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.

1. Generalities

Let ${A}$ be a commutative ring. Ultimately, we are going to think of the cotangent complex of an ${A}$-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 ${\mathbf{Alg}^{A}}$ of $A$-algebras does not exactly admit a nice abelianization functor. Recall:

Definition 1 If ${\mathcal{C}}$ is a category with finite products, then an abelian monoid object in ${\mathcal{C}}$ is an object ${X}$ together with a multiplication morphism ${\mu: X \times X \rightarrow X}$ and a unit ${e: \ast \rightarrow X}$ (where ${\ast}$ is the terminal object). These are required to satisfy the usual commutativity and associativity constraints. For instance,

$\displaystyle \mu \circ ( e \times 1): X \rightarrow \ast \times X \rightarrow X$

should be the identity.

The terminal object in the category ${\mathbf{Alg}^{A}}$ 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 ${\mathbf{Alg}_B^{A}}$

The remedy is going to work in the category ${\mathbf{Alg}_B^{A}}$ of ${A}$-algebras over ${B}$. An object of this category will be a ring ${R}$ together with maps ${i_R: A \rightarrow R, u_R: R \rightarrow B}$; a morphism will be a morphism of rings ${R \rightarrow R'}$ 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 ${\mathbf{Alg}_B^{A}}$ is the fibered product over ${B}$, we see that an abelian group object would consist of an ${A}$-algebra ${R}$ together with the map ${R \rightarrow B}$, along with morphisms of rings

$\displaystyle e: B \rightarrow R, \quad \mu: R \times_B R \rightarrow R$

that satisfy the usual relations.

With this in mind, let us give a simple example. Now let ${M}$ be any ${B}$-module, and consider the object ${X = B \oplus M \in \mathbf{Alg}_B^{A}}$. The structure morphism ${A \rightarrow X }$ just comes from the map ${A \rightarrow B}$, and the structure morphism ${X \rightarrow B}$ is projection on the first factor. Note that this is a ring satisfying ${M^2 = 0}$; we define multiplication via ${(b_1, m_1)(b_2, m_2) = (b_1 b_2, b_1 m_2 + b_2 m_1)}$. The claim is that ${X}$ is an abelian group object.

The section ${e: B \rightarrow X}$ is inclusion on the first factor, and the map

$\displaystyle \mu: X \times_B X \rightarrow X$

sends

$\displaystyle ((b, m_1) , (b, m_2)) \mapsto (b, m_1 + m_2).$

It is straightforward to check that this is indeed a morphism of algebras under ${A}$ and over ${B}$, and that ${\mu, e}$ satisfy the axioms to make ${X}$ into an abelian group object.

Proposition 2 There is a functor from ${B}$-modules to abelian group objects in ${\mathbf{Alg}_B^{A}}$ sending ${M \mapsto B \oplus M}$.

We can also see that ${X = B \oplus M}$ is an abelian group object by considering the functor it represents. For any object ${Y \in \mathbf{Alg}_B^{A}}$, we see that a map ${Y \rightarrow X}$ in this category is determined by a map of ${A}$-modules ${\phi: Y \rightarrow M}$ (since this is a map under ${B}$).

Let ${u_Y: Y \rightarrow B}$ be the structure morphism, so that the map ${Y \rightarrow X}$ is given by ${(u_Y, \phi)}$. This map ${\phi}$ then has to satisfy

$\displaystyle \phi( y_1 y_2) = u_Y(y_1)\phi(y_2) + u_Y(y_2) \phi(y_1).$

In other words, if we make ${M}$ into a ${Y}$-module via the map ${u_Y: Y \rightarrow B}$, then there is a natural equivalence

$\displaystyle \hom_{\mathbf{Alg}_B^{A}}(Y, X) \simeq \mathrm{Der}_A(Y, M).$

This is an abelian group (in an obvious manner), so ${X}$ 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 ${B}$-modules to abelian group objects in ${\mathbf{Alg}_B^{A}}$, ${M \mapsto B \oplus M}$, is an equivalence of categories.

Proof: Let us first check that this functor is fully faithful. It is obviously faithful. Suppose ${B \oplus M \rightarrow B \oplus N}$ is any morphism under ${A}$ and over ${B}$ which is a morphism of abelian group objects. Then this map sends ${M}$ into ${N}$ (because it is a morphism over ${B}$) and the associated map ${f: M \rightarrow N}$ is a group-homomorphism because the map ${B \oplus M \rightarrow B \oplus N}$ is a homomorphism of group objects.

So, by additivity, our map is of the form

$\displaystyle \phi: (b, m) \mapsto (b, f(m))$

where ${f}$ is some group-homomorphism. The claim is that ${f}$ is a homomorphism of ${B}$-modules. But

$\displaystyle \phi( b, bm) = \phi ( (1, m)(b, 0)) = \phi(1, m) \phi(b, 0) = (1, f(m)) (b, 0)$

so that

$\displaystyle (b, f(bm)) = (b, b f(m))$

proving the claim.

Now we need to show that the map is essentially surjective. So let ${A \rightarrow X \rightarrow B}$ be an abelian group object in ${\mathbf{Alg}_B^{A}}$. Then the structure map ${u_X: X \rightarrow B}$ has a section ${e: X \rightarrow B}$ (because ${e}$ is a morphism in ${\mathbf{Alg}_B^{A}}$). Let ${M = \ker u_X}$, so that we have a decomposition

$\displaystyle X = B \oplus M$

as ${A}$-modules. We know that the ${u_X: X \rightarrow B}$ corresponds to projection on the first factor via this decomposition. Now we claim that ${M}$ (which is obviously an ideal) squares to zero.

Let ${\mu: X \times_B X \rightarrow X}$ be the multiplication. Indeed, we have that one of the axioms of a group object is that

$\displaystyle \mu( u_B(x), x) = x, \quad x \in X$

because ${u_B: X \rightarrow B}$ is the unique map to the terminal object. If ${x \in M \subset X}$, then ${u_B(x) = 0}$ so that ${\mu( x,0) = x}$. Similarly ${\mu(0, y) = y}$.

Consequently

$\displaystyle xy = \mu ( x, 0) \mu(0, y) = \mu(0,0) = 0.$

It follows that ${M}$ has square zero. It is now easy to see that ${X}$ is of the given form; note that ${M}$ is automatically a ${B}$-module if ${B \oplus M}$ is a ring. We should just check that the multiplication map

$\displaystyle \mu: X \times_B X \rightarrow X$

sends ${\mu ( b+ m, b +m') ) \mapsto (b, m + m')}$. But ${\mu}$ is a homomorphism, and consequently this is ${\mu (b, b) + \mu(0, m) + \mu(m', 0) = b + m + m'}$, as we wanted.

3. Abelianization

We have now determined the category of abelian group objects in ${\mathbf{Alg}_B^{A}}$. Obviously there is a functor from abelian group objects in any category ${\mathcal{C}}$ to ${\mathcal{C}}$ 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 ${\mathbf{Alg}_B^{A}}$ of an object ${A \stackrel{i_X}{\rightarrow} X \stackrel{u_X}{\rightarrow}B}$ corresponds to the ${B}$-module ${\Omega_{X/A} \otimes_X B}$ (with the above equivalence of categories between ${B}$-modules and abelian group objects in ${\mathbf{Alg}_B^{A}}$).

Proof: Note that there is a morphism

$\displaystyle \psi_X: X \rightarrow B \oplus \Omega_{X/A} \otimes_X B$

sending

$\displaystyle x \mapsto (u_X(x), d x).$

Since the map ${ x \mapsto dx}$ is an ${A}$-derivation, this is indeed a morphism in ${\mathbf{Alg}_B^{A}}$.

Now suppose

$\displaystyle X \rightarrow B \oplus M$

is any map of ${X}$ into the object ${B \oplus M}$, where ${M}$ is a ${B}$-module. The claim is that it factors through ${\psi_X}$. But this is given by a map

$\displaystyle (u_X, \delta)$

where ${\delta: X \rightarrow M}$ is an ${A}$-derivation (${M}$ being considered as an ${X}$-module via ${X \rightarrow B}$). This map ${\delta}$ naturally factors uniquely through ${\Omega_{X/A}}$, and since ${M}$ is a ${B}$-module, it factors uniquely through ${\Omega_{X/A} \otimes_X B}$. From this the result is clear.