]$ is finite.

For $\{x\}$ is a p-free subset of $L$ over $K$ and so it is enough to apply Th. 2, a).

Let K be a field and $L$ an extension of K. If $D$ is a K-derivation of L with values in a vector L-space, then we have $D(x^p) = 0$ for all $x \in L$, and so $D$ is a $K(L^p)$-derivation; it follows that $\Omega_K(L) = \Omega_{K(L^p)}(L)$. Since $L$ is a p-radical extension of height $\leq 1$ of $K(L^p)$, we can apply the above results. For example, from Th. 1, c) we deduce the following: let $(x_i)_{i \in I}$ be a family of elements of $L$; for $(dx_i)_{i \in I}$ to be a basis of the vector L-space $\Omega_K(L)$ it is necessary and sufficient that $(x_i)_{i \in I}$ should be a p-basis of L over $K(L^p)$. Similarly, Cor. 2 of Prop. 5 implies:

The proof requires several preliminary lemmas.

For each $i \in \mathbf{F}_p$ denote by $P_i(X)$ the polynomial $- \prod_{j \neq i} (X - j)$. We have

by Formula (2) ($V$, p. 94). On deriving the quoted formula (2), we find

The formulae (8) and (9) show that the endomorphism $P_i(u)$ of $V$ maps $V$ into $V_i$ and that $\sum_{i \in \mathbf{F}_p} P_i(u) = 1$, whence $V = \sum_{i \in \mathbf{F}_p} V_i$. It remains to show that the sum is direct. For each $i \in \mathbf{F}_p$ let $v_i \in V_i$; it is clear that $P_i(u)$ annihilates $v_j$ for all $j \neq i$ and we have $P(u)v_i = av_i$ with $a = -\prod_{n \in \mathbf{F}_p^*} n \neq 0$. The relation $\sum_{i \in \mathbf{F}_p} v_i = 0$ thus implies $v_i = 0$ for all $i \in \mathbf{F}_p$ and so the sum is direct.

It is clear that K is a subfield of E containing $E^p$. Denote by $K_i$ the kernel of $D - i$ for $i \in \mathbf{F}_p$. We have $K_0 = K$ and Lemma 1 shows that $E = \bigoplus_{i \in \mathbf{F}_p} K_i$. Let $i \in \mathbf{F}_p$ and let $u$ be in $K_i$; we have

whence $xu \in K_{i+1}$. Since x is non-zero, multiplication by x is an automorphism of the vector K-space E which maps $K_i$ onto $K_{i+1}$ for all $i \in \mathbf{F}_p$. Since $[K_0 : K] = 1$, we have $[K_i : K] = 1$ for all $i \in \mathbf{F}_p$ whence $[E : K] = p$.

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 $K = I(g(K))$.