

Anyway, we have an exact sequence








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}}.