hcju/stacksqa huggingface dataset:
Assume $R \to S$ is formally smooth. Consider the commutative diagram $$ \xymatrix{ S \ar[r] \ar@{-->}[rd] & P/J \\ R \ar[r] \ar[u] &  P/J^2\ar[u] } $$ By assumption the dotted arrow exists. This proves that $\sigma$ exists. \medskip\noindent Conversely, suppose we have a $\sigma$ as in the lemma. Let a solid diagram $$ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } $$ as in Definition \ref{definition-formally-smooth} be given. Because $P$ is formally smooth by Lemma \ref{lemma-polynomial-ring-formally-smooth}, there exists an $R$-algebra homomorphism $\psi : P \to A$ which lifts the map $P \to S \to A/I$. Clearly $\psi(J) \subset I$ and since $I^2 = 0$ we conclude that $\psi(J^2) = 0$. Hence $\psi$ factors as $\overline{\psi} : P/J^2 \to A$. The desired dotted arrow is the composition $\overline{\psi} \circ \sigma : S \to A$.
Assume $S$ is formally smooth over $R$. By Lemma \ref{lemma-characterize-formally-smooth} this means there exists an $R$-algebra map $S \to P/J^2$ which is a right inverse to the canonical map $P/J^2 \to S$. By Lemma \ref{lemma-differential-seq-split} we see that the sequence is split. \medskip\noindent Assume the exact sequence of the lemma is split exact. Choose a splitting $\sigma : \Omega_{S/R} \to \Omega_{P/R} \otimes_R S$. For each $\lambda \in S$ choose $x_\lambda \in P$ which maps to $\lambda$. Next, for each $\lambda \in S$ choose $f_\lambda \in J$ such that $$ \text{d}f_\lambda = \text{d}x_\lambda - \sigma(\text{d}\lambda) $$ in the middle term of the exact sequence. We claim that $s : \lambda \mapsto x_\lambda - f_\lambda \mod J^2$ is an $R$-algebra homomorphism $s : S \to P/J^2$. To prove this we will repeatedly use that if $h \in J$ and $\text{d}h = 0$ in $\Omega_{P/R} \otimes_R S$, then $h \in J^2$. Let $\lambda, \mu \in S$. Then $\sigma(\text{d}\lambda + \text{d}\mu - \text{d}(\lambda + \mu)) = 0$. This implies $$ \text{d}(x_\lambda + x_\mu - x_{\lambda + \mu} - f_\lambda - f_\mu + f_{\lambda + \mu}) = 0 $$ which means that $x_\lambda + x_\mu - x_{\lambda + \mu} - f_\lambda - f_\mu + f_{\lambda + \mu} \in J^2$, which in turn means that $s(\lambda) + s(\mu) = s(\lambda + \mu)$. Similarly, we have $\sigma(\lambda \text{d}\mu + \mu \text{d}\lambda - \text{d}\lambda \mu) = 0$ which implies that $$ \mu(\text{d}x_\lambda - \text{d}f_\lambda) + \lambda(\text{d}x_\mu - \text{d}f_\mu) - \text{d}x_{\lambda\mu} + \text{d}f_{\lambda\mu} = 0 $$ in the middle term of the exact sequence. Moreover we have $$ \text{d}(x_\lambda x_\mu) = x_\lambda \text{d}x_\mu + x_\mu \text{d}x_\lambda = \lambda \text{d}x_\mu + \mu \text{d} x_\lambda $$ in the middle term again. Combined these equations mean that $x_\lambda x_\mu - x_{\lambda\mu} - \mu f_\lambda - \lambda f_\mu + f_{\lambda\mu} \in J^2$, hence $(x_\lambda - f_\lambda)(x_\mu - f_\mu) - (x_{\lambda\mu} - f_{\lambda\mu}) \in J^2$ as $f_\lambda f_\mu \in J^2$, which means that $s(\lambda)s(\mu) = s(\lambda\mu)$. If $\lambda \in R$, then $\text{d}\lambda = 0$ and we see that $\text{d}f_\lambda = \text{d}x_\lambda$, hence $\lambda - x_\lambda + f_\lambda \in J^2$ and hence $s(\lambda) = \lambda$ as desired. At this point we can apply Lemma \ref{lemma-characterize-formally-smooth} to conclude that $S/R$ is formally smooth.
Assume (1), (2) and (3). Let $P = R[\{x_t\}_{t \in T}] \to S$ be a surjection of $R$-algebras with kernel $J$. Thus $0 \to J \to P \to S \to 0$ is a short exact sequence of flat $R$-modules. This implies that $I \otimes_R S = IS$, $I \otimes_R P = IP$ and $I \otimes_R J = IJ$ as well as $J \cap IP = IJ$. We will use throughout the proof that $$ \Omega_{(S/IS)/(R/I)} = \Omega_{S/R} \otimes_S (S/IS) = \Omega_{S/R} \otimes_R R/I = \Omega_{S/R} / I\Omega_{S/R} $$ and similarly for $P$ (see Lemma \ref{lemma-differentials-base-change}). By Lemma \ref{lemma-characterize-formally-smooth-again} the sequence \begin{equation} \label{equation-split} 0 \to J/(IJ + J^2) \to \Omega_{P/R} \otimes_P S/IS \to \Omega_{S/R} \otimes_S S/IS \to 0 \end{equation} is split exact. Of course the middle term is $\bigoplus_{t \in T} S/IS \text{d}x_t$. Choose a splitting $\sigma : \Omega_{P/R} \otimes_P S/IS \to J/(IJ + J^2)$. For each $t \in T$ choose an element $f_t \in J$ which maps to $\sigma(\text{d}x_t)$ in $J/(IJ + J^2)$. This determines a unique $S$-module map $$ \tilde \sigma : \Omega_{P/R} \otimes_R S = \bigoplus S\text{d}x_t \longrightarrow J/J^2 $$ with the property that $\tilde\sigma(\text{d}x_t) = f_t$. As $\sigma$ is a section to $\text{d}$ the difference $$ \Delta = \text{id}_{J/J^2} - \tilde \sigma \circ \text{d} $$ is a self map $J/J^2 \to J/J^2$ whose image is contained in $(IJ + J^2)/J^2$. In particular $\Delta((IJ + J^2)/J^2) = 0$ because $I^2 = 0$. This means that $\Delta$ factors as $$ J/J^2 \to J/(IJ + J^2) \xrightarrow{\overline{\Delta}} (IJ + J^2)/J^2 \to J/J^2 $$ where $\overline{\Delta}$ is a $S/IS$-module map. Using again that the sequence (\ref{equation-split}) is split, we can find a $S/IS$-module map $\overline{\delta} : \Omega_{P/R} \otimes_P S/IS \to (IJ + J^2)/J^2$ such that $\overline{\delta} \circ d$ is equal to $\overline{\Delta}$. In the same manner as above the map $\overline{\delta}$ determines an $S$-module map $\delta : \Omega_{P/R} \otimes_P S \to J/J^2$. After replacing $\tilde \sigma$ by $\tilde \sigma + \delta$ a simple computation shows that $\Delta = 0$. In other words $\tilde \sigma$ is a section of $J/J^2 \to \Omega_{P/R} \otimes_P S$. By Lemma \ref{lemma-characterize-formally-smooth-again} we conclude that $R \to S$ is formally smooth.
We are going to use that smooth is equivalent to finite presentation and formally smooth, see Proposition \ref{proposition-smooth-formally-smooth}. Write $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ and denote $I = (f_1, \ldots, f_m)$. Choose a right inverse $\sigma : S \to R[x_1, \ldots, x_n]/I^2$ to the projection to $S$ as in Lemma \ref{lemma-characterize-formally-smooth}. Choose $h_i \in R[x_1, \ldots, x_n]$ such that $\sigma(x_i \bmod I) = h_i \bmod I^2$. The fact that $\sigma$ is an $R$-algebra homomorphism $R[x_1, \ldots, x_n]/I \to R[x_1, \ldots, x_n]/I^2$ is equivalent to the condition that $$ f_j(h_1, \ldots, h_n) = \sum\nolimits_{j_1 j_2} a_{j_1 j_2} f_{j_1} f_{j_2} $$ for certain $a_{kl} \in R[x_1, \ldots, x_n]$. Let $R_0 \subset R$ be the subring generated over $\mathbf{Z}$ by all the coefficients of the polynomials $f_j, h_i, a_{kl}$. Set $S_0 = R_0[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$, with $I_0 = (f_1, \ldots, f_m)$. Let $\sigma_0 : S_0 \to R_0[x_1, \ldots, x_n]/I_0^2$ defined by the rule $x_i \mapsto h_i \bmod I_0^2$; this works since the $a_{lk}$ are defined over $R_0$ and satisfy the same relations. Thus by Lemma \ref{lemma-characterize-formally-smooth} the ring $S_0$ is formally smooth over $R_0$.
If $R \to S$ is formally smooth, then $R' \to S'$ is formally smooth by Lemma \ref{lemma-base-change-fs}. To prove the converse, assume $R' \to S'$ is formally smooth. Note that $N \otimes_R R' = N \otimes_S S'$ for any $S$-module $N$. In particular $S \to S'$ is faithfully flat also. Choose a polynomial ring $P = R[\{x_i\}_{i \in I}]$ and a surjection of $R$-algebras $P \to S$ with kernel $J$. Note that $P' = P \otimes_R R'$ is a polynomial algebra over $R'$. Since $R \to R'$ is flat the kernel $J'$ of the surjection $P' \to S'$ is $J \otimes_R R'$. Hence the split exact sequence (see Lemma \ref{lemma-characterize-formally-smooth-again}) $$ 0 \to J'/(J')^2 \to \Omega_{P'/R'} \otimes_{P'} S' \to \Omega_{S'/R'} \to 0 $$ is the base change via $S \to S'$ of the corresponding sequence $$ J/J^2 \to \Omega_{P/R} \otimes_P S \to \Omega_{S/R} \to 0 $$ see Lemma \ref{lemma-differential-seq}. As $S \to S'$ is faithfully flat we conclude two things: (1) this sequence (without ${}'$) is exact too, and (2) $\Omega_{S/R}$ is a projective $S$-module. Namely, $\Omega_{S'/R'}$ is projective as a direct sum of the free module $\Omega_{P'/R'} \otimes_{P'} S'$ and $\Omega_{S/R} \otimes_S {S'} = \Omega_{S'/R'}$ by what we said above. Thus (2) follows by descent of projectivity through faithfully flat ring maps, see Theorem \ref{theorem-ffdescent-projectivity}. Hence the sequence $0 \to J/J^2 \to \Omega_{P/R} \otimes_P S \to \Omega_{S/R} \to 0$ is exact also and we win by applying Lemma \ref{lemma-characterize-formally-smooth-again} once more.
By Lemma \ref{lemma-finite-presentation-fs-Noetherian} we can extend the diagram to a commutative diagram $$ \xymatrix{ S_0 \ar[r] & S \ar[r] \ar@{-->}[rd] & A/I \\ R_0 \ar[r] \ar[u] & R \ar[r] \ar[u] & A \ar[u] } $$ with $R_0 \to S_0$ smooth, $R_0$ of finite type over $\mathbf{Z}$, and $S = S_0 \otimes_{R_0} R$. Let $x_1, \ldots, x_n \in S_0$ be generators of $S_0$ over $R_0$. Let $a_1, \ldots, a_n$ be elements of $A$ which map to the same elements in $A/I$ as the elements $x_1, \ldots, x_n$. Denote $A_0 \subset A$ the subring generated by the image of $R_0$ and the elements $a_1, \ldots, a_n$. Set $I_0 = A_0 \cap I$. Then $A_0/I_0 \subset A/I$ and $S_0 \to A/I$ maps into $A_0/I_0$. Thus it suffices to find the dotted arrow in the diagram $$ \xymatrix{ S_0 \ar[r] \ar@{-->}[rd] & A_0/I_0 \\ R_0 \ar[r] \ar[u] & A_0 \ar[u] } $$ The ring $A_0$ is of finite type over $\mathbf{Z}$ by construction. Hence $A_0$ is Noetherian, whence $I_0$ is nilpotent, see Lemma \ref{lemma-Noetherian-power}. Say $I_0^n = 0$. By Proposition \ref{proposition-smooth-formally-smooth} we can successively lift the $R_0$-algebra map $S_0 \to A_0/I_0$ to $S_0 \to A_0/I_0^2$, $S_0 \to A_0/I_0^3$, $\ldots$, and finally $S_0 \to A_0/I_0^n = A_0$.
By Lemma \ref{lemma-differential-seq-split} applied to $R \to S \to R$ we see that $I/I^2 = \Omega_{S/R} \otimes_{S, \sigma} R$. Since by definition of a smooth morphism the module $\Omega_{S/R}$ is finite locally free over $S$ we deduce that (1) holds. If $I/I^2$ is free, then choose $f_1, \ldots, f_d \in I$ whose images in $I/I^2$ form an $R$-basis. Consider the $R$-algebra map defined by $$ \Psi : R[[x_1, \ldots, x_d]] \longrightarrow S^\wedge, \quad x_i \longmapsto f_i. $$ Denote $P = R[[x_1, \ldots, x_d]]$ and $J = (x_1, \ldots, x_d) \subset P$. We write $\Psi_n : P/J^n \to S/I^n$ for the induced map of quotient rings. Note that $S/I^2 = \varphi(R) \oplus I/I^2$. Thus $\Psi_2$ is an isomorphism. Denote $\sigma_2 : S/I^2 \to P/J^2$ the inverse of $\Psi_2$. We will prove by induction on $n$ that for all $n > 2$ there exists an inverse $\sigma_n : S/I^n \to P/J^n$ of $\Psi_n$. Namely, as $S$ is formally smooth over $R$ (by Proposition \ref{proposition-smooth-formally-smooth}) we see that in the solid diagram $$ \xymatrix{ S \ar@{..>}[r] \ar[rd]_{\sigma_{n - 1}} & P/J^n \ar[d] \\  & P/J^{n - 1} } $$ of $R$-algebras we can fill in the dotted arrow by some $R$-algebra map $\tau : S \to P/J^n$ making the diagram commute. This induces an $R$-algebra map $\overline{\tau} : S/I^n \to P/J^n$ which is equal to $\sigma_{n - 1}$ modulo $J^n$. By construction the map $\Psi_n$ is surjective and now $\overline{\tau} \circ \Psi_n$ is an $R$-algebra endomorphism of $P/J^n$ which maps $x_i$ to $x_i + \delta_{i, n}$ with $\delta_{i, n} \in J^{n -1}/J^n$. It follows that $\Psi_n$ is an isomorphism and hence it has an inverse $\sigma_n$. This proves the lemma.
Consider the exact sequence $$ \mathfrak m/\mathfrak m^2 \to \Omega_{S/k} \otimes_S \kappa(\mathfrak m) \to \Omega_{\kappa(\mathfrak m)/k} \to 0 $$ of Lemma \ref{lemma-differential-seq}. We would like to show that the first map is an isomorphism. Since $k$ is algebraically closed the composition $k \to \kappa(\mathfrak m)$ is an isomorphism by Theorem \ref{theorem-nullstellensatz}. So the surjection $S \to \kappa(\mathfrak m)$ splits as a map of $k$-algebras, and Lemma \ref{lemma-differential-seq-split} shows that the sequence above is exact on the left. Since $\Omega_{\kappa(\mathfrak m)/k} = 0$, we win.
If $S$ is smooth at $\mathfrak q$ over $k$, then there exists a $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is standard smooth over $k$, see Lemma \ref{lemma-smooth-syntomic}. A standard smooth algebra over $k$ has a module of differentials which is free of rank equal to the dimension, see Lemma \ref{lemma-standard-smooth} (use that a relative global complete intersection over a field has dimension equal to the number of variables minus the number of equations). Thus we see that (1) implies (3). To finish the proof of the lemma it suffices to show that (2) implies (1) and that it implies that $S_{\mathfrak q}$ is regular. \medskip\noindent Assume (2). By Nakayama's Lemma \ref{lemma-NAK} we see that $\Omega_{S/k, \mathfrak q}$ can be generated by $\leq \dim_x X$ elements. We may replace $S$ by $S_g$ for some $g \in S$, $g \not \in \mathfrak q$ such that $\Omega_{S/k}$ is generated by at most $\dim_x X$ elements. Let $K \supset k$ be an algebraically closed field extension such that there exists a $k$-algebra map $\psi : \kappa(\mathfrak q) \to K$. Consider $S_K = K \otimes_k S$. Let $\mathfrak m \subset S_K$ be the maximal ideal corresponding to the surjection $$ \xymatrix{ S_K = K \otimes_k S \ar[r] & K \otimes_k \kappa(\mathfrak q) \ar[r]^-{\text{id}_K \otimes \psi} & K. } $$ Note that $\mathfrak m \cap S = \mathfrak q$, in other words $\mathfrak m$ lies over $\mathfrak q$. By Lemma \ref{lemma-dimension-at-a-point-preserved-field-extension} the dimension of $X_K = \Spec(S_K)$ at the point corresponding to $\mathfrak m$ is $\dim_x X$. By Lemma \ref{lemma-dimension-closed-point-finite-type-field} this is equal to $\dim((S_K)_{\mathfrak m})$. By Lemma \ref{lemma-differentials-base-change} the module of differentials of $S_K$ over $K$ is the base change of $\Omega_{S/k}$, hence also generated by at most $\dim_x X = \dim((S_K)_{\mathfrak m})$ elements. By Lemma \ref{lemma-characterize-smooth-kbar} we see that $S_K$ is smooth at $\mathfrak m$ over $K$. By Lemma \ref{lemma-flat-base-change-locus-smooth} this implies that $S$ is smooth at $\mathfrak q$ over $k$. This proves (1). Moreover, we know by Lemma \ref{lemma-characterize-smooth-kbar} that the local ring $(S_K)_{\mathfrak m}$ is regular. Since $S_{\mathfrak q} \to (S_K)_{\mathfrak m}$ is flat we conclude from Lemma \ref{lemma-flat-under-regular} that $S_{\mathfrak q}$ is regular.
Denote $R = S_{\mathfrak q}$ and denote its maximal by $\mathfrak m$ and its residue field $\kappa$. By Lemma \ref{lemma-computation-differential} and \ref{lemma-differential-seq} we see that there is a short exact sequence $$ 0 \to \mathfrak m/\mathfrak m^2 \to \Omega_{R/k} \otimes_R \kappa \to \Omega_{\kappa/k} \to 0 $$ Note that $\Omega_{R/k} = \Omega_{S/k, \mathfrak q}$, see Lemma \ref{lemma-differentials-localize}. Moreover, since $\kappa$ is separable over $k$ we have $\dim_{\kappa} \Omega_{\kappa/k} = \text{trdeg}_k(\kappa)$. Hence we get $$ \dim_{\kappa} \Omega_{R/k} \otimes_R \kappa = \dim_\kappa \mathfrak m/\mathfrak m^2 + \text{trdeg}_k (\kappa) \geq \dim R + \text{trdeg}_k (\kappa) = \dim_{\mathfrak q} S $$ (see Lemma \ref{lemma-dimension-at-a-point-finite-type-field} for the last equality) with equality if and only if $R$ is regular. Thus we win by applying Lemma \ref{lemma-characterize-smooth-over-field}.
For $a \in S$ write $\text{d}(a) = \theta(a)\text{d}f + c(a)$ for some $\theta(a) \in S$ and $c(a) \in C$. Consider the $R$-derivation $S \to S$, $a \mapsto \theta(a)$. Note that $\theta(f) = 1$. \medskip\noindent If $f^n = 0$ with $n > 1$ minimal, then $0 = \theta(f^n) = n f^{n - 1}$ contradicting the minimality of $n$. We conclude that $f$ is not nilpotent. \medskip\noindent Suppose $fa = 0$. If $f$ is a unit then $a = 0$ and we win. Assume $f$ is not a unit. Then $0 = \theta(fa) = f\theta(a) + a$ by the Leibniz rule and hence $a \in (f)$. By induction suppose we have shown $fa = 0 \Rightarrow a \in (f^n)$. Then writing $a = f^nb$ we get $0 = \theta(f^{n + 1}b) = (n + 1)f^nb + f^{n + 1}\theta(b)$. Hence $a = f^n b = -f^{n + 1}\theta(b)/(n + 1) \in (f^{n + 1})$. Since in the Noetherian local ring $S$ we have $\bigcap (f^n) = 0$, see Lemma \ref{lemma-intersect-powers-ideal-module-zero} we win.
In characteristic zero any field extension is separable and hence the equivalence of (1) and (3) follows from Lemma \ref{lemma-separable-smooth}. Also (1) implies (2) by definition of smooth algebras. Assume that $\Omega_{S/k, \mathfrak q}$ is free over $S_{\mathfrak q}$. We are going to use the notation and observations made in the proof of Lemma \ref{lemma-separable-smooth}. So $R = S_{\mathfrak q}$ with maximal ideal $\mathfrak m$ and residue field $\kappa$. Our goal is to prove $R$ is regular. \medskip\noindent If $\mathfrak m/\mathfrak m^2 = 0$, then $\mathfrak m = 0$ and $R \cong \kappa$. Hence $R$ is regular and we win. \medskip\noindent If $\mathfrak m/ \mathfrak m^2 \not = 0$, then choose any $f \in \mathfrak m$ whose image in $\mathfrak m/ \mathfrak m^2$ is not zero. By Lemma \ref{lemma-computation-differential} we see that $\text{d}f$ has nonzero image in $\Omega_{R/k}/\mathfrak m\Omega_{R/k}$. By assumption $\Omega_{R/k} = \Omega_{S/k, \mathfrak q}$ is finite free and hence by Nakayama's Lemma \ref{lemma-NAK} we see that $\text{d}f$ generates a direct summand. We apply Lemma \ref{lemma-characteristic-zero} to deduce that $f$ is a nonzerodivisor in $R$. Furthermore, by Lemma \ref{lemma-differential-seq} we get an exact sequence $$ (f)/(f^2) \to \Omega_{R/k} \otimes_R R/fR \to \Omega_{(R/fR)/k} \to 0 $$ This implies that $\Omega_{(R/fR)/k}$ is finite free as well. Hence by induction we see that $R/fR$ is a regular local ring. Since $f \in \mathfrak m$ was a nonzerodivisor we conclude that $R$ is regular, see Lemma \ref{lemma-regular-mod-x}.
Assume $R \to S$ is smooth at $(0)$. We may replace $S$ by $S_g$ for some nonzero $g \in S$ and assume that $R \to S$ is smooth. Then $K \to S \otimes_R K$ is smooth (Lemma \ref{lemma-base-change-smooth}). Moreover, for any field extension $K \subset K'$ the ring map $K' \to S \otimes_R K'$ is smooth as well. Hence $S \otimes_R K'$ is a regular ring by Lemma \ref{lemma-characterize-smooth-over-field}, in particular reduced. It follows that $S \otimes_R K$ is a geometrically reduced over $K$. Hence $L$ is geometrically reduced over $K$, see Lemma \ref{lemma-geometrically-reduced-permanence}. Hence $L/K$ is separable by Lemma \ref{lemma-characterize-separable-field-extensions}. \medskip\noindent Conversely, assume that $L/K$ is separable. We may assume $R \to S$ is of finite presentation, see Lemma \ref{lemma-generic-finite-presentation}. It suffices to prove that $K \to S \otimes_R K$ is smooth at $(0)$, see Lemma \ref{lemma-flat-base-change-locus-smooth}. This follows from Lemma \ref{lemma-separable-smooth}, the fact that a field is a regular ring, and the assumption that $L/K$ is separable.
Let $R \to S$ be \'etale. Choose a presentation $S = R[x_1, \ldots, x_n]/I$. As $R \to S$ is \'etale we know that $$ \text{d} : I/I^2 \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} S\text{d}x_i $$ is an isomorphism, in particular $I/I^2$ is a free $S$-module. Thus by Lemma \ref{lemma-huber} we may assume (after possibly changing the presentation), that $I = (f_1, \ldots, f_c)$ such that the classes $f_i \bmod I^2$ form a basis of $I/I^2$. It follows immediately from the fact that the displayed map above is an isomorphism that $c = n$ and that $\det(\partial f_j/\partial x_i)$ is invertible in $S$.
In each case we use the corresponding result for smooth ring maps with a small argument added to show that $\Omega_{S/R}$ is zero. \medskip\noindent Proof of (1). The ring map $R \to R_f$ is smooth and $\Omega_{R_f/R} = 0$. \medskip\noindent Proof of (2). The composition $A \to C$ of smooth maps $A \to B$ and $B \to C$ is smooth, see Lemma \ref{lemma-compose-smooth}. By Lemma \ref{lemma-exact-sequence-differentials} we see that $\Omega_{C/A}$ is zero as both $\Omega_{C/B}$ and $\Omega_{B/A}$ are zero. \medskip\noindent Proof of (3). Let $R \to S$ be \'etale and $R \to R'$ be arbitrary. Then $R' \to S' = R' \otimes_R S$ is smooth, see Lemma \ref{lemma-base-change-smooth}. Since $\Omega_{S'/R'} = S' \otimes_S \Omega_{S/R}$ by Lemma \ref{lemma-differentials-base-change} we conclude that $\Omega_{S'/R'} = 0$. Hence $R' \to S'$ is \'etale. \medskip\noindent Proof of (4). Assume the hypotheses of (4). By Lemma \ref{lemma-locally-smooth} we see that $R \to S$ is smooth. We are also given that $\Omega_{S_{g_i}/R} = (\Omega_{S/R})_{g_i} = 0$ for all $i$. Then $\Omega_{S/R} = 0$, see Lemma \ref{lemma-cover}. \medskip\noindent Proof of (5). The result for smooth maps is Lemma \ref{lemma-flat-base-change-locus-smooth}. In the proof of that lemma we used that $\NL_{S/R} \otimes_S S'$ is homotopy equivalent to $\NL_{S'/R'}$. This reduces us to showing that if $M$ is a finitely presented $S$-module the set of primes $\mathfrak q'$ of $S'$ such that $(M \otimes_S S')_{\mathfrak q'} = 0$ is the inverse image of the set of primes $\mathfrak q$ of $S$ such that $M_{\mathfrak q} = 0$. This follows from Lemma \ref{lemma-support-base-change}. \medskip\noindent Proof of (6). Follows directly from the corresponding result for smooth ring maps (Lemma \ref{lemma-smooth-syntomic}). \medskip\noindent Proof of (7). Follows from Lemma \ref{lemma-characterize-smooth-over-field} and the definitions. \medskip\noindent Proof of (8). Lemma \ref{lemma-finite-presentation-fs-Noetherian} gives the result for smooth ring maps. The resulting smooth ring map $R_0 \to S_0$ satisfies the hypotheses of Lemma \ref{lemma-relative-dimension-CM}, and hence we may replace $S_0$ by the factor of relative dimension $0$ over $R_0$. \medskip\noindent Proof of (9). Follows from (8) since $R_0 \to A$ will factor through $A_i$ for some $i$ by Lemma \ref{lemma-characterize-finite-presentation}. \medskip\noindent Proof of (10). Follows from (9), (1), and (2) since $S^{-1}A$ is a filtered colimit of principal localizations of $A$.
If $k \to k'$ is a finite separable field extension then we can write $k' = k(\alpha) \cong k[x]/(f)$. Here $f$ is the minimal polynomial of the element $\alpha$. Since $k'$ is separable over $k$ we have $\gcd(f, f') = 1$. This implies that $\text{d} : k'\cdot f \to k' \cdot \text{d}x$ is an isomorphism. Hence $k \to k'$ is \'etale. \medskip\noindent Conversely, suppose that $k \to S$ is \'etale. Let $\overline{k}$ be an algebraic closure of $k$. Then $S \otimes_k \overline{k}$ is \'etale over $\overline{k}$. Suppose we have the result over $\overline{k}$. Then $S \otimes_k \overline{k}$ is reduced and hence $S$ is reduced. Also, $S \otimes_k \overline{k}$ is finite over $\overline{k}$ and hence $S$ is finite over $k$. Hence $S$ is a finite product $S = \prod k_i$ of fields, see Lemma \ref{lemma-finite-dimensional-algebra} and Proposition \ref{proposition-dimension-zero-ring}. The result over $\overline{k}$ means $S \otimes_k \overline{k}$ is isomorphic to a finite product of copies of $\overline{k}$, which implies that each $k \subset k_i$ is finite separable, see for example Lemmas \ref{lemma-characterize-separable-field-extensions} and \ref{lemma-geometrically-reduced-finite-purely-inseparable-extension}. Thus we have reduced to the case $k = \overline{k}$. In this case Lemma \ref{lemma-characterize-smooth-kbar} (combined with $\Omega_{S/k} = 0$) we see that $S_{\mathfrak m} \cong k$ for all maximal ideals $\mathfrak m \subset S$. This implies the result because $S$ is the product of the localizations at its maximal ideals by Lemma \ref{lemma-finite-dimensional-algebra} and Proposition \ref{proposition-dimension-zero-ring} again.
First we may replace $S$ by $S_g$ for some $g \in S$, $g \not \in \mathfrak q$ and assume that $R \to S$ is \'etale. Then the lemma follows from Lemma \ref{lemma-etale-over-field} by unwinding the fact that $S \otimes_R \kappa(\mathfrak p)$ is \'etale over $\kappa(\mathfrak p)$.
Let $R \to S$ be an \'etale ring map. By definition $R \to S$ is of finite type. For any prime $\mathfrak p \subset R$ the fibre ring $S \otimes_R \kappa(\mathfrak p)$ is \'etale over $\kappa(\mathfrak p)$ and hence a finite products of fields finite separable over $\kappa(\mathfrak p)$, in particular finite over $\kappa(\mathfrak p)$. Thus $R \to S$ is quasi-finite by Lemma \ref{lemma-quasi-finite}.