Kahler differentials

I now want to talk about some of the material in Hartshorne, II.8.  First, we need some preliminaries from commutative algebra.

Let be a commutative ring, an -algebra, and a -module. Then an -derivation of in is a linear map satisfying for and

The set of all such derivations forms a -module . If we regard this as a set, clearly, we have a contravariant functor

because if is a homomorphism of -algebras, we can pull back a derivation.

Before proceeding, I should say something about the canonical example. Let be a smooth manifold and the local ring (of germs of smooth functions) at . Then becomes an -module if the germ acts by multiplication by . More precisely, we have an exact sequence

for the maximal ideal of functions vanishing at , and this is the way is an -module.

Anyway, an -derivation is just a tangent vector at .

Now back to the algebraic theory. It turns out that the functor is representable. In other words, for each -algebra , there is a -module such that

the isomorphism being functorial. In addition, there must be a “universal” derivation (corresponding to the identity in the above functorial isomorphism), that any derivation factors through.

The construction of is straightforward. We define it as the -module generated by symbols , modulo the relations for , , and . It is now clear that we have a functorial isomorphism as above. Now, is called the module of Kahler differentials of over .

I will now prove some basic properties of Kahler differentials.

Tensor products

First, linearizes tensor products. In detail, if are -algebras, then

To see this, we will consider what a -derivation of in a -module looks like. It induces maps given by , which are themselves -derivations. Conversely, given a pair of such -derivations , we can define an -derivation on by

In particular, we find that for any -module ,

In particular, we have a functorial isomorphism:

Since is a -module as well, the last functor is naturally isomorphic to

and since belongs to the category of -modules, Yoneda’s lemma completes the proof.

First exact sequence: chains of rings

Differentials behave well with respect to changing rings:

Theorem 1

Let be a sequence of homomorphisms of commutative rings. Then there is an exact sequence of -modules:

 

 

The first map sends to , where is the ring-homomorphism. The second sends .

Anyway, to prove this, we’ll use a categorical argument. Let be a -module. I claim there is an exact sequence

Here the maps are the obvious one: a -derivation of is automatically an -derivation, and an -derivation of restricts to an -derivation of . The image of consists of derivations that vanish on , so it is the kernel of the last map, which establishes exactness. This is an exact sequence

or equivalently, an exact sequence

which as one may check, comes from the sequence defined above

which is therefore exact, being arbitrary.

The second exact sequence: quotients

Now suppose is an ideal in the -algebra .

 

Theorem 2

There is an exact sequence of modules

 

Here the first map is .

Note that is a -module. Again, we shall prove this by abstract nonsense. Let be an -module. Then, I claim there is an exact sequence

The first map here, , is the obvious pull-back of a derivation. The second is slightly more subtle: note that any derivation of in vanishes on because is annihilated by . Also, the restriction is actually a -homomorphism because

for the same reason. So, the sequence is defined. We will now prove it exact.

Exactness at the first step is clear. Also, it is clear that the composition of the last two maps is zero. Finally, any derivation of that restricts to on is a derivation of , evidently.

Now, writing , we see that this becomes an exact sequence, functorial in ,

which is seen to be induced by the sequence

which is therefore exact.

Next time, I’ll say something about how differentials deal with localization, maybe something on separability, and then onto sheaves of differentials.