Differentials of a separably generated field extension

Set $K^{\mathrm{\prime }}:=k(T)=k(t_1,\dots ,t_r)$ and let $\alpha \in K$ with $K=K^{\mathrm{\prime }}(\alpha )$ be separable over $K^{\mathrm{\prime }}$, as supplied by [F2]. Let $P\in K^{\mathrm{\prime }}[X]$ be the minimal polynomial of $\alpha$ over $K^{\mathrm{\prime }}$; by [F3] it is monic and irreducible, and by [F1] and the definition of a separable element $P$ is separable over $K^{\mathrm{\prime }}$. Hence $\gcd (P,P^{\mathrm{\prime }})=1$ by [F4], $P^{\mathrm{\prime }}\mathrm{\ne }0$, and the class of $P^{\mathrm{\prime }}$ is invertible modulo $P$, so $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ in the field $K=K^{\mathrm{\prime }}[X]\mathrm{/}(P)$.

${\mathrm{\Omega }}_{K^{\mathrm{\prime }}\mathrm{/}k}$ has $K^{\mathrm{\prime }}$-basis $\mathrm{d}{t}_1,\dots ,\mathrm{d}{t}_r$. Indeed $k[T]\to K^{\mathrm{\prime }}$, $T_i\mapsto t_i$, identifies $k[T]$ with the polynomial ring on the algebraically independent elements $t_i$ by [F1], so by [F5] (with $A=k$ and $n=r$) the module ${\mathrm{\Omega }}_{k[T]\mathrm{/}k}$ is free on $\mathrm{d}{T}_1,\dots ,\mathrm{d}{T}_r$; since a nonzero element of $k[T]$ maps to a nonzero element of $K^{\mathrm{\prime }}=k(T)$, the localisation isomorphism of [F6] applies with $U=k[T]\setminus \{0\}$ and exhibits ${\mathrm{\Omega }}_{K^{\mathrm{\prime }}\mathrm{/}k}\cong K^{\mathrm{\prime }}{\otimes }_{k[T]}{\mathrm{\Omega }}_{k[T]\mathrm{/}k}$ with the images of $\mathrm{d}{T}_1,\dots ,\mathrm{d}{T}_r$ as a $K^{\mathrm{\prime }}$-basis, and those images are exactly $\mathrm{d}{t}_1,\dots ,\mathrm{d}{t}_r$.

${\mathrm{\Omega }}_{K\mathrm{/}{K}^{\mathrm{\prime }}}=0$: for every $K$-module $M$ and every $K^{\mathrm{\prime }}$-derivation $D\text{\hspace{0.17em}⁣}:K\to M$ we have $D(\alpha )=0$, because $0=D(P(\alpha ))={\sum }_{i}D(c_i){\alpha }^i+P^{\mathrm{\prime }}(\alpha )D(\alpha )=P^{\mathrm{\prime }}(\alpha )D(\alpha )$ with $c_i\in K^{\mathrm{\prime }}$ the coefficients of $P$ and $D(c_i)=0$, and $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ by step 1.1; then $D=0$ on all of $K=K^{\mathrm{\prime }}[\alpha ]$, since a derivation vanishing on $K^{\mathrm{\prime }}$ and on $\alpha$ vanishes on every polynomial in $\alpha$. By [F8] this says ${\mathrm{Hom}}_K({\mathrm{\Omega }}_{K\mathrm{/}{K}^{\mathrm{\prime }}},M)=0$ for all $M$, hence ${\mathrm{\Omega }}_{K\mathrm{/}{K}^{\mathrm{\prime }}}=0$. Applying the transitivity sequence of [F7] to $k\to K^{\mathrm{\prime }}\to K$, the first map $\tau \text{\hspace{0.17em}⁣}:K{\otimes }_{K^{\mathrm{\prime }}}{\mathrm{\Omega }}_{K^{\mathrm{\prime }}\mathrm{/}k}\to {\mathrm{\Omega }}_{K\mathrm{/}k}$ is therefore surjective, and by step 1.2 the elements $\tau (1\otimes \mathrm{d}{t}_i)=\mathrm{d}{t}_i$ generate ${\mathrm{\Omega }}_{K\mathrm{/}k}$ as a $K$-module.

Independence of the generators. For each $j$ the $K^{\mathrm{\prime }}$-linear functional on ${\mathrm{\Omega }}_{K^{\mathrm{\prime }}\mathrm{/}k}$ with $\mathrm{d}{t}_i\mapsto {\delta }_{ij}$ exists by step 1.2 and corresponds by [F8] to a $k$-derivation ${\mathrm{\partial }}_j\text{\hspace{0.17em}⁣}:K^{\mathrm{\prime }}\to K^{\mathrm{\prime }}$ with ${\mathrm{\partial }}_j(t_i)={\delta }_{ij}$. Write $P={\sum }_i{c}_i{X}^i$ and put ${\beta }_j:=-({\sum }_i{\mathrm{\partial }}_j(c_i){\alpha }^i)P^{\mathrm{\prime }}(\alpha {)}^{-1}\in K$, which is defined by step 1.1. On the polynomial ring $K^{\mathrm{\prime }}[X]$ define $T_j(F):={\sum }_i{\mathrm{\partial }}_j(b_i){\alpha }^i+F^{\mathrm{\prime }}(\alpha ){\beta }_j$ for $F={\sum }_i{b}_i{X}^i$, so that $T_j(cF)=c{T}_j(F)+{\mathrm{\partial }}_j(c)F(\alpha )$ and $T_j(FG)=F(\alpha )T_j(G)+G(\alpha )T_j(F)$ for all $c\in K^{\mathrm{\prime }}$ and $F,G\in K^{\mathrm{\prime }}[X]$: these identities follow from additivity of ${\mathrm{\partial }}_j$ and the product rules for ${\mathrm{\partial }}_j$ and for the formal derivative, and they say that $T_j$ is a derivation along the evaluation $K^{\mathrm{\prime }}[X]\to K$, $X\mapsto \alpha$, with $T_j(X)={\beta }_j$. Moreover $T_j(P)={\sum }_i{\mathrm{\partial }}_j(c_i){\alpha }^i+P^{\mathrm{\prime }}(\alpha ){\beta }_j=0$, so $T_j$ vanishes on the ideal $(P)$; since $K=K^{\mathrm{\prime }}[X]\mathrm{/}(P)$ by [F3], $T_j$ descends to a well-defined $k$-derivation $D_j\text{\hspace{0.17em}⁣}:K\to K$ extending ${\mathrm{\partial }}_j$, and by [F8] to a $K$-linear map ${\mathrm{\Omega }}_{K\mathrm{/}k}\to K$ sending $\mathrm{d}{t}_i$ to $D_j(t_i)={\mathrm{\partial }}_j(t_i)={\delta }_{ij}$. If now ${\sum }_i{c}_i\mathrm{d}{t}_i=0$ with $c_i\in K$, applying that map gives $c_j=0$ for each $j$, so $\mathrm{d}{t}_1,\dots ,\mathrm{d}{t}_r$ are linearly independent over $K$ and, with step 2.1, form a $K$-basis of ${\mathrm{\Omega }}_{K\mathrm{/}k}$.

The second part. Assume $\mathrm{char}k=0$ and choose, using [F13], a maximal algebraically independent subset $T_0\subseteq K$ over $k$ and a maximal algebraically independent subset $B\subseteq L$ over $K$. By [F11] the extensions $K\mathrm{/}k(T_0)$ and $L\mathrm{/}K(B)$ are algebraic; by [F9] every field of characteristic $0$ is perfect and by [F10] every algebraic extension of a perfect field is separable, so $K\mathrm{/}k(T_0)$ and $L\mathrm{/}K(B)$ are separable algebraic. Because $K\mathrm{/}k$ is finitely generated, [F12] makes $K\mathrm{/}k(T_0)$ finite, so $T_0$ is a separating transcendence basis of $K\mathrm{/}k$ in the sense of [F1], and step 3.1 exhibits a finite $K$-basis $\mathrm{d}{t}_1,\dots ,\mathrm{d}{t}_r$ of ${\mathrm{\Omega }}_{K\mathrm{/}k}$.

Extension of derivations. Let $D\text{\hspace{0.17em}⁣}:K\to L$ be a $k$-derivation. First extend $D$ to $K(B)$: each element of $K(B)$ lies in $K(B_0)$ for some finite $B_0\subseteq B$, and on the polynomial ring $K[B_0]$, for which ${\mathrm{\Omega }}_{K[B_0]\mathrm{/}K}$ is free on the $\mathrm{d}b$, $b\in B_0$, by [F5], the prescription $b\mapsto 0$ for $b\in B_0$, together with $D$ on $K$, defines a unique $k$-derivation of $K[B_0]$ extending $D$; it extends uniquely to the fraction field $K(B_0)$ by the quotient rule $\mathrm{d}(c\mathrm{/}d)=(d\mathrm{d}c-c\mathrm{d}d)\mathrm{/}{d}^2$ of [F6], and for $B_0\subseteq B_1$ the extension on $K(B_1)$ restricts to the one on $K(B_0)$ by uniqueness, so a derivation $D_1\text{\hspace{0.17em}⁣}:K(B)\to L$ extending $D$ is well defined on the union of the fields $K(B_0)$. Next let $x\in L$; then $K(B)(x)\mathrm{/}K(B)$ is finite separable by step 4.1, hence simple, with minimal polynomial $Q$ of $x$ over $K(B)$ that is separable by [F10], so $Q^{\mathrm{\prime }}(x)\mathrm{\ne }0$ by [F3] and [F4]. Substituting $K(B)$ for $K^{\mathrm{\prime }}$, $K(B)(x)$ for $K$ and $Q$ for $P$ in the construction of step 3.1 produces an extension of $D_1$ to $K(B)(x)$, and any two extensions of $D_1$ to $K(B)(x)$ agree, because their difference vanishes on $K(B)$ and takes at $x$ a value killed by $Q^{\mathrm{\prime }}(x)\mathrm{\ne }0$. Declaring the value at $x$ to be that unique value defines $D_2(x)$ for every $x\in L$; it is a derivation because for $x,y\in L$ the field $K(B)(x,y)$ is finite separable over $K(B)$ by [F10] and [F12], carries an extension of $D_1$ by the same construction, and on it the derivation laws hold while its restrictions to $K(B)(x)$ and $K(B)(y)$ agree with the unique extensions, so the values $D_2$ assigns are additive and satisfy Leibniz.

Injectivity. Keep the notation of step 4.1, let $\rho \text{\hspace{0.17em}⁣}:L{\otimes }_K{\mathrm{\Omega }}_{K\mathrm{/}k}\to {\mathrm{\Omega }}_{L\mathrm{/}k}$ be the natural map of [F6], and let $x={\sum }_i{c}_i(1\otimes \mathrm{d}{t}_i)\in \ker \rho$, the elements $\mathrm{d}{t}_i$ being a $K$-basis of ${\mathrm{\Omega }}_{K\mathrm{/}k}$ by step 4.1. For each $j$ the functional $\mathrm{d}{t}_i\mapsto {\delta }_{ij}$ is a $K$-linear map ${\mathrm{\Omega }}_{K\mathrm{/}k}\to L$, hence by [F8] equals $E_j\circ \mathrm{d}$ for a $k$-derivation $E_j\text{\hspace{0.17em}⁣}:K\to L$; by step 5.1 there is a $k$-derivation ${\tilde{E}}_j\text{\hspace{0.17em}⁣}:L\to L$ extending $E_j$, and by [F8] it induces an $L$-linear ${\phi }_j\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{L\mathrm{/}k}\to L$ with ${\phi }_j(\rho (1\otimes \mathrm{d}{t}_i))={\tilde{E}}_j(t_i)=E_j(t_i)={\delta }_{ij}$. Then $0={\phi }_j(\rho (x))=c_j$ for every $j$, so $x=0$. Hence $\rho$ is injective, which is the second assertion, and the theorem is proved.