Proposition 20

Suppose that $A' = A \otimes_K K'$, with $\eta': K' \to A'$ and $u: A \to A'$ the canonical homomorphisms. Then the $A'$-linear mapping

is an isomorphism.

By virtue of the fact that in diagram (42) the vertical arrows are bijective, it reduces to proving that, for every $A'$-module $E'$, the homomorphism $C(u): D \mapsto D \circ u$ in diagram (42) is bijective (II, § 2, no. 1, Theorem 1). Now $\mathrm{Hom}(u, 1_{E'}): \mathrm{Hom}_{K'}(A \otimes_K K', E') \to \mathrm{Hom}_K(A, E')$ is an isomorphism (II, § 5, no. 1, Proposition 1) and $C(u)$ is its restriction to $D_{K'}(A', E')$ and hence is injective; moreover, if $f: A' \to E'$ is a $K'$-linear mapping such that

is a $K$-derivation, it follows immediately from the fact that $f$ is $K'$-linear and the fact that $f((x \otimes 1)(y \otimes 1)) = (y \otimes 1)f(x \otimes 1) + (x \otimes 1)f(y \otimes 1)$ for $x, y$ in $A$, that $f$ is a $K'$-*derivation*, the elements $x \otimes 1$ for $x \in A$ forming a generating system of the $K'$-module $A'$; this completes the proof that $C(u)$ is bijective.

From now on we confine our attention to the case where $\rho: K \to K'$ is the identity mapping of $K$; every $K$-algebra homomorphism $u: A \to B$ is therefore mapped to a $B$-linear mapping

On the other hand, we can consider the $B$-module of $A$-*differentials* $\Omega_A(B)$ since $B$ is an $A$-algebra by means of $u$; the canonical derivation $d_{B/A}: B \to \Omega_A(B)$ being a fortiori a $K$-derivation, it factorizes uniquely into

where $\Omega_u$ is a $B$-linear mapping (no. 11, Proposition 18). For every $B$-module $E$, there is a commutative diagram

where $j_u$ is the canonical injection (no. 7); this follows immediately from Proposition 18 of no. 11.