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 1 If
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 2 If
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 3 If
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 4 Let
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 5 Let
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 6
Given a towerwith
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 7
Given a towerwith
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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