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

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

Where:
$S\otimes_R S\to S$ 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 $S$ (essentially a simplicial resolution of $S$ by free algebras over $A$) (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 $QS$?

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

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

Where:
$S\otimes_R S\to S$ 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 $QS$?

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

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

Where:
$S\otimes_R S\to S$ 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 simplicial modules of $S$ (with all of the homotopical baggage they entail) 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 homotopy equivalent to a "trivial simplicial extension"?

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

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

Where:
$S\otimes_R S\to S$ 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 $S$ (essentially a simplicial resolution of $S$ by free algebras over $A$) (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 $QS$?