hcju/stacksqa huggingface dataset:
We prove each point, in order. \medskip\noindent Ad (1). Follows from Lemmas \ref{lemma-differentials-base-change} and \ref{lemma-base-change-finiteness}. \medskip\noindent Ad (2). Follows from Lemmas \ref{lemma-exact-sequence-differentials} and \ref{lemma-base-change-finiteness}. \medskip\noindent Ad (3). Follows by direct computation of $\Omega_{R_f/R}$ which we omit. \medskip\noindent Ad (4). We have $\Omega_{(R/I)/R} = 0$, see Lemma \ref{lemma-trivial-differential-surjective}, and the ring map $R \to R/I$ is of finite type. If $I$ is a finitely generated ideal then $R \to R/I$ is of finite presentation. \medskip\noindent Ad (5). See discussion following Definition \ref{definition-etale}. \medskip\noindent Ad (6). In this case $\Omega_{S/R}$ is a finite $S$-module (see Lemma \ref{lemma-differentials-finitely-generated}) and hence there exists a $g \in S$, $g \not \in \mathfrak q$ such that $(\Omega_{S/R})_g = 0$. By Lemma \ref{lemma-differentials-localize} this means that $\Omega_{S_g/R} = 0$ and hence $R \to S_g$ is unramified as desired. \medskip\noindent Ad (7). Use Nakayama's lemma (Lemma \ref{lemma-NAK}) to see that the condition is equivalent to the condition of (6). \medskip\noindent Ad (8) and (9). These are equivalent in the same manner that (6) and (7) are equivalent. Moreover $\Omega_{S \otimes_R \kappa(\mathfrak p)/\kappa(\mathfrak p)} = \Omega_{S/R} \otimes_S (S \otimes_R \kappa(\mathfrak p))$ by Lemma \ref{lemma-differentials-base-change}. Hence we see that (9) is equivalent to (7) since the $\kappa(\mathfrak q)$ vector spaces in both are canonically isomorphic. \medskip\noindent Ad (10). Follows from Lemmas \ref{lemma-cover} and \ref{lemma-differentials-localize}. \medskip\noindent Ad (11). Follows from (6) and (7) and the fact that the spectrum of $S$ is quasi-compact. \medskip\noindent Ad (12). Write $S = R[x_1, \ldots, x_n]/(g_1, \ldots, g_m)$. As $\Omega_{S/R} = 0$ we can write $$ \text{d}x_i = \sum h_{ij}\text{d}g_j + \sum a_{ijk}g_j\text{d}x_k $$ in $\Omega_{R[x_1, \ldots, x_n]/R}$ for some $h_{ij}, a_{ijk} \in R[x_1, \ldots, x_n]$. Choose a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ containing all the coefficients of the polynomials $g_i, h_{ij}, a_{ijk}$. Set $S_0 = R_0[x_1, \ldots, x_n]/(g_1, \ldots, g_m)$. This works. \medskip\noindent Ad (13). Write $S = R[x_1, \ldots, x_n]/I$. As $\Omega_{S/R} = 0$ we can write $$ \text{d}x_i = \sum h_{ij}\text{d}g_{ij} + \sum g'_{ik}\text{d}x_k $$ in $\Omega_{R[x_1, \ldots, x_n]/R}$ for some $h_{ij} \in R[x_1, \ldots, x_n]$ and $g_{ij}, g'_{ik} \in I$. Choose a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ containing all the coefficients of the polynomials $g_{ij}, h_{ij}, g'_{ik}$. Set $S_0 = R_0[x_1, \ldots, x_n]/(g_{ij}, g'_{ik})$. This works.
Let $J = \Ker(S \otimes_R S \to S)$. By assumption $J/J^2 = 0$, see Lemma \ref{lemma-differentials-diagonal}. Since $S$ is of finite type over $R$ we see that $J$ is finitely generated, namely by $x_i \otimes 1 - 1 \otimes x_i$, where $x_i$ generate $S$ over $R$. We win by Lemma \ref{lemma-ideal-is-squared-union-connected}.
We may first replace $S$ by $S_g$ for some $g \in S$, $g \not \in \mathfrak q$ and assume that $R \to S$ is unramified. The base change $S \otimes_R \kappa(\mathfrak p)$ is unramified over $\kappa(\mathfrak p)$ by Lemma \ref{lemma-unramified}. By Lemma \ref{lemma-characterize-smooth-over-field} it is smooth hence \'etale over $\kappa(\mathfrak p)$. Hence we see that $S \otimes_R \kappa(\mathfrak p) = (R \setminus \mathfrak p)^{-1} S/\mathfrak pS$ is a product of finite separable field extensions of $\kappa(\mathfrak p)$ by Lemma \ref{lemma-etale-over-field}. This implies the lemma.
By Lemma \ref{lemma-unramified} (8) it suffices to show that $\Omega_{S \otimes_R \kappa(\mathfrak p) / \kappa(\mathfrak p)}$ is zero when localized at $\mathfrak q$. Hence we may replace $S$ by $S \otimes_R \kappa(\mathfrak p)$ and $R$ by $\kappa(\mathfrak p)$. In other words, we may assume that $R = k$ is a field and $S$ is a finite type $k$-algebra. In this case the hypotheses imply that $S_{\mathfrak q} \cong \kappa(\mathfrak q)$. Thus $(\Omega_{S/k})_{\mathfrak q} = \Omega_{S_\mathfrak q/k} = \Omega_{\kappa(\mathfrak q)/k}$ is zero as desired (the first equality is Lemma \ref{lemma-differentials-localize}).
Parts (2) and (3) are equivalent by definition. The implication (1) $\Rightarrow$ (3) follows from the fact that \'etale ring maps are of finite presentation, Lemma \ref{lemma-etale} (flatness of \'etale maps), and Lemma \ref{lemma-unramified} (\'etale maps are unramified). Conversely, the characterization of \'etale ring maps in Lemma \ref{lemma-characterize-etale} and the structure of unramified ring maps in Lemma \ref{lemma-unramified-at-prime} shows that (3) implies (1). (This uses that $R \to S$ is \'etale if $R \to S$ is \'etale at every prime $\mathfrak q \subset S$, see Lemma \ref{lemma-etale}.)
Assume (a) and (b). Condition (b) implies that $\Omega_{A/k[x_1, \ldots, x_n]} = 0$ and hence $\varphi$ is unramified. Thus it suffices to prove that $\varphi$ is flat, see Lemma \ref{lemma-etale-flat-unramified-finite-presentation}. Let $\mathfrak m \subset A$ be a maximal ideal. Set $X = \Spec(A)$ and denote $x \in X$ the closed point corresponding to $\mathfrak m$. Then $\dim(A_\mathfrak m)$ is $\dim_x X$, see Lemma \ref{lemma-dimension-closed-point-finite-type-field}. Thus by Lemma \ref{lemma-characterize-smooth-over-field} we see that if (a) and (b) hold, then $A_\mathfrak m$ is a regular local ring for every maximal ideal $\mathfrak m$. Then $k[x_1, \ldots, x_n]_{\varphi^{-1}(\mathfrak m)} \to A_\mathfrak m$ is flat by Lemma \ref{lemma-CM-over-regular-flat} (and the fact that a regular local ring is CM, see Lemma \ref{lemma-regular-ring-CM}). Thus $\varphi$ is flat by Lemma \ref{lemma-flat-localization}. \medskip\noindent Assume $\varphi$ is \'etale. Then $\Omega_{A/k[x_1, \ldots, x_n]} = 0$ and hence (b) holds. On the other hand, \'etale ring maps are flat (Lemma \ref{lemma-etale}) and quasi-finite (Lemma \ref{lemma-etale-quasi-finite}). Hence for every maximal ideal $\mathfrak m$ of $A$ we my apply Lemma \ref{lemma-dimension-base-fibre-equals-total} to $k[x_1, \ldots, x_n]_{\varphi^{-1}(\mathfrak m)} \to A_\mathfrak m$ to see that $\dim(A_\mathfrak m) = n$ and hence (a) holds.
We may replace $(R \to S, \mathfrak p, \mathfrak q)$ with any base change $(R' \to R'\otimes_R S, \mathfrak p', \mathfrak q')$ by an \'etale ring map $R \to R'$ with a prime $\mathfrak p'$ lying over $\mathfrak p$, and a choice of $\mathfrak q'$ lying over both $\mathfrak q$ and $\mathfrak p'$. Note also that given $R \to R'$ and $\mathfrak p'$ a suitable $\mathfrak q'$ can always be found. \medskip\noindent The assumption that $R \to S$ is of finite type means that we may apply Lemma \ref{lemma-etale-makes-quasi-finite-finite-variant}. Thus we may assume that $S = A_1 \times \ldots \times A_n \times B$, that each $R \to A_i$ is finite with exactly one prime $\mathfrak r_i$ lying over $\mathfrak p$ such that $\kappa(\mathfrak p) \subset \kappa(\mathfrak r_i)$ is purely inseparable and that $R \to B$ is not quasi-finite at any prime lying over $\mathfrak p$. Then clearly $\mathfrak q = \mathfrak r_i$ for some $i$, since an unramified morphism is quasi-finite (see Lemma \ref{lemma-unramified-quasi-finite}). Say $\mathfrak q = \mathfrak r_1$. By Lemma \ref{lemma-unramified-at-prime} we see that $\kappa(\mathfrak p) \subset \kappa(\mathfrak r_1)$ is separable hence the trivial field extension, and that $\mathfrak p(A_1)_{\mathfrak r_1}$ is the maximal ideal. Also, by Lemma \ref{lemma-unique-prime-over-localize-below} (which applies to $R \to A_1$ because a finite ring map satisfies going up by Lemma \ref{lemma-integral-going-up}) we have $(A_1)_{\mathfrak r_1} = (A_1)_{\mathfrak p}$. It follows from Nakayama's Lemma \ref{lemma-NAK} that the map of local rings $R_{\mathfrak p} \to (A_1)_{\mathfrak p} = (A_1)_{\mathfrak r_1}$ is surjective. Since $A_1$ is finite over $R$ we see that there exists a $f \in R$, $f \not \in \mathfrak p$ such that $R_f \to (A_1)_f$ is surjective. After replacing $R$ by $R_f$ we win.
We may apply Lemma \ref{lemma-etale-makes-quasi-finite-finite-variant}. Thus, after an \'etale base change, we may assume that $S = A_1 \times \ldots \times A_n \times B$, that each $R \to A_i$ is finite with exactly one prime $\mathfrak r_i$ lying over $\mathfrak p$ such that $\kappa(\mathfrak p) \subset \kappa(\mathfrak r_i)$ is purely inseparable, and that $R \to B$ is not quasi-finite at any prime lying over $\mathfrak p$. Since $R \to S$ is quasi-finite (see Lemma \ref{lemma-unramified-quasi-finite}) we see there is no prime of $B$ lying over $\mathfrak p$. By Lemma \ref{lemma-unramified-at-prime} we see that $\kappa(\mathfrak p) \subset \kappa(\mathfrak r_i)$ is separable hence the trivial field extension, and that $\mathfrak p(A_i)_{\mathfrak r_i}$ is the maximal ideal. Also, by Lemma \ref{lemma-unique-prime-over-localize-below} (which applies to $R \to A_i$ because a finite ring map satisfies going up by Lemma \ref{lemma-integral-going-up}) we have $(A_i)_{\mathfrak r_i} = (A_i)_{\mathfrak p}$. It follows from Nakayama's Lemma \ref{lemma-NAK} that the map of local rings $R_{\mathfrak p} \to (A_i)_{\mathfrak p} = (A_i)_{\mathfrak r_i}$ is surjective. Since $A_i$ is finite over $R$ we see that there exists a $f \in R$, $f \not \in \mathfrak p$ such that $R_f \to (A_i)_f$ is surjective. After replacing $R$ by $R_f$ we win.
Part (2) implies part (1) since $S$ as in part (1) is a finite product of its localizations at the primes lying over $\mathfrak m$. Part (2) follows from Lemma \ref{lemma-characterize-henselian} part (10) since any finite $S$-algebra is also a finite $R$-algebra. If $R \to S$ and $\mathfrak q$ are as in (3), then $S_{\mathfrak q}$ is a local ring of a finite $R$-algebra by Lemma \ref{lemma-characterize-henselian} part (11). Hence (3) follows from (1). Part (4) follows from part (3).
Denote $\mathcal{C} \to \mathcal{D}$ the functor of categories of the statement. Suppose that $R \to S$ is finite \'etale. Then we may write $$ S = A_1 \times \ldots \times A_n $$ with $A_i$ local and finite \'etale over $S$, use either Lemma \ref{lemma-mop-up} or Lemma \ref{lemma-characterize-henselian} part (10). In particular $A_i/\mathfrak mA_i$ is a finite separable field extension of $\kappa$, see Lemma \ref{lemma-etale-at-prime}. Thus we see that every object of $\mathcal{C}$ and $\mathcal{D}$ decomposes canonically into irreducible pieces which correspond via the given functor. Next, suppose that $S_1$, $S_2$ are finite \'etale over $R$ such that $\kappa_1 = S_1/\mathfrak mS_1$ and $\kappa_2 = S_2/\mathfrak mS_2$ are fields (finite separable over $\kappa$). Then $S_1 \otimes_R S_2$ is finite \'etale over $R$ and we may write $$ S_1 \otimes_R S_2 = A_1 \times \ldots \times A_n $$ as before. Then we see that $\Hom_R(S_1, S_2)$ is identified with the set of indices $i \in \{1, \ldots, n\}$ such that $S_2 \to A_i$ is an isomorphism. To see this use that given any $R$-algebra map $\varphi : S_1 \to S_2$ the map $\varphi \times 1 : S_1 \otimes_R S_2 \to S_2$ is surjective, and hence is equal to projection onto one of the factors $A_i$. But in exactly the same way we see that $\Hom_\kappa(\kappa_1, \kappa_2)$ is identified with the set of indices $i \in \{1, \ldots, n\}$ such that $\kappa_2 \to A_i/\mathfrak mA_i$ is an isomorphism. By the discussion above these sets of indices match, and we conclude that our functor is fully faithful. Finally, let $\kappa \subset \kappa'$ be a finite separable field extension. By Lemma \ref{lemma-make-etale-map-prescribed-residue-field} there exists an \'etale ring map $R \to S$ and a prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ such that $\kappa \subset \kappa(\mathfrak q)$ is isomorphic to the given extension. By part (1) we may write $S = A_1 \times \ldots \times A_n \times B$. Since $R \to S$ is quasi-finite we see that there exists no prime of $B$ over $\mathfrak m$. Hence $S_{\mathfrak q}$ is equal to $A_i$ for some $i$. Hence $R \to A_i$ is finite \'etale and produces the given residue field extension. Thus the functor is essentially surjective and we win.
First write $S = A_1 \times \ldots \times A_n \times B$ as in Lemma \ref{lemma-mop-up}. Now we see that $R \to A_i$ is finite unramified and $A_i$ local. Hence the maximal ideal of $A_i$ is $\mathfrak mA_i$ and its residue field $A_i / \mathfrak m A_i$ is a finite separable extension of $\kappa$, see Lemma \ref{lemma-unramified-at-prime}. However, the condition that $R$ is strictly henselian means that $\kappa$ is separably algebraically closed, so $\kappa = A_i / \mathfrak m A_i$. By Nakayama's Lemma \ref{lemma-NAK} we conclude that $R \to A_i$ is surjective as desired.
Let $R \to S$ be a finite ring map. By Lemma \ref{lemma-characterize-henselian} it suffices to show that $S$ is a product of local rings. By Lemma \ref{lemma-finite-finite-fibres} $S$ has finitely many primes $\mathfrak m_1, \ldots, \mathfrak m_r$ which all lie over $\mathfrak m$. There are no inclusions among these primes, see Lemma \ref{lemma-integral-no-inclusion}, hence they are all maximal. Every element of $\mathfrak m_1 \cap \ldots \cap \mathfrak m_r$ is nilpotent by Lemma \ref{lemma-Zariski-topology}. It follows $S$ is the product of the localizations of $S$ at the primes $\mathfrak m_i$ by Lemma \ref{lemma-product-local}.
Consider $A \otimes_R S$. This is an \'etale algebra over $S$, see Lemma \ref{lemma-etale}. Moreover, the kernel $$ \mathfrak q' = \Ker(A \otimes_R S \to \kappa(\mathfrak q) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak m_S) \to \kappa(\mathfrak m_S)) $$ of the map using the map given in (3) is a prime ideal lying over $\mathfrak m_S$ with residue field equal to the residue field of $S$. Hence by Lemma \ref{lemma-characterize-henselian} there exists a unique splitting $\tau : A \otimes_R S \to S$ with $\tau^{-1}(\mathfrak m_S) = \mathfrak q'$. Set $f$ equal to the composition $A \to A \otimes_R S \to S$.
The first solution set is canonically isomorphic to the set $$ \Hom_R(R[x_1, \ldots, x_n]/(P_1, \ldots, P_n), R). $$ As $R$ is henselian the map $R \to R/\mathfrak m_R$ induces a bijection between this set and the set of solutions in the residue field $R/\mathfrak m_R$, see Lemma \ref{lemma-characterize-henselian}. The same is true for $S$. Now since $R[x_1, \ldots, x_n]/(P_1, \ldots, P_n)$ is \'etale over $R$ and $R/\mathfrak m_R$ is separably algebraically closed we see that $R/\mathfrak m_R[x_1, \ldots, x_n]/(\overline{P_1}, \ldots, \overline{P_n})$ is a finite product of copies of $R/\mathfrak m_R$. Hence the tensor product $$ R/\mathfrak m_R[x_1, \ldots, x_n]/(\overline{P_1}, \ldots, \overline{P_n}) \otimes_{R/\mathfrak m_R} S/\mathfrak m_S = S/\mathfrak m_S[x_1, \ldots, x_n]/(\overline{P_1'}, \ldots, \overline{P_n'}) $$ is also a finite product of copies of $S/\mathfrak m_S$ with the same index set. This proves the lemma.
Let $M$ be a countably generated and Mittag-Leffler $R$-module. We claim that for any element $x \in M$ there exists a direct sum decomposition $M = N \oplus K$ with $x \in N$, the module $N$ finitely presented, and $K$ Mittag-Leffler. \medskip\noindent Suppose the claim is true. Choose generators $x_1, x_2, x_3, \ldots$ of $M$. By the claim we can inductively find direct sum decompositions $$ M = N_1 \oplus N_2 \oplus \ldots \oplus N_n \oplus K_n $$ with $N_i$ finitely presented, $x_1, \ldots, x_n \in N_1 \oplus \ldots \oplus N_n$, and $K_n$ Mittag-Leffler. Repeating ad infinitum we see that $M = \bigoplus N_i$. \medskip\noindent We still have to prove the claim. Let $x \in M$. By Lemma \ref{lemma-ML-countable} there exists an endomorphism $\alpha : M \to M$ such that $\alpha$ factors through a finitely presented module, and $\alpha (x) = x$. Say $\alpha$ factors as $$ \xymatrix{ M \ar[r]^\pi & P \ar[r]^i & M } $$ Set $a = \pi \circ \alpha \circ i : P \to P$, so $i \circ a \circ \pi = \alpha^3$. By Lemma \ref{lemma-charpoly-module} there exists a monic polynomial $P \in R[T]$ such that $P(a) = 0$. Note that this implies formally that $\alpha^2 P(\alpha) = 0$. Hence we may think of $M$ as a module over $R[T]/(T^2P)$. Assume that $x \not = 0$. Then $\alpha(x) = x$ implies that $0 = \alpha^2P(\alpha)x = P(1)x$ hence $P(1) = 0$ in $R/I$ where $I = \{r \in R \mid rx = 0\}$ is the annihilator of $x$. As $x \not = 0$ we see $I \subset \mathfrak m_R$, hence $1$ is a root of $\overline{P} = P \bmod \mathfrak m_R \in R/\mathfrak m_R[T]$. As $R$ is henselian we can find a factorization $$ T^2P = (T^2 Q_1) Q_2 $$ for some $Q_1, Q_2 \in R[T]$ with $Q_2 = (T - 1)^e \bmod \mathfrak m_R R[T]$ and $Q_1(1) \not = 0 \bmod \mathfrak m_R$, see Lemma \ref{lemma-characterize-henselian}. Let $N = \Im(\alpha^2Q_1(\alpha) : M \to M)$ and $K = \Im(Q_2(\alpha) : M \to M)$. As $T^2Q_1$ and $Q_2$ generate the unit ideal of $R[T]$ we get a direct sum decomposition $M = N \oplus K$. Moreover, $Q_2$ acts as zero on $N$ and $T^2Q_1$ acts as zero on $K$. Note that $N$ is a quotient of $P$ hence is finitely generated. Also $x \in N$ because $\alpha^2Q_1(\alpha)x = Q_1(1)x$ and $Q_1(1)$ is a unit in $R$. By Lemma \ref{lemma-direct-sum-ML} the modules $N$ and $K$ are Mittag-Leffler. Finally, the finitely generated module $N$ is finitely presented as a finitely generated Mittag-Leffler module is finitely presented, see Example \ref{example-ML} part (1).
We will use the criterion of Lemma \ref{lemma-when-colimit}. Let $A \to P \to C$ be a factorization of $A \to C$ with $P$ of finite presentation over $A$. Write $B = \colim_{i \in I} B_i$ where $I$ is a directed set and where $B_i$ is an \'etale $A$-algebra.  Write $C = \colim_{j \in J} C_j$ where $J$ is a directed set and where $C_j$ is an \'etale $B$-algebra. We can factor $P \to C$ as $P \to C_j \to C$ for some $j$ by Lemma \ref{lemma-characterize-finite-presentation}. By Lemma \ref{lemma-etale} we can find an $i \in I$ and an \'etale ring map $B_i \to C'_j$ such that $C_j = B \otimes_{B_i} C'_j$. Then $C_j = \colim_{i' \geq i} B_{i'} \otimes_{B_i} C'_j$ and again we see that $P \to C_j$ factors as $P \to B_{i'} \otimes_{B_i} C'_j \to C$. As $A \to C' = B_{i'} \otimes_{B_i} C'_j$ is \'etale as compositions and tensor products of \'etale ring maps are \'etale. Hence we have factored $P \to C$ as $P \to C' \to C$ with $C'$ \'etale over $A$ and the criterion of Lemma \ref{lemma-when-colimit} applies.
Write $A_i = \colim_{j \in J_i} A_j$ where $J_i$ is a directed set and $A_j$ is an \'etale $R$-algebra. For each $i \leq i'$ and $j \in J_i$ there exists an $j' \in J_{i'}$ and an $R$-algebra map $\varphi_{jj'} : A_j \to A_{j'}$ making the diagram $$ \xymatrix{ A_i \ar[r] & A_{i'} \\ A_j \ar[u] \ar[r]^{\varphi_{jj'}} & A_{j'} \ar[u] } $$ commute. This is true because $R \to A_j$ is of finite presentation so that Lemma \ref{lemma-characterize-finite-presentation} applies. Let $\mathcal{J}$ be the category with objects $\coprod_{i \in I} J_i$ and morphisms triples $(j, j', \varphi_{jj'})$ as above (and obvious composition law). Then $\mathcal{J}$ is a filtered category and $A = \colim_\mathcal{J} A_j$. Details omitted.
Write $A = \colim A_i$ and $B = \colim B_j$ as filtered colimits with $A_i$ and $B_j$ \'etale over $R$. For each $i$ we can find a $j$ such that $A_i \to B$ factors through $B_j$, see Lemma \ref{lemma-characterize-finite-presentation}. The factorization $A_i \to B_j$ is \'etale by Lemma \ref{lemma-map-between-etale}. Since $A \to A \otimes_{A_i} B_j$ is \'etale (Lemma \ref{lemma-etale}) it suffices to prove that $B = \colim A \otimes_{A_i} B_j$ where the colimit is over pairs $(i, j)$ and factorizations $A_i \to B_j \to B$ of $A_i \to B$ (this is a directed system; details omitted). This is clear because colimits commute with tensor products and hence $\colim A \otimes_{A_i} B_j = A \otimes_A B = B$.
Categories, Lemma \ref{categories-lemma-directed-category-system} says that this is really just a question about a colimit of henselian local rings over a directed set. Let $(R_i, \varphi_{ii'})$ be such a system with each $\varphi_{ii'}$ local. Then $R = \colim_i R_i$ is local, and its residue field $\kappa$ is $\colim \kappa_i$ (argument omitted). Suppose that $f \in R[T]$ is monic and that $a_0 \in \kappa$ is a simple root of $\overline{f}$. Then for some large enough $i$ there exists an $f_i \in R_i[T]$ mapping to $f$ and an $a_{0, i} \in \kappa_i$ mapping to $a_0$. Since $\overline{f_i}(a_{0, i}) \in \kappa_i$, resp.\ $\overline{f_i'}(a_{0, i}) \in \kappa_i$ maps to $0 = \overline{f}(a_0) \in \kappa$, resp.\ $0 \not = \overline{f'}(a_0) \in \kappa$ we conclude that $a_{0, i}$ is a simple root of $\overline{f_i}$. As $R_i$ is henselian we can find $a_i \in R_i$ such that $f_i(a_i) = 0$ and $a_{0, i} = \overline{a_i}$. Then the image $a \in R$ of $a_i$ is the desired solution. Thus $R$ is henselian.
This is proved by exactly the same proof as used for Lemma \ref{lemma-henselization}. The only difference is that, instead of pairs, one uses triples $(S, \mathfrak q, \alpha)$ where $R \to S$ \'etale, $\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$, and $\alpha : \kappa(\mathfrak q) \to \kappa^{sep}$ is an embedding of extensions of $\kappa$.
By Lemma \ref{lemma-henselization-different} we see that $R^h$, resp.\ $S^h$ are filtered colimits of \'etale $R$, resp.\ $S$-algebras. Hence we see that $R^h \otimes_R S$ is a filtered colimit of \'etale $S$-algebras $A_i$ (Lemma \ref{lemma-etale}). By Lemma \ref{lemma-colimits-of-etale} we see that $S^h$ is a filtered colimit of \'etale $R^h \otimes_R S$-algebras. Since moreover $S^h$ is a henselian local ring with residue field equal to $\kappa(\mathfrak q)$, the statement follows from the uniqueness result of Lemma \ref{lemma-uniqueness-henselian}.
Note that $R_{\mathfrak p}^h \otimes_R S$ is quasi-finite over $R_{\mathfrak p}^h$ at the prime ideal corresponding to $\mathfrak q$, see Lemma \ref{lemma-four-rings}. Hence the localization $S'$ of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see Lemma \ref{lemma-finite-over-henselian}. As a localization $S'$ is a filtered colimit of \'etale $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma \ref{lemma-henselian-functorial-improve} we see that $S_\mathfrak q^h$ is the henselization of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^h$ by the uniqueness result of Lemma \ref{lemma-uniqueness-henselian}.
The residue field of $R_{\mathfrak p}^{sh}$ is the separable algebraic closure of $\kappa(\mathfrak p)$ in $\kappa^{sep}$. Note that $R_{\mathfrak p}^{sh} \otimes_R S$ is quasi-finite over $R_{\mathfrak p}^{sh}$ at the prime ideal corresponding to $\mathfrak q$, see Lemma \ref{lemma-four-rings}. Hence the localization $S'$ of $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see Lemma \ref{lemma-finite-over-henselian}. Note that the residue field of $S'$ is $\kappa^{sep}$ since it contains both the separable algebraic closure of $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$. Furthermore, as a localization $S'$ is a filtered colimit of \'etale $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma \ref{lemma-strictly-henselian-functorial-improve} we see that $S_{\mathfrak q}^{sh}$ is a strict henselization of $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^{sh}$ by the uniqueness result of Lemma \ref{lemma-uniqueness-henselian}.
Let $\mathfrak p$ be an embedded associated prime of $M$. Then there exists another associated prime $\mathfrak q$ of $M$ such that $\mathfrak p \supset \mathfrak q$. In particular this implies that $\dim(\text{Supp}(M_{\mathfrak p})) \geq 1$ (since $\mathfrak q$ is in the support as well). On the other hand $\mathfrak pR_{\mathfrak p}$ is associated to $M_{\mathfrak p}$ (Lemma \ref{lemma-associated-primes-localize}) and hence $\text{depth}(M_{\mathfrak p}) = 0$ (see Lemma \ref{lemma-ideal-nonzerodivisor}). In other words $(S_1)$ does not hold. Conversely, if $(S_1)$ does not hold then there exists a prime $\mathfrak p$ such that $\dim(\text{Supp}(M_{\mathfrak p})) \geq 1$ and $\text{depth}(M_{\mathfrak p}) = 0$. Then we see (arguing backwards using the lemmas cited above) that $\mathfrak p$ is an embedded associated prime.