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! (more…)

I’ll now say a few words on formal smoothness. This happens to be closely related to the theory of the cotangent complex (namely, the cotangent complex provides a clean criterion for when a morphism is formally smooth). Ultimately, I would like to aim first for the result that a formally smooth morphism of finite presentation is flat, and thus to characterize such morphisms via the geometric idea of “smoothness” (even though the algebraic version of formally smooth is pure commutative algebra).

1. What is formal smoothness?

The idea of a smooth morphism in algebraic geometry is one that is surjective on the tangent space, at least if one is working with smooth varieties over an algebraically closed field. So this means that one should be able to lift tangent vectors, which are given by maps from the ring into ${k[\epsilon]/\epsilon^2}$.

This makes the following definition seem more plausible:

Definition 1 Let ${B}$ be an ${A}$-algebra. Then ${B}$ is formally smooth if given any ${A}$-algebra ${D}$ and ideal ${I \subset D }$ of square zero, the map

$\displaystyle \hom_A(B, D) \rightarrow \hom_A(B, D/I)$

is a surjection.

So this means that in any diagram

there exists a dotted arrow making the diagram commute. (more…)

Finally, we’re going to come to the Kahler criterion for regularity.  As far as algebraic geometry is concerned, it states that a variety over an algebraically closed field of characteristic zero is nonsingular precisely when the sheaf of differentials on it (to be defined shortly) is locally free of rank equal to the dimension.

Theorem 1 Suppose ${A}$ is a local domain which is a localization of a finitely generated ${k}$-algebra for ${k}$ a field of characteristic zero, with residue field ${k}$. Then ${A}$ is a regular local ring if and only if ${\Omega_{A/k}}$ is a free ${A}$-module of rank ${\dim A}$.

First, I claim ${\Omega_{A/k}}$ is finitely generated. This follows because the corresponding claim is true for a polynomial ring, we have a conormal sequence implying it for finitely generated algebras over a field, and taking differentials commutes with localization.

Let ${K}$ be the residue field of ${A}$. I claim

$\displaystyle \boxed{ \dim k \otimes \Omega_{A/k} = \dim( \mathfrak{m}/\mathfrak{m}^2 ) , \quad \dim (K \otimes \Omega_{A/k}) = \dim A.}$

Then, the theorem will follow from the next lemma:

Lemma 2

Let ${M}$ be a finitely generated module over the local noetherian domain ${A}$, with residue field ${k}$ and quotient field ${K}$. Then ${M}$ is free iff$\displaystyle \dim K \otimes M = \dim k \otimes M$ (more…)

Now, I need to discuss what happens for Kahler differentials ${\Omega_{E/k}}$ when ${E,k}$ are fields. For simplicity, I am going to assume ${k}$ 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 ${p}$-bases and whatnot.

So, first suppose ${E/k}$ is a finite algebraic extension.

Proposition 1 If ${k}$ is of characteristic zero and ${E}$ is a finite algebraic extension, then ${\Omega_{E/k} = 0}$.

To do this, we will show that ${d\beta = 0}$ for all ${\beta \in E}$. Now ${\beta}$ is separable over ${k}$ by characteristic zero, so there is a polynomial ${P \in k[X]}$ such that ${P(\beta)=0, P'(\beta) \neq 0}$. As a result

$\displaystyle 0= d(P(\beta)) = P'(\beta) d \beta,$

by the product rule many times, whence ${d \beta = 0}$, proving the result.

Corollary 2 If ${E/k}$ is algebraic, then ${\Omega_{E/k}=0}$.

This is now easy to prove. Given ${\alpha \in E}$, we must prove ${d\alpha = 0 \in \Omega_{E/k}}$. Now ${\alpha}$ is contained in some finite-dimensional subextension ${F \subset E}$, and since ${d\alpha = 0 \in \Omega_{F/k}}$, a fortiori the same is true in ${\Omega_{E/k}}$. (more…)

Ok, so now I want to talk about how Kahler differentials say something about regularity. Let ${A}$ be a local ring with maximal ideal ${\mathfrak{m}}$ and residue field ${k}$. Suppose moreover that ${A}$ is a ${k}$-algebra, i.e. that there is a section ${k \rightarrow A}$. For instance, a localization of a polynomial ring over an algebraically closed field at a maximal ideal counts, by the Nullstellensatz.

Anyway, we have an exact sequence

$\displaystyle \mathfrak{m} / \mathfrak{m}^2 \rightarrow k \otimes_A \ \Omega_{A/k} \rightarrow 0$

because ${\Omega_{k/k}=0}$, clearly. I claim now that in this case, the first map is injective (hence an isomorphism). Indeed, these are vector spaces over ${k}$, so it will be enough to prove the map ${\hom( \Omega_{A/k}, k) \rightarrow \hom(\mathfrak{m}/\mathfrak{m}^2, k)}$ is surjective. But given any map ${f: \mathfrak{m}/\mathfrak{m}^2 \rightarrow k}$, we can define a ${k}$-derivation ${D: A \rightarrow k}$ as follows: if ${x \in A}$, write ${x = a+b, a \in k, b \in \mathfrak{m}}$. Then ${D(x) := f(b)}$. One immediately checks that ${D}$ is a ${k}$-derivation, which proves that

$\displaystyle \mathfrak{m} / \mathfrak{m}^2 \simeq k \otimes_A \Omega_{A/k}.$

There is a more general criterion of when the first map in the “second exact sequence” (which I really should call the conormal sequence, so will do that) is a split injection. See the book by Eisenbud.

So, why do we care about this? Well, determining ${\dim_k(\mathfrak{m}/\mathfrak{m}^2)}$ is what tells us whether the ring ${A}$ is a regular local ring. As for why we care about regular local rings—the geometric interpretation of that is nonsingularity. Regular local rings are also UFDs, so as a result, when you have a nonsingular variety, it turns out that Weil divisors (this is the older notion of subvarieties of codimension 1) turn out to be equivalent to the fancier, more modern, and sheafier Cartier divisors.

But, ${\Omega_{A/k}}$ right now doesn’t appear very friendly. We will have to do somewhat more to see what is going on.

First, let’s take a look at what happens when ${A = O_x}$, the local ring of germs of smooth functions on a manifold: we find ${\Omega_{O_x/\mathbb{R}} \otimes \mathbb{R} \simeq m_x/m_x^2 \simeq T_x^*(M)}$, the cotangent space. (more…)

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

Let ${A}$ be a commutative ring, ${B}$ an ${A}$-algebra, and ${M}$ a ${B}$-module. Then an ${A}$-derivation of ${B}$ in ${M}$ is a linear map ${D: B \rightarrow M}$ satisfying ${D(a)=0}$ for ${a \in A}$ and

$\displaystyle D(bb') = (Db) b' + b (Db').$

The set of all such derivations forms a ${B}$-module ${\mathrm{Der}_A(B,M)}$. If we regard this as a set, clearly, we have a contravariant functor

$\displaystyle \mathrm{Der}_A(B, -): B-\mathrm{mod} \rightarrow \mathrm{Set}$

because if ${B \rightarrow B'}$ is a homomorphism of ${A}$-algebras, we can pull back a derivation.

Before proceeding, I should say something about the canonical example. Let ${M}$ be a smooth manifold and ${O_x}$ the local ring (of germs of smooth functions) at ${x \in M}$. Then ${\mathbb{R}}$ becomes an ${O_x}$-module if the germ ${f}$ acts by multiplication by ${f(x)}$. More precisely, we have an exact sequence

$\displaystyle 0 \rightarrow m_x \rightarrow O_x \rightarrow \mathbb{R} \rightarrow 0$

for ${m_x \subset O_x}$ the maximal ideal of functions vanishing at ${x}$, and this is the way ${\mathbb{R}}$ is an ${O_x}$-module.

Anyway, an ${\mathbb{R}}$-derivation ${O_x \rightarrow \mathbb{R}}$ is just a tangent vector at ${x}$.

Now back to the algebraic theory. It turns out that the functor ${\mathrm{Der}_A}$ is representable. In other words, for each ${A}$-algebra ${B}$, there is a ${B}$-module ${\Omega_{B/A}}$ such that

$\displaystyle \hom_B( \Omega_{B/A}, M) \simeq \mathrm{Der}_A(B,M) ,$

the isomorphism being functorial. In addition, there must be a “universal” derivation ${d: B \rightarrow \Omega_{B/A}}$ (corresponding to the identity ${\Omega_{B/A} \rightarrow \Omega_{B/A}}$ in the above functorial isomorphism), that any derivation factors through.

The construction of ${\Omega_{B/A}}$ is straightforward. We define it as the ${B}$-module generated by symbols ${db, b\in B}$, modulo the relations ${da = 0}$ for ${a \in A}$, ${d(b+b') = db + db'}$, and ${d(bb') = b' db + b db'}$. It is now clear that we have a functorial isomorphism as above. Now, ${\Omega_{B/A}}$ is called the module of Kahler differentials of ${B}$ over ${A}$. (more…)