In the stacks project, there is a 'naive version' of cotangent complex $NL_{S/R}$ for a ring morphism $R\rightarrow S$, given by the chan complex $(I/I^{2}\rightarrow \Omega_{R[S]/R}\otimes_{R[S]}S)$ where $I$ is the kernel of the morphism $R[S]\rightarrow S$.
I have several questions related with this 'naive cotangent complex':
$\textbf{Question 1:}$ In stacks, there is a claim that this naive chain complex is the $\textbf{1-truncation of cotangent complex $L_{S/R}$}$.
It really confuses me how this claim is proved: Let's first start with the simplicial module induced by the simplicial resolution of $S$ as $R$ algebra: $\cdots\Omega_{R[R[R[S]]]/R}\otimes_{R[R[R[S]]]}S\stackrel{d^{2}_{0},d^{2}_{1},d^{2}_{2}}{\longrightarrow}\Omega_{R[R[S]]/R}\otimes_{R[R[S]]}S\stackrel{d^{1}_{0},d^{1}_{1}}{\longrightarrow}\Omega_{R[S]/R}\otimes_{R[S]}S$, and this simplicial resolution corresponds to a chain complex by normalized chain functor, which looks like $\cdots \cap_{i=1,2}ker(d^{1}_{i})\stackrel{d_{0}^{2}}{\longrightarrow}\cap_{i=1}ker(d^{1}_{1})\stackrel{d_{1}^{0}}{\longrightarrow}\Omega_{R[S]/R}\otimes_{R[S]}S$. For the 1 truncation of this chain complex, by definition of truncation of chain complex, should be the $\cdots0\rightarrow coker(d_{0}^{2})\rightarrow \Omega_{R[S]/R}\otimes_{R[S]}S\rightarrow 0$.
Let's start with the calculation of 0 th homology group of the above chain complex, actually $H_{0}(L_{S/R})$ should be $\Omega_{S/R}$, and this is kind of alluded by the fact that $\Omega_{R[S]/R}$ is free $R[S]$ module generated by $ds$ with $s\in S$. Thus $H_{0}$ is the cokernel of $d_{1}^{0}:\cap_{i=1}ker(d^{1}_{1})\rightarrow\Omega_{R[S]/R}\otimes_{R[S]}S$ while $\Omega_{R[S]/R}\otimes_{R[S]}R\cong \oplus_{s\in S}R$ is again free $R$ module generated by basis $s$ (direct sum commutes with tensor product), so it suffices to show that the image of $d_{1}^{0}$ is in the form of $\{d[fg]\otimes 1-[f]d[g]\otimes 1-[g]d[f]\otimes 1\}$ along with $\{d[f+g]\otimes 1-d[f]\otimes 1-d[g]\otimes 1\}$ (under the transition of coefficient under tensor product, this is the relation imposed on Kähler differential form).
Indeed, this is true for the 0 the homology group: starting from canonical surjective ring morphism $R[S]\rightarrow S$, if we replace $S$ by $R[S]$, then we get $d_{1}^{0}$ whose kernel looks like $[[f]][[g]]-[[f]][[g]]$ and $[[f]+[g]]-[[f]]-[[g]]$, while $d_{1}^{1}$ is induced by $R[-]$ acting on $R[S]\rightarrow S$, with the kernel looking like $[[fg]]-[[f][g]]$ and $[[f]+[g]]-[[f+g]]$. And $\Omega_{R[S]/R}$ can be expressed in the form of $d(r[s])$ with a basis $d[s]$ plus $r$ linear relations and $d[s_{1}s_{2}]=s_{1}d[s_{2}]+s_{2}d[s_{1}]$, so is $\Omega_{R[R[S]]/R}$. Thus by the above description, the image of $d_{1}^{0}$ on $ker(d_{1}^{1})$ is exactly the ordinary relations imposed on the $\{ds\}_{s\in S}$.
$\textbf{Question 1.1:}$ Seriously, except this way, is there any simpler way to get the result $H^{0}(L_{S/R})\cong \Omega_{S/R}$?
$\textbf{Question 1.2:}$ The computation complexity grows when we try to apply the same procedure for the calculation of $H^{1}(L_{R/S})$, so how do we prove that $coker(d_{0}^{2}:\cap_{i=1,2}ker(d_{i}^{1})\rightarrow ker(d_{1}^{1}))$ is isomorphic to $I/I^{2}$?
I doubt this is a very simple thing, and there must be some better way other than using the tedious simplicial resolution along with normalized chain complex functor to get the result.
In the classical setting, the 'étale ring morphism' $F:A\rightarrow B$' is defined to be of finite presentation and the 'naive cotangent complex' is quasi-isomorphic to zero, while in the setting of simplicial commutative rings, 'étale ring morphism' is defined to be of finite presentation (in the simplicial sense: $\pi_{0}$ is of finite presentation and cotangent complex is perfect, which is quasi-isomorphic to bounded chain complex of projective modules), and also formally étale: the cotangent complex $L_{B/A}\cong 0$.
So it makes me confused that if we regard a discrete commutative ring morphism $f:R\rightarrow S$ as simplicial ring morphism, then the two definitions of étale ring morphism should coincide, but the discrete definition only requires 'naive cotangent complex $\tau_{\leq 1}L_{S/R}$ to be quasi-isomorphic to 0 while the simplicial version requires the whole cotangent complex $L_{S/R}$ to be quasi-isomorphic to 0, so my question is:
$\textbf{Question 2:}$How to prove the 'naive cotangent complex' version of 'étale ring morphism' condition are $\textbf{equivalent}$ to the 'cotangent complex' version for the classical ring morphism? And is it true that the 1-truncation of cotangent complex (that is, the 'naive cotangent complex') kind of $\textbf{'containing all the meaningful information'}$ for the classical rings as implied in the definition of 'étale morphism'?
I really think these are some fundamental facts and everyone talks about them as self-evident truths, but I really get stuck here. Any clues or references are very welcome!
