Ok, so now I want to talk about how Kahler differentials say something about regularity. Let ${A}$ be a local ring with maximal ideal ${\mathfrak{m}}$ and residue field ${k}$. Suppose moreover that ${A}$ is a ${k}$-algebra, i.e. that there is a section ${k \rightarrow A}$. For instance, a localization of a polynomial ring over an algebraically closed field at a maximal ideal counts, by the Nullstellensatz.

Anyway, we have an exact sequence

$\displaystyle \mathfrak{m} / \mathfrak{m}^2 \rightarrow k \otimes_A \ \Omega_{A/k} \rightarrow 0$

because ${\Omega_{k/k}=0}$, clearly. I claim now that in this case, the first map is injective (hence an isomorphism). Indeed, these are vector spaces over ${k}$, so it will be enough to prove the map ${\hom( \Omega_{A/k}, k) \rightarrow \hom(\mathfrak{m}/\mathfrak{m}^2, k)}$ is surjective. But given any map ${f: \mathfrak{m}/\mathfrak{m}^2 \rightarrow k}$, we can define a ${k}$-derivation ${D: A \rightarrow k}$ as follows: if ${x \in A}$, write ${x = a+b, a \in k, b \in \mathfrak{m}}$. Then ${D(x) := f(b)}$. One immediately checks that ${D}$ is a ${k}$-derivation, which proves that

$\displaystyle \mathfrak{m} / \mathfrak{m}^2 \simeq k \otimes_A \Omega_{A/k}.$

There is a more general criterion of when the first map in the “second exact sequence” (which I really should call the conormal sequence, so will do that) is a split injection. See the book by Eisenbud.

So, why do we care about this? Well, determining ${\dim_k(\mathfrak{m}/\mathfrak{m}^2)}$ is what tells us whether the ring ${A}$ is a regular local ring. As for why we care about regular local rings—the geometric interpretation of that is nonsingularity. Regular local rings are also UFDs, so as a result, when you have a nonsingular variety, it turns out that Weil divisors (this is the older notion of subvarieties of codimension 1) turn out to be equivalent to the fancier, more modern, and sheafier Cartier divisors.

But, ${\Omega_{A/k}}$ right now doesn’t appear very friendly. We will have to do somewhat more to see what is going on.

First, let’s take a look at what happens when ${A = O_x}$, the local ring of germs of smooth functions on a manifold: we find ${\Omega_{O_x/\mathbb{R}} \otimes \mathbb{R} \simeq m_x/m_x^2 \simeq T_x^*(M)}$, the cotangent space.

Localization and Kahler differentials

Kahler differentials behave about as well as one could expect with respect to localization:

Proposition 1 Let ${B}$ be an ${A}$-algebra, ${S \subset B}$ a multiplicative subset. Then$\displaystyle \Omega_{S^{-1}B/A} \simeq S^{-1} \Omega_{B/A}.$

I will show that both sides represent the same functor.

So, given a ${S^{-1}B}$-module ${M}$, I claim that any ${A}$-derivation ${B \rightarrow M}$ extends uniquely to a derivation ${S^{-1}B \rightarrow M}$. This will imply that we have functorial isomorphisms for ${S^{-1}B}$-modules ${M}$:

$\displaystyle \hom_{S^{-1}B} ( S^{-1} \Omega_{B/A}, M) \simeq \hom_B( \Omega_{B/A}, M) \simeq \hom_{S^{-1}B}( \Omega_{S^{-1}B/A}, M),$

and we can use Yonda’s lemma.

Given ${D}$, we extend it to ${S^{-1}B}$ via ${D(s^{-1}b) := s^{-2} ( b Ds - s Db)}$. We will check that it is well-defined, whence linearity will follow (in view of the product rule). Suppose that ${s^{-1}b = 0}$ in ${S^{-1}B}$. We will show that ${ s^{-2} ( b Ds - s Db) = 0}$ in ${M}$. Note that on ${M}$, elements of ${S}$ act by automorphisms. The first condition means that there is some ${t}$ with ${tb = 0}$ in ${B}$. Therefore,

$\displaystyle t( b Ds - sDb) = ts Db .$

We will now show that if ${b \in B}$ is annihilated by ${s}$, then ${Db =0 \in M}$. For this,

$\displaystyle 0 = s( D(bs) )= s( sDb + b Ds ) = s^2 Db,$

whence ${Db=0}$. As a result it follows that ${D}$ maps everything equivalent to ${0}$ into ${0}$. Also, the map ${(b,s) \rightarrow s^{-2} ( b Ds - s Db)}$ is linear on the abelian group of pairs ${(b,s)}$ with the usual addition ${(b,s) + (b',s') := (bs' + b's, ss')}$, so therefore, ${D}$ factors through as a derivation of ${S^{-1}B}$.

This proves the proposition.

So, how does this help in our case? We want to compute the rank of ${K \otimes_A \Omega_{A/k}}$, where ${K}$ is the quotient field of ${A}$. This will turn out to be the dimension of ${A}$ in certain special cases, as we will see soon. And this will yield the connection with regularity; we will compare the ranks of ${K \otimes \Omega_{A/k}}$ with ${k \otimes \Omega_{A/k}}$.