Theorem 4

Let P be a perfect field, L an extension of P and K a subextension of L; we suppose that L is a finitely generated extension of K. Then the canonical L-linear mapping $a : \Omega_p(K) \otimes_K L \to \Omega_p(L)$ has index equal to the transcendence degree of L over K.

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 $\operatorname{tr} . \deg_K L = 0$.