Question

As is well known, every differential calculus $(\Omega,d)$ over an algebra $A$ is a quotient of the universal calculus $(\Omega_A,d)$, by some ideal $I$. In the classical case, when $A$ is the coordinate ring of a variety $V(J)$ (for some ideal of polynomials $J$), and $(\Omega,d)$ is its ordinary calculus, how is $I$ related to $J$?

Answer 1

In the classical case, if $\Omega(A)$ is the kernel of the multiplication map $m:A\otimes A\to A$, then—since $A$ is commutative, so that $m$ is not only a map of $A$-bimodules but also a morphism of $k$-algebras,—it turns out that $\Omega(A)$ is an ideal of $A\otimes A$, not only a sub-$A$-bimodule. In particular, you can compute its square $(\Omega(A))^2$. Then the classical module of Kähler differentials $\Omega^1_{A/k}$ is the quotient $\Omega(A)/\Omega(A)^2$.

(This is the construction used by Grothendieck in EGA IV, for example)