## Classical Calculi as Universal Quotients

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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 $J$?

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## 1 Answer

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)

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One way to think of this is that $\omega(A)$ is universal for derivations into arbitrary $A$-bimodules; the "classical" K&auml;hler module $\omega(A)/\omeg(A)^2$ is universal for derivations into symmetric (a.k.a. commutative) $A$-bimodules. So the change of category means one works with a different notion of "universal derivation". – Yemon Choi Nov 18 2009 at 4:08
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