hcju/stacksqa huggingface dataset:
Let $f \in A$, $f \not \in \mathfrak p$. Suppose that $A'' \to A_f$ satisfies (a), (b), (c) for the induced map $E \otimes_A A_f \to \NL_{A_f/\Lambda}$, see Algebra, Lemma \ref{algebra-lemma-localize-NL}. Then we can set $A' = A'' \times_{A_f} A$ and get a solution. Namely, it is clear that $A' \to A$ satisfies (a) because $\Ker(A' \to A) = \Ker(A'' \to A) = I$. Pick $f'' \in A''$ lifting $f$. Then the localization of $A'$ at $(f'', f)$ is isomorphic to $A''$ (for example by More on Algebra, Lemma \ref{more-algebra-lemma-diagram-localize}). Thus (b) and (c) are clear for $A'$ too. In this way we see that we may replace $A$ by the localization $A_f$ (finitely many times). In particular (after such a replacement) we may assume that $\mathfrak p$ is a maximal ideal of $A$, see Morphisms, Lemma \ref{morphisms-lemma-point-finite-type}. \medskip\noindent Choose a presentation $A = \Lambda[x_1, \ldots, x_n]/J$. Then $\NL_{A/\Lambda}$ is (canonically) homotopy equivalent to $$ J/J^2 \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} A\text{d}x_i, $$ see Algebra, Lemma \ref{algebra-lemma-NL-homotopy}. After localizing if necessary (using Nakayama's lemma) we can choose generators $f_1, \ldots, f_m$ of $J$ such that $f_j \otimes 1$ form a basis for $J/J^2 \otimes_A k$. Moreover, after renumbering, we can assume that the images of $\text{d}f_1, \ldots, \text{d}f_r$ form a basis for the image of $J/J^2 \otimes k \to \bigoplus k\text{d}x_i$ and that $\text{d}f_{r + 1}, \ldots, \text{d}f_m$ map to zero in $\bigoplus k\text{d}x_i$. With these choices the space $$ H^{-1}(\NL_{A/\Lambda} \otimes^{\mathbf{L}}_A k) = H^{-1}(\NL_{A/\Lambda} \otimes_A k) $$ has basis $f_{r + 1} \otimes 1, \ldots, f_m \otimes 1$. Changing basis once again we may assume that the image of $H^{-1}(\xi \otimes^{\mathbf{L}} k)$ is contained in the $k$-span of $f_{r + 1} \otimes 1, \ldots, f_{m - 1} \otimes 1$. Set $$ A' = \Lambda[x_1, \ldots, x_n]/(f_1, \ldots, f_{m - 1}, \mathfrak pf_m) $$ By construction $A' \to A$ satisfies (a). Since $\text{d}f_m$ maps to zero in $\bigoplus k\text{d}x_i$ we see that (b) holds. Finally, by construction the induced map $E \to \NL_{A/A'} = I[1]$ induces the zero map $H^{-1}(E \otimes_A^\mathbf{L} k) \to I \otimes_A k$. By Lemma \ref{lemma-compute-ext-into-field} we see that the composition is zero.
Let $C$ be the cone of $\xi$ so that we have a distinguished triangle $$ E \to \NL_{A/\Lambda} \to C \to E[1] $$ in $D^{-}(A)$. By Lemma \ref{lemma-characterize-versal} the assumption that $x$ is versal at $u_0$ implies that $H^{-1}(C \otimes^\mathbf{L} k) = 0$. By More on Algebra, Lemma \ref{more-algebra-lemma-cut-complex-in-two} there exists an $f \in A$ not contained in the prime corresponding to $u_0$ such that $H^{-1}(C \otimes^\mathbf{L}_A M) = 0$ for any $A_f$-module $M$. Using Lemma \ref{lemma-characterize-versal} again we see that we have versality for all finite type points of the open $D(f) \subset U$.
Let $z = x|_{\Spec(k)}$. Then (RS*) implies that $$ \textit{Lift}(x, A[k]) = \textit{Lift}(z, k[k]) \quad\text{and}\quad \textit{Lift}(y, B[k]) = \textit{Lift}(z, k[k]) $$ because $A[k] = A \times_k k[k]$ and $B[k] = B \times_k k[k]$. Hence the properties of a naive obstruction theory imply that the functoriality map $E_x \to E_y$ induces bijections $\Ext^i_A(E_x, k) \to \text{Ext}^i_B(E_y, k)$ for $i = -1, 0$. By Lemma \ref{lemma-compute-ext-into-field} our maps $H^i(E_x \otimes_A^{\mathbf{L}} k) \to H^i(E_y \otimes_B^{\mathbf{L}} k)$, $i = 0, 1$ induce isomorphisms on dual vector spaces hence are isomorphisms.
Let $U$ be a scheme locally of finite type over $S$, let $x$ be an object of $\mathcal{X}$ over $U$, and let $u_0$ be a finite type point of $U$ such that $x$ is versal at $u_0$. We may first shrink $U$ to an affine scheme such that $u_0$ is a closed point and such that $U \to S$ maps into an affine open $\Spec(\Lambda)$. Say $U = \Spec(A)$. Let $\xi_x : E_x \to \NL_{A/\Lambda}$ be the obstruction map. At this point we may apply Lemma \ref{lemma-openness} to conclude.
We may assume that $K^\bullet$ is a finite complex of finite projective $A$-modules. Thus the derived tensor product with $K^\bullet$ is the same as simply tensoring with $K^\bullet$. Let $E^\bullet$ be the dual perfect complex to $K^\bullet$, see More on Algebra, Lemma \ref{more-algebra-lemma-dual-perfect-complex}. (So $E^n = \Hom_A(K^{-n}, A)$ with differentials the transpose of the differentials of $K^\bullet$.) Let $E \in D^{-}(A)$ denote the object represented by the complex $E^\bullet[-1]$. Let $\xi \in H^1(\text{Tot}(K^\bullet \otimes_A \NL_{A/\Lambda}))$ be the element constructed in Lemma \ref{lemma-dual-obstruction} and denote $\xi : E = E^\bullet[-1] \to \NL_{A/\Lambda}$ the corresponding map (loc.cit.). We claim that the pair $(E, \xi)$ satisfies all the assumptions of Lemma \ref{lemma-openness} which finishes the proof. \medskip\noindent Namely, assumption (i) of Lemma \ref{lemma-openness} follows from conclusion (1) of Lemma \ref{lemma-dual-obstruction} and the fact that $H^2(K^\bullet \otimes_A^\mathbf{L} -) = \Ext^1(E, -)$ by loc.cit. Assumption (ii) of Lemma \ref{lemma-openness} follows from conclusion (2) of Lemma \ref{lemma-dual-obstruction} and the fact that $H^1(K^\bullet \otimes_A^\mathbf{L} -) = \Ext^0(E, -)$ by loc.cit. Assumption (iii) of Lemma \ref{lemma-openness} is clear.
Let $F : (\textit{Noetherian}/S)_\tau^{opp} \to \textit{Sets}$ be a functor which is both limit preserving and a sheaf. By Topologies, Lemmas \ref{topologies-lemma-extend} and \ref{topologies-lemma-extend-sheaf-general} there exists a unique functor $F' : (\Sch/S)_\tau^{opp} \to \textit{Sets}$ which is limit preserving, a sheaf, and restricts to $F$. In fact, the construction of $F'$ in Topologies, Lemma \ref{topologies-lemma-extend} is functorial in $F$ and this construction is a quasi-inverse to restriction. Some details omitted.
Let $V \subset X$ be an affine open subscheme which maps into an affine open $U \subset S$. We may write $V = \lim V_i$ as a directed limit of affine schemes $V_i$ of finite presentation over $U$, see Algebra, Lemma \ref{algebra-lemma-ring-colimit-fp}. By assumption, the arrow $V \to X$ factors as $V \to V_i \to X$ for some $i$. After increasing $i$ we may assume $V_i \to X$ factors through $V$ as the inverse image of $V \subset X$ in $V_i$ eventually becomes equal to $V_i$ by Limits, Lemma \ref{limits-lemma-descend-opens}. Then the identity morphism $V \to V$ factors through $V_i$ for some $i$ in the category of schemes over $U$. Thus $V \to U$ is of finite presentation; the corresponding algebra fact is that if $B$ is an $A$-algebra such that $\text{id} : B \to B$ factors through a finitely presented $A$-algebra, then $B$ is of finite presentation over $A$ (nice exercise). Hence $X$ is locally of finite presentation over $S$.
Let $V$ be a locally Noetherian scheme over $S$. Let $x : V \to X$ be a morphism over $S$. Then we get an element $(Z, u', \hat x)$ in $F(V)$ as follows \begin{enumerate} \item $Z \subset V$ is the inverse image of $T$ by $x$, \item $u' : V \setminus Z \to U' = U$ is the restriction of $x$ to $V \setminus Z$, \item $\hat x : V_{/Z} \to W$ is the composition of $x_{/Z} : V_{/Z} \to X_{/T}$ with the isomorphism $a : X_{/T} \to W$. \end{enumerate} This triple satisfies the compatibility condition because we can take $V' = V \times_{x, X} X'$, we can take $\hat x'$ the completion of the projection $x' : V' \to X'$. \medskip\noindent Conversely, suppose given an element $(Z, u', \hat x)$ of $F(V)$. We claim there is a unique morphism $x : V \to X$ compatible with $u'$ and $\hat x$. Namely, let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. Then Algebraization of Formal Spaces, Proposition \ref{restricted-proposition-glue-modification} is exactly the result we need to find a unique morphism $x : V \to X$ agreeing with $\hat x$ over $V_{/Z}$ and with $x'$ over $V'$ (and a fortiori agreeing with $u'$ over $V \setminus Z$). \medskip\noindent We omit the verification that the two constructions above define inverse bijections between their respective domains.
Recall that the condition only involves the evaluation $F(V)$ of the functor $F$ on schemes $V$ over $S$ which are spectra of Artinian local rings and the restriction maps $F(V_2) \to F(V_1)$ for morphisms $V_1 \to V_2$ of schemes over $S$ which are spectra of Artinian local rings. Thus let $V/S$ be the spetruim of an Artinian local ring. If $\xi = (Z, u', \hat x) \in F(V)$ then either $Z = \emptyset$ or $Z = V$ (set theoretically). In the first case we see that $\hat x$ is a morphism from the empty formal algebraic space into $W$. In the second case we see that $u'$ is a morphism from the empty scheme into $X'$ and we see that $\hat x : V \to W$ is a morphism into $W$. We conclude that $$ F(V) = U'(V) \amalg W(V) $$ and moreover for $V_1 \to V_2$ as above the induced map $F(V_2) \to F(V_1)$ is compatible with this decomposition. Hence it suffices to prove that both $U'$ and $W$ satisfy the Rim-Schlessinger condition. For $U'$ this follows from Lemma \ref{lemma-algebraic-stack-RS}. To see that it is true for $W$, we write $W = \colim W_n$ as in Formal Spaces, Lemma \ref{formal-spaces-lemma-structure-locally-noetherian}. Say $V = \Spec(A)$ with $(A, \mathfrak m)$ an Artinian local ring. Pick $n \geq 1$ such that $\mathfrak m^n = 0$. Then we have $W(V) = W_n(V)$. Hence we see that the Rim-Schlessinger condition for $W$ follows from the Rim-Schlessinger condition for $W_n$ for all $n$ (which in turn follows from Lemma \ref{lemma-algebraic-stack-RS}).
In the proof of Lemma \ref{lemma-rs} we have seen that $F(V) = U'(V) \amalg W(V)$ if $V$ is the spectrum of an Artinian local ring. The tangent spaces are computed entirely from evaluations of $F$ on such schemes over $S$. Hence it suffices to prove that the tangent spaces of the functors $U'$ and $W$ are finite dimensional. For $U'$ this follows from Lemma \ref{lemma-finite-dimension}. Write $W = \colim W_n$ as in the proof of Lemma \ref{lemma-rs}. Then we see that the tangent spaces of $W$ are equal to the tangent spaces of $W_2$, as to get at the tangent space we only need to evaluate $W$ on spectra of Artinian local rings $(A, \mathfrak m)$ with $\mathfrak m^2 = 0$. Then again we see that the tangent spaces of $W_2$ have finite dimension by Lemma \ref{lemma-finite-dimension}.
We have to show the following. Given a scheme $V$ locally of finite type over $S$, given $\xi \in F(V)$, and given a finite type point $v_0 \in V$ such that $\xi$ is versal at $v_0$, after replacing $V$ by an open neighbourhood of $v_0$ we have that $\xi$ is versal at every finite type point of $V$. Write $\xi = (Z, u', \hat x)$. \medskip\noindent First case: $v_0 \not \in Z$. Then we can first replace $V$ by $V \setminus Z$. Hence we see that $\xi = (\emptyset, u', \emptyset)$ and the morphism $u' : V \to X'$ is versal at $v_0$. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-lifting-along-artinian-at-point} this means that $u' : V \to X'$ is smooth at $v_0$. Since the set of a points where a morphism is smooth is open, we can after shrinking $V$ assume $u'$ is smooth. Then the same lemma tells us that $\xi$ is versal at every point as desired. \medskip\noindent Second case: $v_0 \in Z$. Write $W = \colim W_n$ as in Formal Spaces, Lemma \ref{formal-spaces-lemma-structure-locally-noetherian}. By Lemma \ref{lemma-openness-smoothness} we may assume $\hat x : V_{/Z} \to W$ is a smooth morphism of formal algebraic spaces. It follows immediately that $\xi = (Z, u', \hat x)$ is versal at all finite type points of $Z$. Let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. We see that $\hat x' : V'_{/Z} \to X'_{/T'}$ is smooth as a base change of $\hat x$. Since $\hat x'$ is the completion of $x' : V' \to X'$ this implies that $x' : V' \to X'$ is smooth at all points of $(V' \to V)^{-1}(Z) = |x'|^{-1}(T') \subset |V'|$ by the already used More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-lifting-along-artinian-at-point}. Since the set of smooth points of a morphism is open, we see that the closed set of points $B \subset |V'|$ where $x'$ is not smooth does not meet $(V' \to V)^{-1}(Z)$. Since $V' \to V$ is proper and hence closed, we see that $(V' \to V)(B) \subset V$ is a closed subset not meeting $Z$. Hence after shrinking $V$ we may assume $B = \emptyset$, i.e., $x'$ is smooth. By the discussion in the previous paragraph this exactly means that $\xi$ is versal at all finite type points of $V$ not contained in $Z$ and the proof is complete.
Lemma \ref{lemma-get-smooth} applies to $F$. Using this we choose, for every finite type field $k$ over $S$ and $x_0 \in F(\Spec(k))$, an affine scheme $U_{k, x_0}$ of finite type over $S$ and a smooth morphism $U_{k, x_0} \to F$ such that there exists a finite type point $u_{k, x_0} \in U_{k, x_0}$ with residue field $k$ such that $x_0$ is the image of $u_{k, x_0}$. Then $$ U = \coprod\nolimits_{k, x_0} U_{k, x_0} \longrightarrow F $$ is smooth\footnote{Set theoretical remark: This coproduct is (isomorphic) to an object of $(\Sch/S)_{fppf}$ as we have a bound on the index set by axiom [-1], see Sets, Lemma \ref{sets-lemma-what-is-in-it}.}. To finish the proof it suffices to show this map is surjective, see Bootstrap, Lemma \ref{bootstrap-lemma-spaces-etale-smooth-cover} (this is where we use axiom [0]). By Criteria for Representability, Lemma \ref{criteria-lemma-check-property-limit-preserving} it suffices to show that $U \times_F V \to V$ is surjective for those $V \to F$ where $V$ is an affine scheme locally of finite presentation over $S$. Since $U \times_F V \to V$ is smooth the image is open. Hence it suffices to show that the image of $U \times_F V \to V$ contains all finite type points of $V$, see Morphisms, Lemma \ref{morphisms-lemma-enough-finite-type-points}. Let $v_0 \in V$ be a finite type point. Then $k = \kappa(v_0)$ is a finite type field over $S$. Denote $x_0$ the composition $\Spec(k) \xrightarrow{v_0} V \to F$. Then $(u_{k, x_0}, v_0) : \Spec(k) \to U \times_F V$ is a point mapping to $v_0$ and we win.
We first prove that $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces. To do this it suffices to show that $$ \mathcal{Y} = \mathcal{X} \times_{\Delta, \mathcal{X} \times \mathcal{X}, y} (\Sch/V)_{fppf} $$ is representable by an algebraic space for any affine scheme $V$ locally of finite presentation over $S$ and object $y$ of $\mathcal{X} \times \mathcal{X}$ over $V$, see Criteria for Representability, Lemma \ref{criteria-lemma-check-representable-limit-preserving}\footnote{The set theoretic condition in Criteria for Representability, Lemma \ref{criteria-lemma-check-representable-limit-preserving} will hold: the size of the algebraic space $Y$ representing $\mathcal{Y}$ is suitably bounded. Namely, $Y \to S$ will be locally of finite type and $Y$ will satisfy axiom [-1]. Details omitted.}. Observe that $\mathcal{Y}$ is fibred in setoids (Stacks, Lemma \ref{stacks-lemma-isom-as-2-fibre-product}) and let $Y : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$, $T \mapsto \Ob(\mathcal{Y}_T)/\cong$ be the functor of isomorphism classes. We will apply Proposition \ref{proposition-spaces-diagonal-representable} to see that $Y$ is an algebraic space. \medskip\noindent Note that $\Delta_\mathcal{Y} : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ (and hence also $Y \to Y \times Y$) is representable by algebraic spaces by condition (1) and Criteria for Representability, Lemma \ref{criteria-lemma-second-diagonal}. Observe that $Y$ is a sheaf for the \'etale topology by Stacks, Lemmas \ref{stacks-lemma-stack-in-setoids-characterize} and \ref{stacks-lemma-2-fibre-product-gives-stack-in-setoids}, i.e., axiom [0] holds. Also $Y$ is limit preserving by Lemma \ref{lemma-fibre-product-limit-preserving}, i.e., we have [1]. Note that $Y$ has (RS), i.e., axiom [2] holds, by Lemmas \ref{lemma-algebraic-stack-RS} and \ref{lemma-fibre-product-RS}. Axiom [3] for $Y$ follows from Lemmas \ref{lemma-finite-dimension} and \ref{lemma-fibre-product-tangent-spaces}. Axiom [4] follows from Lemmas \ref{lemma-effective} and \ref{lemma-fibre-product-effective}. Axiom [5] for $Y$ follows directly from openness of versality for $\Delta_\mathcal{X}$ which is part of axiom [5] for $\mathcal{X}$. Thus all the assumptions of Proposition \ref{proposition-spaces-diagonal-representable} are satisfied and $Y$ is an algebraic space. \medskip\noindent At this point it follows from Lemma \ref{lemma-diagonal-representable} that $\mathcal{X}$ is an algebraic stack.
Recall that the sites $(\Sch/S)_{fppf}$ and $(\Sch/S)_\etale$ have the same underlying category, see discussion in Section \ref{section-noetherian}. Similarly the sites $(\textit{Noetherian}/S)_\etale$ and $(\textit{Noetherian}/S)_{fppf}$ have the same underlying categories. By axioms [0] and [1] the functor $F$ is a sheaf and limit preserving. Let $F' : (\Sch/S)_\etale^{opp} \to \textit{Sets}$ be the unique extension of $F$ which is a sheaf (for the \'etale topology) and which is limit preserving, see Lemma \ref{lemma-canonical-extension}. Then $F'$ satisfies axioms [0] and [1] as given in Section \ref{section-axioms-functors}. By Lemma \ref{lemma-representable} we see that $\Delta' : F' \to F' \times F'$ is representable (by schemes). On the other hand, it is immediately clear that $F'$ satisfies axioms [-1], [2], [3], [4], [5] of Section \ref{section-axioms-functors} as each of these involves only evaluating $F'$ at objects of $(\textit{Noetherian}/S)_\etale$ and we've assumed the corresponding conditions for $F$. Whence $F'$ is an algebraic space by Proposition \ref{proposition-spaces-diagonal-representable}.
This theorem is a copy of Quot, Theorem \ref{quot-theorem-coherent-algebraic-general}. The reason we have this copy here is that with the material in this section we get a second proof (as discussed at the beginning of this section). Namely, we argue exactly as in the proof of Quot, Theorem \ref{quot-theorem-coherent-algebraic} except that we substitute Lemma \ref{lemma-coherent-defo-thy-general} for Quot, Lemma \ref{quot-lemma-coherent-defo-thy}.
This lemma is true because $M$ is finitely presented even without the assumption that $M \otimes_R R'$ is finitely presented for every $R'$ as in the statement of the lemma. This follows from More on Flatness, Proposition \ref{flat-proposition-flat-finite-type-finite-presentation-domain}. Originally this lemma had an erroneous proof (thanks to Ofer Gabber for finding the gap) and was used in an alternative proof of the proposition cited. To reinstate this lemma, we need a correct argument in case $R$ is a local normal domain using only results from the chapters on commutative algebra; please email \href{mailto:stacks.project@gmail.com}{stacks.project@gmail.com} if you have an argument.
Pick $h \in \mathfrak m$ such that $f, h$ is a regular sequence in $A$ (this follows from Algebra, Lemmas \ref{algebra-lemma-criterion-normal} and \ref{algebra-lemma-depth-drops-by-one}). We will prove the lemma with $M = \text{length}_A(A/(f, h))$ and with $N$ any integer such that $\mathfrak m^N \subset (f, h)$. Such an integer $N$ exists because $\sqrt{(f, h)} = \mathfrak m$. Note that $M = \text{length}_A(A/(f + g, h))$ for all $g \in \mathfrak m^N$ because $(f, h) = (f + g, h)$. This moreover implies that $f + g, h$ is a regular sequence in $A$ too, see Algebra, Lemma \ref{algebra-lemma-reformulate-CM}. Now suppose that $\text{div}(f + g ) = \sum m_j (\mathfrak q_j)$. Then consider the map $$ c : A/(f + g) \longrightarrow \prod A/\mathfrak q_j^{(m_j)} $$ where $\mathfrak q_j^{(m_j)}$ is the symbolic power, see Algebra, Section \ref{algebra-section-symbolic-power}. Since $A$ is normal, we see that $A_{\mathfrak q_i}$ is a discrete valuation ring and hence $$ A_{\mathfrak q_i}/(f + g) = A_{\mathfrak q_i}/\mathfrak q_i^{m_i} A_{\mathfrak q_i} = (A/\mathfrak q_i^{(m_i)})_{\mathfrak q_i} $$ Since $V(f + g, h) = \{\mathfrak m\}$ this implies that $c$ becomes an isomorphism on inverting $h$ (small detail omitted). Since $h$ is a nonzerodivisor on $A/(f + g)$ we see that the length of $A/(f + g, h)$ equals the Herbrand quotient $e_A(A/(f + g), 0, h)$ as defined in Chow Homology, Section \ref{chow-section-periodic-complexes}. Similarly the length of $A/(h, \mathfrak q_j^{(m_j)})$ equals $e_A(A/\mathfrak q_j^{(m_j)}, 0, h)$. Then we have \begin{align*} M & = \text{length}_A(A/(f + g, h) \\ & = e_A(A/(f + g), 0, h) \\ & = \sum\nolimits_i e_A(A/\mathfrak q_j^{(m_j)}, 0, h) \\ & = \sum\nolimits_i \sum\nolimits_{m = 0, \ldots, m_j - 1} e_A(\mathfrak q_j^{(m)}/\mathfrak q_j^{(m + 1)}, 0, h) \end{align*} The equalities follow from Chow Homology, Lemmas \ref{chow-lemma-additivity-periodic-length} and \ref{chow-lemma-finite-periodic-length} using in particular that the cokernel of $c$ has finite length as discussed above. It is straightforward to prove that $e_A(\mathfrak q^{(m)}/\mathfrak q^{(m + 1)}, 0, h)$ is at least $1$ by Nakayama's lemma. This finishes the proof of the lemma.
According to Local Cohomology, Lemma \ref{local-cohomology-lemma-sitting-in-degrees} we have to show this for the finite $A$-module $E^i = \text{Ext}^{-i}_A(M, \omega_A^\bullet)$ for $i \leq s$. The support $Z$ of $E^0 \oplus \ldots \oplus E^s$ is closed in $\Spec(A)$ and does not contain any prime as in (2). Hence it is contained in $V(JI^n)$ for some $J$ as in the statement of the lemma.
The assumptions of Algebraic and Formal Geometry, Lemma \ref{algebraization-lemma-algebraize-local-cohomology-bis} by the more general Algebraic and Formal Geometry, Lemma \ref{algebraization-lemma-bootstrap-bis-bis}. Then the conclusion of Algebraic and Formal Geometry, Lemma \ref{algebraization-lemma-algebraize-local-cohomology-bis} gives the second statement.
Say $t^d + \sum_{i < d} \varphi(f_i)t^i = 0$ with $f_i \in R[x]$. Let $\ell$ be the maximum degree in $x$ of all the $f_i$. Multiply the equation by $\varphi(a)^\ell$ to get $\varphi(a)^\ell t^d + \sum_{i < d} \varphi(a^\ell f_i)t^i = 0$. Note that each $\varphi(a^\ell f_i)$ is in the image of $\varphi_a$. The result follows from Algebra, Lemma \ref{algebra-lemma-make-integral-trivial}.
We prove this by induction on $n$. As $u_n = \varphi(a_n)$ we conclude from Algebra, Lemma \ref{algebra-lemma-make-integral-trivial} that $u_nt$ is integral over $R$. Of course $u_n = \varphi(a_n)$ is integral over $R$. Then $u_{n - 1} = u_n t  + \varphi(a_{n - 1})$ is integral over $R$ (see Algebra, Lemma \ref{algebra-lemma-integral-closure-is-ring}) and we have $$ \varphi(a_0) + \varphi(a_1)t + \ldots + \varphi(a_{n - 1})t^{n - 1} + u_{n - 1}t^{n - 1} = 0. $$ Hence by the induction hypothesis applied to the map $S' \to S$ where $S'$ is the integral closure of $R$ in $S$ and the displayed equation we see that $u_{n-1}, \ldots, u_1$ and $u_{n-1}t, \ldots, u_1t$ are all in $S'$ too. The statement on the ideals is immediate from the shape of the elements and the fact that $u_1t + \varphi(a_0) = 0$.
The $R$-module $S_n$ has a presentation $$ R[X, Y]_{n-d} \to R[X, Y]_n \to S_n \to 0. $$ Thus by Algebra, Lemma \ref{algebra-lemma-cokernel-flat} it is enough to show that multiplication by $F$ induces an injective map $\kappa(\mathfrak p)[X, Y] \to \kappa(\mathfrak p)[X, Y]$ for all primes $\mathfrak p$. This is clear from the assumption that $F$ does not map to the zero polynomial mod $\mathfrak p$. The assertion on ranks is clear from this as well.