I'm reading Daniel Quillen's paper "Homology of commutative rings," in which he proves:


A finitely presented morphism of rings $A \to B$ is 


1. Formally etale iff $L\_{B/A}$ (this denotes the cotangent complex) is homotopy-equivalent to zero
2. Formally smooth iff $\Omega\_{B/A}$ is projective and $L\_{B/A}$ is homotopy-equivalent to it (i.e. acyclic outside degree zero).
3. (It's also true, and elementary (not in the paper), that formally unramified iff $\Omega\_{B/A} = 0$.)


I've heard that these results are true even without finitely presented hypotheses. In fact, I understand that the fpqc localness of projectivity is what one uses to show that formal smoothness is, in fact, a local property (cf. 2 above). I also know how to show that if $A \to B$ is formally smooth, then the differentials are a projective $B$-module (take a quotient by some polynomial ring, $B = C/I$, and show that the sequence $I/I^2 \to \Omega\_{C/A} \otimes\_C B \to \Omega\_{B/A} \to 0$ is actually split exact).
In fact, one can show that if $C$ is a formally smooth $A$-algebra and $B = C/I$, then $B$ is smooth iff the conormal sequence above is split exact.


So I am guessing that there is an extension of the conormal sequence to the cotangent complex, and if this works will prove 2 without finitely presented hypotheses (and thus 1 as well). How does this "long exact sequence" work?



Formally smooth morphisms, the cotangent complex, and an extension of the conormal sequence
We deduce this from the Jacobi-Zariski exact sequence as follows:


Some notation first: Let $H\_i(A,B,W)$ where B is an A-algebra and W is a B-module denote $\pi\_i(L\_{B/A}\otimes\_B W)$. 


Then given an $A$-algebra $B$, a $B$-algebra $C$, and a $C$-module $W$, we have a long-exact sequence, called the Jacobi-Zariski sequence:
$$\dots \to H\_n(A,B,W)\to H\_n(A,C,W) \to H\_n(B,C,W)\to H\_{n-1}(A,B,W)\to \dots \to H\_0(B,C,W)\to 0$$. 


Now, let $C$ be an $A$-algebra, let $B$ be a polynomial ring over $A$ with an ideal $I$ such that $B/I\cong C$. 


Then we have the following long-exact sequence: 


$$\dots \to H\_n(A,B,C)\to H\_n(A,C,C) \to H\_n(B,C,C)\to H\_{n-1}(A,B,C)\to \dots \to H\_0(B,C,C)\to 0$$.


Since $C$ is a quotient of $B$, it is formally unramified over $B$, so $H\_0(B,C,C)\cong \Omega\_{C/B}=0$. We also see that $I/I^2$ is precisely $H\_1(B,C,C)$ by Proposition 1 of Chapter VI.a of the book *Homologie des algèbres commuatatives* by Michel André. This gives us a long-exact sequence 


$$\dots \to H\_n(A,B,C)\to H\_n(A,C,C) \to H\_n(B,C,C)\to H\_{n-1}(A,B,C)\to \dots $$ 
$$\to H\_1(A,B,C) \to H\_1(A,C,C)\to I/I^2 \to H\_0(A,B,C)\to H\_0(A,C,C)\to 0$$.


We see that the conormal exact sequence is the truncation at $I/I^2$, and that this sequence splits if $H\_0(A,C,C)$ is projective and $H\_1(A,C,C)$ is $0$. Conversely, if the sequence is split-exact, then $H\_0(A,C,C)$ is a direct summand of $H\_0(A,B,C)$, which is projective (it might be free, but I don't remember), and the kernel of $I/I^2\to H\_0(A,B,C)$ is $0$, which combined with the fact that $H\_1(A,B,C)=0$ implies that $H\_1(A,C,C)=0$. 


