Recall that an $R$-algebra $R\to S$ is called formally smooth (resp. formally unramified resp. formally étale) if given any lifting problem of the form

$$\begin{matrix} R&\to &T\\ \downarrow&{}^?\nearrow&\downarrow\\ S&\to&T/J\end{matrix}$$

where $J$ is a square-zero nilpotent ideal of $T$, there exists at least one (resp. at most one, resp. exactly one) lift $S\to T$ making the diagram commute.

In Quillen's paper "Homology of Commutative Rings", he cites the following construction (Not sure who came up with it, but my copy of the paper has "Hopf?" written next to the relevant proposition):

Given an $R$-algebra $S$ and an $S$-module $M$, we let $S\oplus M$ be the commutative ring on the underlying $S$-module $S\oplus M$ with the multiplication given by the composite

$$(S\oplus M)\otimes_R (S\oplus M)\cong (S\otimes_R S) \oplus (S\otimes_R M) \oplus (M\otimes_R S) \oplus (M\otimes_R M)\to S\times M\cong S\oplus M$$

$S\otimes_R S\to S$ is the multiplication of $S$
$S\otimes_R M\to M$ is the left action of $S$ on $M$
$M\otimes_R S\to M$ is the right action of $S$ on $M$
$M\otimes_R M\to M$ is the zero map $M\otimes_R M\to 0\to M$

This ring is commutative, unital, and is equipped with a canonical $R$-linear projection $\phi:S\oplus M\to S$ with kernel $S$-linearly isomorphic to $M$. Also notice that $ker(\phi)$ is square-zero.

Interestingly, given any $R$-algebra $A$ extending $S$, i.e. $A\in (R\operatorname{-Alg}\downarrow S)$, it is easily verified that $\operatorname{Hom}_{(R\operatorname{-Alg}\downarrow S)}(A,S\oplus M)\cong \operatorname{Der}_R(A,Res_{A\to S}(M))$, that is, lifts $A\to S\oplus M$ are in bijective correspondence with derivations $A\to Res_{A\to S}(M)$ where $Res_{A\to S}$ is the restriction of scalars by the map $A\to S$.

In particular, it seems, at least to my untrained eye, that there's a connection between this construction and the definition of formally smooth (resp. formally unramified resp. formally étale) algebras. The lifts we see, at least when $T$ is of the form $U\oplus M$ for some $U$-module $M$ correspond exactly to $R$-derivations.

Quillen then proves that the exact objects corresponding to the modules of the form $S\oplus M$ are the abelian group objects of the category $(R\operatorname{-Alg}\downarrow S)$. In particular (back to our original notation), this does not necessarily hit all square-zero extensions of $T/J$, since it misses those square-zero extensions that don't admit a section $T/J\to T$. Since the module of relative Kähler differentials carries all of the data of these derivations, this appears to explain exactly why Kähler differentials are only sufficient to characterize formally unramified algebras.

So far, I follow.

So here's the question: Why, morally, do we need to look at the $S$-modules over the cofibrant replacement of $S$ (we can give a functorial cofibrant replacement by looking at a simplicial resolution of $S$ by defining a simplicial ring consisting of free algebras over $A$ in every degree (c.f. André 1974)) (with all of the homotopical baggage we need to characterize the model structure) to capture the lifting data from the rest of the square-zero extensions that we would need to characterize formal smoothness (resp. formal étaleness)?

That is, similar to how every trivial extension is of the form $S\oplus M$ for an $S$-module M, can we give a homotopical version of the trivial extension such that the nilpotent extensions of $S$ are exactly those that that are trivial over some cofibrant replacement $QS$?

  • $\begingroup$ (I think that the reason why we can characterize unramified morphisms in this way is as follows (I don't remember the argument, but I could work it out): I think we can always find a $T/J$-module $M$ such that $T$ is a subring of $T/J\oplus M$, so if the claim holds for square-zero extensions of $T/J$ representing modules, it implies the general case by the fact that inclusions are monic.) $\endgroup$ – Harry Gindi Nov 10 '10 at 9:07
  • $\begingroup$ (I'm 99% sure that the module in question is $T/J\otimes_T \Omega_{T/R}$) $\endgroup$ – Harry Gindi Nov 10 '10 at 9:35
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    $\begingroup$ The construction on $R\oplus M$ is not due to Quillen. It goes back at least to Dorroh in the 30s, was named the 'trivial extension' by Hochschild in the 50s, and was popularized in commutative algebra by Nagata's book. $\endgroup$ – Graham Leuschke Nov 10 '10 at 11:45
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    $\begingroup$ You can put his name back in with a suitable invocation of "André-Quillen". $\endgroup$ – S. Carnahan Nov 10 '10 at 15:06
  • $\begingroup$ I may have neglected to point out that an extension of the form $S\oplus M$ is called a trivial extension of $S$, and that these can be identified exactly with the nilpotent extensions of $S$ admitting a section that is a morphism of $A$-algebras. See $EGA0_{IV}.18$ or M. André's book Homologie des Algèbres Commutatives Ch. 16. $\endgroup$ – Harry Gindi Nov 13 '10 at 22:14

Your question is:

Why, morally, do we need to look at the S-modules over the cofibrant replacement of S to capture the lifting data from the rest of the square-zero extensions that we would need to characterize formal smoothness (resp. formal étaleness)?

Here's how I think about it, though I don't know if it addresses what's bothering you about this.

You'd like to understand surjective maps of rings $f:T\to T'$ whose kernel $M$ (an $S$-module) is square-zero, but where $f$ might not admit a section.

There's a universal example of such an extension (in $R$-algebras over $S$); it lives in simplicial rings, i.e., in $(s(R\operatorname{-Alg})\downarrow S)$. Namely, form the simplicial $S$-module $BM$ by shifting (so $\pi_1BM=M$, $\pi_iBM=0$ for $i\neq1$), then consider the simplicial ring $S\oplus BM$, formed with trivial multiplication on the $BM$ part.

Now let $g: S\to S\oplus BM$ be the evident inclusion map into the first factor. The "kernel" of this map of simplicial rings is the homotopy fiber of $g$ (calculated in simplicial $S$-modules), and this homotopy fiber is equivalent to $M$ (concentrated in dimension $0$). So $g$ has "kernel" $M$.

The claim is that isomorphism classes of square-zero extensions (possibly not split) over $S$ is the same as homotopy classes of maps $S\to S\oplus BM$ in $(s(R\operatorname{-Alg})\downarrow S)$. Calculating homotopy classes requires using a cofibrant model for the domain $S$. (In fact, there are no nontrivial maps $S\to S\oplus BM$ at all.)

Given such a map $f:QS\to S\oplus BM$ from a cofibrant replacement, pull back $g$ along it to get an extension $g'\colon E\to QS$. The simplcial ring $E$ has homotopy concentrated in degree $0$, and in fact there's an exact sequence $0\to M\to \pi_0 E\to \pi_0 QS=S$.

(Note that $(s(R\operatorname{-Alg})\downarrow QS)$ and $(s(R\operatorname{-Alg})\downarrow S)$ are Quillen equivalent model categories, so in fact giving an "extension" over $QS$ is the same as giving one for $S$.)

This story is morally just like the one you can tell for extension of $R$-modules: to compute extensions of $M$ by $N$, note that there's a universal extension by $N$ in the derived category of $R$-modules, of the form $N\to C\to N[1]$ (where $H_*C=0$). So extensions over $M$ are $\mathrm{Hom}_{D(R)}(M, N[1])$, and you can compute this group by replacing $M$ with a projective resolution (i.e., cofibrant replacement).

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  • $\begingroup$ Super awesome answer. Thanks a million. It addressed exactly what I wanted to know! $\endgroup$ – Harry Gindi Nov 13 '10 at 23:19
  • $\begingroup$ Dear Charles, are you familiar with any books or papers that go into some more detail concerning these universal extensions? $\endgroup$ – Harry Gindi Nov 20 '10 at 12:20

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