Now, I need to discuss what happens for Kahler differentials {\Omega_{E/k}} when {E,k} are fields. For simplicity, I am going to assume {k} is characteristic zero. This is very much a cop-out. But for the sake of time, I am going to do it this way, and to please the Bourbakistas  do things generally, I can always come back to talk about {p}-bases and whatnot.

So, first suppose {E/k} is a finite algebraic extension.

Proposition 1 If {k} is of characteristic zero and {E} is a finite algebraic extension, then {\Omega_{E/k} = 0}.


To do this, we will show that {d\beta = 0} for all {\beta \in E}. Now {\beta} is separable over {k} by characteristic zero, so there is a polynomial {P \in k[X]} such that {P(\beta)=0, P'(\beta) \neq 0}. As a result

\displaystyle 0= d(P(\beta)) = P'(\beta) d \beta,

by the product rule many times, whence {d \beta = 0}, proving the result.

Corollary 2 If {E/k} is algebraic, then {\Omega_{E/k}=0}.


This is now easy to prove. Given {\alpha \in E}, we must prove {d\alpha = 0 \in \Omega_{E/k}}. Now {\alpha} is contained in some finite-dimensional subextension {F \subset E}, and since {d\alpha = 0 \in \Omega_{F/k}}, a fortiori the same is true in {\Omega_{E/k}}. (more…)