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'}.

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