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 fancily, I could also say that “Kahler differentials commute with colimits” in a certain sense.

For the next result on transcendental extensions, we will need to have a computation of Kahler differentials for polynomial rings

Proposition 3 If ${S = R[X_i]}$ is the polynomial ring in variables ${X_i, i \in I}$ over ${R}$, then ${\Omega_{S/R} \simeq \bigoplus_I S}$, where the generators of the free module ${S^n}$ may be given by ${dX_i, i \in I}$.

${I}$ may very well be infinite.

To do this, we just need to describe what a derivation out of ${S}$ looks like. If we can show that a derivation ${D}$ out of ${S}$ into some ${S}$-module can be given by arbitrarily precribing ${n}$ values for each ${X_i}$, we will be done (by Yoneda’s lemma). But this is easy to check. If ${D(X_i)}$ is given, then ${D}$ at any polynomial is determined by the product rule. This determination can be checked to satisfy the product rule, so ${D}$ as defined becomes a derivation.

Now we look at the case of a transcendental extension.

Proposition 4 Let ${E/k}$ be an extension where ${k}$ is of characteristic zero. Then ${dt_i, i \in I}$ for ${t_i \in E}$ form an ${E}$-basis for ${\Omega_{E/k}}$ if and only if ${\{t_i\}}$ is a transcendence basis of ${E/k}$.

To start off, we’ll compute Kahler differentials for a pure transcendental extension.

Lemma 5 Let ${E = k(t_i)}$ for ${t_i}$ algebraically independent over ${k}$. Then ${\Omega_{E/k}}$ has the ${dt_i}$ as a basis.

Indeed, ${E}$ is the quotient field of the polynomial ring ${R = k[X_i]}$, and we have already found ${\Omega_{R/k}}$ to be ${R}$-free on ${dX_i}$. Now since Kahler differentials commute with localization, we’re done.

Now back to the proposition. If ${\{ t_i\}}$ is a transcendence basis, let ${F}$ be generated over ${k}$ by it; then there is a tower ${k \subset F \subset E}$ and a corresponding first exact sequence

$\displaystyle E \otimes_F \Omega_{F/k} \rightarrow \Omega_{E/k} \rightarrow \Omega_{E/F} \rightarrow 0$

where the term ${\Omega_{E/F}=0}$ because ${E/F}$ is algebraic.

We can encapsulate this result in another lemma:

Lemma 6

Given a tower ${k \subset F \subset E}$ with ${E/F}$ algebraic, there is a surjection$\displaystyle E \otimes_F \Omega_{F/k} \rightarrow \Omega_{E/k}$

from the first exact sequence.

However, this only shows in the preceding proposition that the ${dt_i}$ span ${\Omega_{E/k}}$, not that they are a basis. For this we need a strengthening of that lemma:

Lemma 7

Given a tower ${k \subset F \subset E}$ with ${E/F}$ algebraic, there is an isomorphism$\displaystyle E \otimes_F \Omega_{F/k} \rightarrow \Omega_{E/k}$

from the first exact sequence.

I will prove this in the case where ${E/F}$ is finite; the general case follows by taking colimits. (I’m going to handwave away the step that this is completely justified.)

In this case, we can write ${E = F[X]/(P)}$, and the conormal sequence yields

$\displaystyle (P)/(P^2) \rightarrow \Omega_{F[X]/k}/ P \Omega_{F[X]/k} \rightarrow \Omega_{E/k} \rightarrow 0.$

Now by the tensor product property, we have

$\displaystyle \Omega_{F[X]/k} \simeq F[X] \otimes_F \Omega_{F/k} \oplus F[X] ,$

where ${F[X]}$ corresponds to multiples of ${dX}$, which implies that

$\displaystyle \Omega_{F[X]/k}/ P \Omega_{F[X]/k} \simeq E \otimes_F \Omega_{F /k} \oplus E .$

The image of ${(P)/(P^2)}$ in ${\Omega_{F[X]/k}/ P \Omega_{F[X]/k}}$ is generated by ${P'(\alpha) d\alpha}$, where ${\alpha}$ is the image of ${X}$ in ${E}$. This is a nonzero element of the second part of the direct sum, and generates that one-dimensional component. Whence

$\displaystyle E \otimes_F \Omega_{F/k} \rightarrow \Omega_{E/k}$

is an isomorphism.

Now let’s go back to the proposition again. We have proved that the ${dt_i}$ are a basis for ${\Omega_{E/k}}$. We need to show the converse; assume ${\{dt_i, i \in I\}}$ forms a basis, and we will prove the ${t_i}$ are a transcendence basis. First, let ${F}$ be the subfield generated by the ${t_i}$. We have an exact sequence

$\displaystyle E \otimes_F \Omega_{F/k} \rightarrow \Omega_{E/k} \rightarrow \Omega_{E/F} \rightarrow 0$

and since the image of ${E \otimes_F \Omega_{F/k}}$ contains all the ${dt_i}$, it follows that the first map is surjective, and ${\Omega_{E/F}=0}$. But what has alraedy been proved implies that ${\Omega_{E/F}=0}$ means ${E/F}$ is algebraic.

Finally, it is necessary to show algebraic independence. But if ${t_q}$ were in the algebraic closure of ${t_1', \dots, t_n'}$, say, selected from the rest of the ${t_i}$, then the same argument applied to ${k(t_1', \dots, t_n', t_q)}$ would show that ${dt_q}$ was a linear combination of ${dt_1', \dots, dt_n'}$.