If this is so, the derivation D is unique and satisfies

Put $E = K[X_i]_{i \in I}$ and write $\varphi$ for the K-homomorphism of E onto A which maps $X_i$ to $x_i$ for all $i \in I$; we thus have $\varphi(f) = f(x)$ for all $f \in E$. We consider M as an E-module by means of the homomorphism $\varphi : E \to A$ and define a mapping $D'$ of E into M by $D'(f) = f^\Delta(x) + \sum_{i \in I} D_i f(x) \cdot m_i$ (note that the family $(D_i f)_{i \in I}$ has finite support for each $f \in E$). It is clear that $D'$ is the unique derivation of E into M extending A and mapping $X_i$ to $m_i$ for each $i \in I$.

Let D be a derivation of A into M such that $D(c \cdot 1) = \Delta(c)$ for all $c \in K$ and $D(x_i) = m_i$ for all $i \in I$. Then $D \circ \varphi$ is a derivation of E into M extending A and mapping $X_i$ to $m_i$ for all $i \in I$. We thus have $D \circ \varphi = D'$, that is, the relation (2) holds. This proves the uniqueness of D ; moreover (1) is a consequence of $f, (x) = 0$ and (2).

Conversely assume that (1) holds; in other words, we have $D'(f_\lambda) = 0$ for all $\lambda \in A$. Let $f \in a$; there exists a family with finite support $(q_\lambda)_{\lambda \in \Lambda}$ in $E$ such that $f = \sum_{\lambda \in \Lambda} q_\lambda \cdot f_\lambda$. We have

whence $D'(f) = 0$ because $f_\lambda(x)$ and $D'(f_\lambda)$ are zero for all $\lambda \in A$. Since $D'$ vanishes on $a$, there exists a $Z$-linear mapping $D$ of $A$ into $M$ such that $D' = D \circ \varphi$; clearly $D$ is the required derivation of $A$ into $M$.

a) Let $V$ be a vector space over $L$, $A$ a derivation of $K$ into $V$ and $(v_i)_{i \in I}$ a family of elements of $V$. There exists a unique derivation $D$ of $L$ into $V$, extending $A$ and such that $D(x_i) = v_i$ for all $i \in I$.

b) The vector $L$-space $\Omega_K(L)$ of $K$-differentials of $L$ has the family $(dx_i)_{i \in I}$ as basis.

Assertion b) has been proved in IV, p. 23 and Assertion a) follows directly from this.

Prop. 3, a) shows that every $P$-derivation of $K$ into a vector space $V$ over $L$ extends to a $P$-derivation of $L$ into $V$; the injectivity of $\alpha$ then follows by Prop. 2. The second assertion of the corollary follows from the exactness of the sequence $(\mathbf{E}_{P,K,L})$ and Prop. 3, b) (taking Formula (4) into account).

a) Every $K$-derivation of $L$ into $V$ is zero.

b) If $A$ is a derivation of $K$ into $V$, there exists a unique derivation $D$ of $L$ into $V$ which extends $A$.

Let $D$ be a $K$-derivation of $L$ into $V$. If $E$ is a subextension of $L$ of finite degree over $K$, then the $K$-algebra $E$ is etale, and so $\Omega_K(E) = 0$ ($V$, p. 33, Th. 3), whence $D|_E = 0$ by the universal property of $\Omega_K(E)$ (III, p. 569). Since $L$ is the union of a family of subextensions of finite degree over $K$, we have $D = 0$, whence a).

Let $A$ be a derivation of $K$ into $V$. If $D'$ and $D''$ are two extensions of $A$ to a derivation of $L$ into $V$, then the difference $D' - D''$ is a $K$-derivation of $L$ into $V$; hence it is zero by a) and so $D' = D''$.

It remains to prove the existence of an extension of $A$. Zorn's Lemma (E, III, p. 20) implies the existence of a maximal extension $D_0$ of $A$ to a derivation defined on a subfield $L_0$ of $L$ containing $K$.

Let $x$ be in $L$ and $g$ the minimal polynomial of $x$ over $L_0$. Since $L$ is algebraic and separable over $K$, $x$ is algebraic and separable over $L_0$ ($V$, p. 39, Prop. 6 and p. 39, Cor. 2); therefore $x$ is a simple root of $g$ ($V$, p. 39, Prop. 5), whence $g'(x) \neq 0$ (IV, p. 17, Prop. 7). If we define $g^{D_0}(x)$ as in $V$, p. 127, there exists thus an element $u$ of $V$ such that $g^{D_0}(x) + g'(x) \cdot u = 0$; by Prop. 1 (V, p. 127) there exists a derivation $D$ of $L_0(x)$ into $V$ extending $D_0$ and such that $D(x) = u$. In view of the maximal character of $(L_0, D_0)$ we thus have $L_0(x) = L_0$, whence $x \in L_0$. Since $x$ was arbitrary we conclude that $L_0 = L$.

The corollary follows from Prop. 4, a) because the vector $L$-space $\Omega_K(L)$ is generated by the image of the canonical $K$-derivation $d_{L/K} : L \to \Omega_K(L)$.

The mapping $\alpha$ is injective by Prop. 2 (V, p. 128) and Prop. 4, b); since $\Omega_K(L)$ reduces to 0 (Cor. 1) the exactness the sequence $(E_{P.K.L})$ implies that $\alpha$ is surjective.

Let A be the derivation of K into L which agrees with D on K. By Prop. 4 (V, p. 129) there exists a derivation D' of L into L extending A. Now we may consider D' and the restriction D'' of D to L as derivations of $L$ into E ; since they agree on K, we have $D' = D''$ by Prop. 4, whence

Put $E = K(x_i)_{i \in I}$; Prop. 3 (V, p. 128) shows that A extends to a unique derivation $D_0$ of E into V such that $D_0(x_i) = u_i$ for all $i \in I$. The field $L$ is an algebraic extension of E and since L is of characteristic 0, L is separable over E (V, p. 37, Cor.). Therefore (V, p. 129, Prop. 4) $D_0$ extends to a unique derivation D of $L$ into V.

For each $y \in L$ let D(y) be the residue class of $dy$ mod U. Then D is an E-derivation of L into $\Omega_K(L)/U$. Since K is of characteristic 0, every algebraic extension of E is separable (V, p. 37, Cor.) ; if $x \in L$ is algebraic over E, we have $Dx = 0$ by Prop. 4 (V, p. 129), that is, $dx \in U$.

If $x \in L$ is transcendental over $E$, there exists an E-derivation $A$ of $E(x)$ into $L$ such that $\Delta(x) = 1$ ($V$, p. 128, Prop. 3); by Th. 1, $A$ extends to an E-derivation D of $L$ into L. Let $\varphi$ be the linear form on $\Omega_K(L)$ such that $D = \varphi \circ d$; we have $\varphi(dy) = 0$ for $y \in E$ and $\varphi(x) = 1$, whence $dx \notin U$.

Lane's criterion (V, p. 124, Remark), the fields $L^p$ and $K$ are linearly disjoint over $K^p$. Since $A$ is a $K^p$-linear mapping of $K$ into $V$, it extends in a unique way to an $L^p$-linear mapping $A'$ of $K[L^p] = K(L^p)$ into $V$. It is clear that $A'$ is a derivation of $K(L^p)$ into $V$ which vanishes on $L^p$; thus it extends (V, p. 103, Cor. 3) to a derivation of $L$ into $V$.

Conversely, suppose that every derivation of $K$ with values in $L$ extends to a derivation of $L$ into $L$. Let $B$ be a p-basis of $K$ (V, p. 99) and let $A$ be the set of families $(\alpha_h)_{h \in R}$ with finite support consisting of integers between $()$ and $p - 1$. For each $b \in B$ there exists a derivation $\Delta_b$ of $K$ into $K$ characterized by $\Delta_b(b') = \delta_{bb'}$ (Kronecker symbol) for each $b' \in B$ (V, p. 103). By hypothesis there exists for each $b \in B$ a derivation $D_b$ of $L$ into $L$ extending $\Delta_b$. We have $D_b(b') = \delta_{bb'}$ for $b, b'$ in $B$, which proves that in $\Omega_{1/p}(L)$ the differentials $db$ (for $b \in B$) are linearly independent over $L$. Therefore (V, p 102, Th. 1), $B$ is p-free over $L^p$. It follows (V, p. 98, Prop. 1 and p. 124, Remark) that the extension $L$ of $K$ is separable.

The first assertion follows from Prop. 2 (V, p. 128) and Th. 3. Conversely, let $P$ be a perfect subfield of $K$; we have $P = P^p \subset K^p$, hence every derivation of $K$ into a vector $L$-space is a $P$-derivation; the second assertion of the corollary now follows from Cor. 2 (V, p. 130) and Th. 3.

Let $N$ be the kernel, $I$ the image and $C$ the cokernel of $f$. We have $C = V/I$ and $I$ is isomorphic to $U/N$; thus we have $[U : L] = [N : L] + [I : L]$ and $[V : L] = [C : L] + [I : L]$, whence the result.

$^1$ Not necessarily commutative. $^2$ On the left.

Put $h = g \circ f$, and denote by N, N', N'' the kernels off, g, h respectively and by C, C', C'' their cokernels. We have $N \subset N'' \subset U$ and $f(N'') = f(U) \cap N'$; hence there exists a linear mapping $\bar{f} : N'' \to N'$ agreeing with f on N'' and with kernel N. The canonical mapping $\pi$ of V onto $C = V / f(U)$ induces a mapping $\pi'$ of N' into C whose kernel is $f(U) \cap N' = \bar{f}(N'')$. By passage to quotients g defines a mapping $\bar{g}$ of $C = V/f(U)$ into $C'' = W/g(f(U))$ whose kernel is clearly $(N' + f(U))/f(U) = \pi'(N')$. Finally the canonical mapping $\rho$ of $C'' = W/g(f(U))$ onto $C' = W/g(V)$ has the kernel $g(V)/g(f(U)) = \bar{g}(C)$. To sum up, we have established the exactness of the sequence

(where i is the canonical injection of N in N").

!Diagram showing the relationships between N, N', N'', U, V, W, C, C', C'', with arrows indicating mappings and kernels

Fig. 1

By hypothesis N, N', C and C' are of finite dimension; hence the same is true of N'' and C. By Cor. 2 (II, p. 295) we then have

whence $\chi(h) = \chi(f) + \chi(g)$.

If $E$ and $F$ are two subextensions of $L$ such that $E \subset F$, we denote by $\alpha(F/E)$ the canonical $F$-linear mapping of $\Omega_p(E) \otimes_E F$ into $\Omega_p(F)$ and when it is defined, the index of $\alpha(F/E)$ will be written $d(F/E)$. If $E, F$ and $G$ are three subextensions of $L$ such that $E \subset F \subset G$, we have a commutative diagram

where $\beta$ is the canonical isomorphism defined in Prop. 2 (II, p. 278). Since the index is clearly invariant under extension by scalars and the index of an isomorphism is zero, Lemma 2 ($V$, p. 133) shows that the index $d(G/E)$ is defined when $d(F/E)$ and $d(G/F)$ are and then

Since $L$ is a finitely generated extension of $K$, Formula (5) and the Cor. of $V$, p. 111 shows that it is enough to consider the case where $x$ exists such that $L = K(x)$; further if $x$ is algebraic over $K$, there exists a power $q$ of the characteristic exponent of $L$ such that $x^q$ is separable algebraic over $K$ ($V$, p. 44, Prop. 13). So it is enough to establish the equality $d(L/K) = \operatorname{tr} . \deg_K L$ in the three special cases below:

a) $x$ is transcendental over $K$: then $a$ is injective and its cokernel is of rank 1 over $L$ ($V$, p. 129, Cor.) ; so we have $d(L/K) = 1$ and also $\operatorname{tr} . \deg_K L = 1$.

b) $x$ is separable algebraic over $K$: then $a$ is bijective ($V$, p. 130, Cor. 2), whence $d(L/K) = 0$; clearly also $\operatorname{tr} . \deg_K L = 0$.

c) The field $L$ is of characteristic $p \neq 0$, $x \notin K$ and $x^p = a$ is in $K$: the cokernel $C$ of $a$ is isomorphic to $\Omega_K(L)$, and since $\{ x \}$ is a $p$-basis of $L$ over $K$, the space $C$ is of dimension 1 over $L$ ($V$, p. 102, Th. 1). Since $a$ is a $p$-th power in $L$, we have $d_{L/pa} = 0$ and the kernel of $a$ contains the subspace $R$ of $U = \Omega_p(K) \otimes_K L$ generated by $d_{K/pa} \otimes 1$. For each $y \in K$ let $\Delta(y)$ be the residue class of $d_{K/p}y \otimes 1$ mod $R$; then $A$ is a $P$-derivation of $K$ into $U/R$ such that $\Delta(a) = 0$. Prop. 5 ($V$, p. 101) shows that $A$ extends to a $P$-derivation $D$ of $L$ into $U/R$. Hence there exists an $L$-linear mapping $\beta : \Omega_p(L) \to U/R$ such that $D = \beta \circ d_{L/P}$ and $\beta \circ a$ is the canonical mapping of $U$ onto $U/R$. This proves $R$ to be the kernel of $a$. Since $P$ is perfect, we have $P(K^P) = K^p$, whence $a \notin P(K^P)$ and finally $d_{K/pa} \neq 0$ ($V$, p. 103, Prop. 6). The kernel and cokernel of $a$ are thus of dimension 1, whence $d(L/K) = 0$, and we also have $\operatorname{tr} . \deg_K L = 0$.

Let P be the prime subfield of K. Let N be the kernel of α, then by exactness of the sequence $(\mathbf{E}_{P,K,L})$ (V, p. 128) and Th. 4, we have $[ \Omega_K(L) : L ] = s + [N : L]$; by V, p. 132, Cor., the extension L of K is separable if and only if $N = 0$. The corollary now follows.

Put $M = K(x_i)_{i \in I}$, and denote by $\alpha$ the canonical mapping of $\Omega_K(M) \otimes_M L$ into $\Omega_K(L)$. If $(x_i)_{i \in I}$ is a separating transcendence basis of $L$ over $K$, the family $(d_{M/K} x_i)_{i \in I}$, is a basis of the vector $M$-space $\Omega_K(M)$ ($V$, p. 128, Prop. 3) and $\alpha$ is an isomorphism of vector $L$-spaces since $L$ is algebraic and separable over $M$ ($V$, p. 130, Cor. 2). Since $\alpha(d_{M/K} x_i \otimes 1) = d_{L/K} x_i$, the family $(d_{L/K} x_i)_{i \in I}$ is thus a basis of $\Omega_K(L)$ over $L$.

Conversely, suppose that $L$ is a separable finitely generated extension of $K$ and that the family $(d_{L/K} x_i)_{i \in I}$ is a basis of the vector space $\Omega_K(L)$ over $L$. By Cor. 1 of $V$, p. 134, the transcendence degree of $L$ over $K$ is equal to the dimension of $\Omega_K(L)$ over $L$, hence to the cardinal of $I$. From the exact sequence $(E_{K,M,L})$ ($V$, p. 128) we have $\Omega_M(L) = 0$; since $L$ is a finitely generated extension of $M$, Cor. 2 of $V$, p. 135 shows that $L$ is algebraic and separable over $M = K(x_i)_{i \in I}$; since the transcendence degree of $L$ over $K$ is finite and equal to the cardinal of $I$, the family $(x_i)_{i \in I}$ is a transcendence basis of $L$ over $K$ ($V$, p. 110, Cor. 1).