Proposition 22

Let $\mathfrak{J}$ be an ideal of the commutative K-algebra A, $B = A/\mathfrak{J}$ and $u : A \to B$ the canonical homomorphism. The sequence of B-linear mappings

is then exact.

Note that $\Omega_K(A) \otimes_A B$ is identified with $\Omega_K(A)/\mathfrak{J}\Omega_K(A)$ and that the image of $\bar{d}$ is the image of $d(\mathfrak{J})$ in this quotient module; the quotient of $\Omega_K(A) \otimes_A B$ by $\operatorname{Im}(\bar{d})$ is therefore identified with the quotient $\Omega_K(A)/N$, where N is the sub-A-module generated by $\mathfrak{J}\Omega_K(A)$ and $d(\mathfrak{J})$. Moreover, the composite mapping

is a K-derivation (no. 7, Proposition 9) and, since it is zero on $\mathfrak{J}$ by definition of N, it defines, when passing to the quotient, a K-derivation $D_0 : B \to \Omega_K(A)/N$.

Taking account of the uniqueness of the solution of a universal mapping problem, it reduces to proving that, for every B-module E and every K-derivation D: B → E, there exists a unique B-linear mapping g: Ω_K(A)/N → E such that D = g ∘ D_0. But, the composite mapping D ∘ u: A → E is a K-derivation (no. 7, Proposition 9) and hence there exists one and only one A-linear mapping f: Ω_K(A) → E such that f ∘ d_{A/K} = D ∘ u. This relation shows already that f is zero on d(Σ); as also ΣE = {0} since E is a B-module, f is zero on ΣΩ_K(A); hence f is zero on N and defines, when passing to the quotient, a B-linear mapping g: Ω_K(A)/N → E such that g ∘ D_0 = D; the uniqueness of g follows from the uniqueness of f.

It must not be thought that, even if u: A → B is an injective homomorphism, Ω_0(u): Ω_K(A) ⊗_A B → Ω_K(B) is injective (Exercise 5). However we have the following proposition: