Proposition 21

Tag 0679, permanent. Cite it as [Bourbaki, Tag 0679].

The sequence of $B$ -linear mappings

\Omega_K(A) \otimes_A B \xrightarrow{\Omega_0(u)} \Omega_K(B) \xrightarrow{\Omega_u} \Omega_A(B) \to 0

is exact.

It reduces to verifying that, for every B-module E, the sequence

0 \to \operatorname{Hom}_B(\Omega_A(B), E) \xrightarrow{\operatorname{Hom}(\Omega_u, 1_E)} \operatorname{Hom}_B(\Omega_K(B), E) \xrightarrow{\operatorname{Hom}(\Omega_0(u), 1_E)} \operatorname{Hom}_B(\Omega_K(A) \otimes_A B, E)

is exact (II, § 2, no. 1, Theorem 1); but by virtue of the fact that in the commutative diagrams (42) and (44) the vertical arrows are isomorphisms, it suffices to show that the sequence

0 \longrightarrow D_A(B, E) \xrightarrow{j_u} D_K(B, E) \xrightarrow{C(u)} D_K(A, E)

is exact, which is just Proposition 11 of no. 7.

We consider now the case where the K-algebra homomorphism $u : A \to B$ is surjective; if $\mathfrak{J}$ is its kernel, B is then isomorphic to $A/\mathfrak{J}$ . We consider the restriction $d|_{\mathfrak{J}} : \mathfrak{J} \to \Omega_K(A)$ of the canonical derivation $d = d_{A/K}$ and the composite A-linear mapping

d' : \mathfrak{J} \xrightarrow{d|_{\mathfrak{J}}} \Omega_K(A) \xrightarrow{i_A} \Omega_K(A) \otimes_A B.

Then $d'(\mathfrak{J}^2) = 0$ , since, for $x, y$ in $\mathfrak{J}$ ,

d'(xy) = d(xy) \otimes 1 = (x.dy + y.dx) \otimes 1 = dy \otimes u(x) + dx \otimes u(y) = 0

since $u(x) = u(y) = 0$ . Hence we derive from $d'$ , by passing to the quotient, an A-linear mapping

\bar{d} : \mathfrak{J}/\mathfrak{J}^2 \to \Omega_K(A) \otimes_A B

and as $\mathfrak{J}$ annihilates the A-module $\mathfrak{J}/\mathfrak{J}^2$ , $\bar{d}$ is a B-linear mapping.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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§ 16. Differential criteria of separability

Proposition 21

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