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?


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 $H_1(A,C,C)=0$.

  • $\begingroup$ Very nice. So I see that the conormal sequence is really a special case of the Jacobi-Zariski sequence (or the cofiber sequence of cotangent complexes). Thanks! $\endgroup$ – Akhil Mathew May 10 '11 at 19:26
  • $\begingroup$ Dear Harry, do you know why the connecting homomorphism in the Jacobi-Zariski sequence in this case is the same as the usual map $I/I^2 \to \Omega_{B/A} \otimes C$? (BCnrd had brought this up.) $\endgroup$ – Akhil Mathew May 12 '11 at 0:43

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