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…)