I’d now like to begin a series of posts on the cotangent complex, following Daniel Quillen’s paper “Homology of Commutative Rings.” (There are also two very nice articles from a 2004 summer school on “Homotopy and algebra” on the subject, those by Goerss-Schemmerhorn and Iyengar, that discuss the topic.) While the cotangent complex can be defined quite cleanly once one has the appropriate categorical setting, it will be useful to spend a brief period formulating that.
Let be a commutative ring. Ultimately, we are going to think of the cotangent complex of an -algebra as a “linearization” or “abelianization.” Viewed more precisely, the cotangent complex will be the derived functor of abelianization (this is the general means of defining “homology” in a model category). It will turn out that abelianization will correspond to taking the module of Kähler differentials, so that the cotangent complex will also be a derived functor of those.
The problem is the category of -algebras does not exactly admit a nice abelianization functor. Recall:
Definition 1 If is a category with finite products, then an abelian monoid object in is an object together with a multiplication morphism and a unit (where is the terminal object). These are required to satisfy the usual commutativity and associativity constraints. For instance,
should be the identity.
The terminal object in the category is the zero ring, and it cannot map on any nonzero ring. So there are no nontrivial abelian group objects in this category!
2. Group objects in
The remedy is going to work in the category of -algebras over . An object of this category will be a ring together with maps ; a morphism will be a morphism of rings making the obvious diagrams commute. This category is going to have a large supply of abelian group objects.
Let us first analyze what such an object would be. Since the product in is the fibered product over , we see that an abelian group object would consist of an -algebra together with the map , along with morphisms of rings
that satisfy the usual relations.
With this in mind, let us give a simple example. Now let be any -module, and consider the object . The structure morphism just comes from the map , and the structure morphism is projection on the first factor. Note that this is a ring satisfying ; we define multiplication via . The claim is that is an abelian group object.
The section is inclusion on the first factor, and the map
It is straightforward to check that this is indeed a morphism of algebras under and over , and that satisfy the axioms to make into an abelian group object.
Proposition 2 There is a functor from -modules to abelian group objects in sending .
We can also see that is an abelian group object by considering the functor it represents. For any object , we see that a map in this category is determined by a map of -modules (since this is a map under ).
Let be the structure morphism, so that the map is given by . This map then has to satisfy
In other words, if we make into a -module via the map , then there is a natural equivalence
This is an abelian group (in an obvious manner), so must be an abelian group object, by general categorical nonsense. (It is in fact the same abelian group structure as that defined before, as one may check.)
We now want to show:
Theorem 3 The functor from -modules to abelian group objects in , , is an equivalence of categories.
Proof: Let us first check that this functor is fully faithful. It is obviously faithful. Suppose is any morphism under and over which is a morphism of abelian group objects. Then this map sends into (because it is a morphism over ) and the associated map is a group-homomorphism because the map is a homomorphism of group objects.
So, by additivity, our map is of the form
where is some group-homomorphism. The claim is that is a homomorphism of -modules. But
proving the claim.
Now we need to show that the map is essentially surjective. So let be an abelian group object in . Then the structure map has a section (because is a morphism in ). Let , so that we have a decomposition
as -modules. We know that the corresponds to projection on the first factor via this decomposition. Now we claim that (which is obviously an ideal) squares to zero.
Let be the multiplication. Indeed, we have that one of the axioms of a group object is that
because is the unique map to the terminal object. If , then so that . Similarly .
It follows that has square zero. It is now easy to see that is of the given form; note that is automatically a -module if is a ring. We should just check that the multiplication map
sends . But is a homomorphism, and consequently this is , as we wanted.
We have now determined the category of abelian group objects in . Obviously there is a functor from abelian group objects in any category to itself; the left adjoint, if it exists, is called abelianization. So the abelianization is the universal way of imbedding an object into an abelian group object.
Theorem 4 The abelianization in of an object corresponds to the -module (with the above equivalence of categories between -modules and abelian group objects in ).
Proof: Note that there is a morphism
Since the map is an -derivation, this is indeed a morphism in .
is any map of into the object , where is a -module. The claim is that it factors through . But this is given by a map
where is an -derivation ( being considered as an -module via ). This map naturally factors uniquely through , and since is a -module, it factors uniquely through . From this the result is clear.