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.

Definition 2 If the above lifting problem has a unique solution for every pair ${(D, J)}$, then ${B}$ is called formally etale as an ${A}$-algebra. If there is at most one solution, then ${B}$ is called formally unramified.

It’s not too hard to check that formally unramified is equivalent to ${\Omega_{B/A} = 0}$. See for instance my notes on the subject. (Ultimately, we are going to show that formal smoothness is equivalent to the cotangent complex being homotopy equivalent to the homology in dimension zero (which is ${\Omega_{B/A}}$) and the module ${\Omega_{B/A}}$ being projective.)

The basic example of a formally smooth ${A}$-algebra is the polynomial ring ${A[x_1, \dots, x_n]}$. For to give a map ${A[x_1, \dots, x_n] \rightarrow D/J}$ is to give ${n}$ elements of ${D/J}$; each of these elements can clearly be lifted to ${D}$. This is analogous to the statement that a free module is projective.

Now, ultimately, we want to show that this somewhat abstract definition of formal smoothness will give us something nice and geometric when ${B}$ is in addition of finite presentation. In particular, in this case we want to show that ${B}$ is flat. To do this, we will need to do a bit of work, but we can argue in a fairly elementary manner. On the one hand, we will need to give a criterion for when a quotient of a formally smooth ring is formally smooth.

2. Quotients of formally smooth rings

So now we need a result from EGA. I haven’t read the proof there, though; arguing directly seems simpler in the case we care about.

Theorem 3 (EGA 0-IV, 22.6.1) Let ${A}$ be a ring, ${B}$ an ${A}$-algebra. Suppose ${B}$ is formally smooth over ${A}$, and let ${I \subset B}$ be an ideal.Then ${C = B/I}$ is a formally smooth ${A}$-algebra if and only if the canonical map

$\displaystyle I/I^2 \rightarrow \Omega_{B/A} \otimes_B C$

has a section. In other words, ${C}$ is formally smooth precisely when the conormal sequence

$\displaystyle I/I^2 \rightarrow \Omega_{B/A} \otimes_B C \rightarrow \Omega_{C/A} \rightarrow 0$

is split exact.

EGA states this in more generality for topological rings, and uses some functors on ring extensions.

Proof: Suppose first ${C}$ is formally smooth over ${A}$. Then we have a map

$\displaystyle B/I^2 \rightarrow C$

given by the quotient. There is a diagram of ${A}$-algebras

and the lifting ${C \rightarrow B/I^2}$ exists by formal smoothness. This is a section of the natural projection ${B/I^2 \rightarrow C = B/I}$, which is a morphism of ${A}$-algebras. In particular, we get a splitting

$\displaystyle B/I^2 = B/I \oplus I/I^2$

from the exact sequence

$\displaystyle 0 \rightarrow I/I^2 \rightarrow B/I^2 \rightarrow C \rightarrow 0.$

Since the section of ${B/I^2 \rightarrow C}$ was a section of rings, we see that the splitting is a splitting of ${A}$-algebras, where ${I/I^2}$ squares to zero.

We are interested in showing that ${I/I^2 \rightarrow \Omega_{B/A} \otimes_B C}$ is a split injection of ${C}$-modules. To see this, we will show that any map out of the former extends to a map out of the latter. Now suppose given a map of ${C}$-modules

$\displaystyle \phi: I/I^2 \rightarrow M$

into a ${C}$-module ${M}$. Then we get a derivation

$\displaystyle \delta: B/I^2 \rightarrow M$

by using the splitting ${B/I^2 = C \oplus I/I^2}$. (Namely, we just extend the map by zero on ${C}$.) Since ${I/I^2}$ is imbedded in ${B/I^2}$ by the canonical injection, this derivation restricts on ${I/I^2}$ to ${\phi}$. In other words there is a commutative diagram

It follows thus that we may define, by pulling back, an ${A}$-derivation ${B \rightarrow M}$ that restricts on ${I}$ to the map ${I \rightarrow I/I^2 \stackrel{\phi}{\rightarrow} M}$. By the universal property of the differentials, this is the same thing as a homomorphism ${\Omega_{B/A} \rightarrow M}$, or equivalently ${\Omega_{B/A} \otimes_B C \rightarrow M}$ since ${M}$ is a ${C}$-module.

It follows that the map

$\displaystyle \hom_C(\Omega_{B/A} \otimes_B C, M) \rightarrow \hom_C(I/I^2, M)$

is a surjection. This proves one half of the result.

Now for the other. Suppose that there is a section. This translates, as above, to saying that any map ${I/I^2 \rightarrow M}$ (of ${C}$-modules) for a ${C}$-module ${M}$ can be extended to an ${A}$-derivation ${B \rightarrow M}$.

Now let ${E}$ be any ${A}$-algebra, and ${J \subset E}$ an ideal of square zero. We suppose given an ${A}$-homomorphism ${C \rightarrow E/J}$ and would like to lift it to ${C \rightarrow E}$; in other words, we must find a lift in the diagram

Let us pull this map back by the surjection ${B \twoheadrightarrow C}$; we get a diagram

In this diagram, we know that a lifting ${\phi: B \rightarrow E}$ does exist because ${B}$ is formally smooth over ${A}$. So we can find a dotted arrow from ${B \rightarrow E}$ in the diagram. The problem is that it might not send ${I = \ker(C \rightarrow B) }$ into zero. If it does, then we’re golden.

In any event, we have a morphism of ${A}$-modules ${ I \rightarrow E}$ given by restricting ${\phi: B \rightarrow E}$. This lands in ${J}$, so we get a map ${I \rightarrow J}$. Note that ${J}$ is an ${E/J}$-module, hence a ${C}$-module, because ${J}$ has square zero. Moreover ${I^2}$ gets sent to zero because ${J^2 = 0}$, and we have a morphism of ${C}$-modules ${I/I^2 \rightarrow J}$. Now by hypothesis, there is an ${A}$-derivation ${\delta: B \rightarrow J}$ such that ${\delta|_I = \phi}$. Since ${J}$ has square zero, it follows that

$\displaystyle \phi - \delta: B \rightarrow E$

is an ${A}$-homomorphism of algebras, and it kills ${I}$. Consequently this factors through ${C}$ and gives the desired lifting ${C \rightarrow E}$.

Corollary 4 If ${A \rightarrow B}$ is formally smooth, then ${\Omega_{B/A}}$ is a projective ${B}$-module.

Proof: Indeed, we can write ${B}$ as a quotient of a polynomial ring ${D}$ over ${A}$; this is formally smooth. Suppose ${B = D/I}$. Then we know that there is a split exact sequence

$\displaystyle 0 \rightarrow I/I^2 \rightarrow \Omega_{D/A} \otimes_D B \rightarrow \Omega_{B/A} \rightarrow 0.$

But the middle term is free as ${D/A}$ is a polynomial ring; hence the last term is projective.