Now, I need to discuss what happens for Kahler differentials when are fields. For simplicity, I am going to assume 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 -bases and whatnot.

So, first suppose is a finite algebraic extension.

Proposition 1If is of characteristic zero and is a finite algebraic extension, then .

To do this, we will show that for all . Now is separable over by characteristic zero, so there is a polynomial such that . As a result

by the product rule many times, whence , proving the result.

Corollary 2If is algebraic, then .

This is now easy to prove. Given , we must prove . Now is contained in some finite-dimensional subextension , and since , a fortiori the same is true in .

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 3If is the polynomial ring in variables over , then , where the generators of the free module may be given by .

may very well be infinite.

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

Now we look at the case of a transcendental extension.

Proposition 4Let be an extension where is of characteristic zero. Then for form an -basis for if and only if is a transcendence basis of .

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

Lemma 5Let for algebraically independent over . Then has the as a basis.

Indeed, is the quotient field of the polynomial ring , and we have already found to be -free on . Now since Kahler differentials commute with localization, we’re done.

Now back to the proposition. If is a transcendence basis, let be generated over by it; then there is a tower and a corresponding first exact sequence

where the term because is algebraic.

We can encapsulate this result in another lemma:

Lemma 6Given a tower with algebraic, there is a surjection

from the first exact sequence.

However, this only shows in the preceding proposition that the span , not that they are a basis. For this we need a strengthening of that lemma:

Lemma 7Given a tower with algebraic, there is an isomorphism

from the first exact sequence.

I will prove this in the case where 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 , and the conormal sequence yields

Now by the tensor product property, we have

where corresponds to multiples of , which implies that

The image of in is generated by , where is the image of in . This is a nonzero element of the second part of the direct sum, and generates that one-dimensional component. Whence

is an isomorphism.

Now let’s go back to the proposition again. We have proved that the are a basis for . We need to show the converse; assume forms a basis, and we will prove the are a transcendence basis. First, let be the subfield generated by the . We have an exact sequence

and since the image of contains all the , it follows that the first map is surjective, and . But what has alraedy been proved implies that means is algebraic.

Finally, it is necessary to show algebraic independence. But if were in the algebraic closure of , say, selected from the rest of the , then the same argument applied to would show that was a linear combination of .

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