hcju/stacksqa huggingface dataset:
To see this is true, it suffices to check that the construction of Lemma \ref{lemma-contract-prestable-to-stable} is compatible with base change (and isomorphisms but that's immediate), see the (abuse of) language for algebraic stacks introduced in Properties of Stacks, Section \ref{stacks-properties-section-conventions}. To see this it suffices to check properties (1) and (2) of Lemma \ref{lemma-contract-prestable-to-stable} are stable under base change. This is immediately clear for (1). For (2) this follows either from the fact that the contractions of Algebraic Curves, Lemmas \ref{curves-lemma-contracting-rational-tails} and \ref{curves-lemma-contracting-rational-bridges} are stable under ground field extensions, or because the conditions characterizing the morphisms on fibres in Algebraic Curves, Lemma \ref{curves-lemma-characterize-contraction-to-stable} are preserved under ground field extensions.
The statement means that the morphism $\overline{\mathcal{M}}_g \to \Spec(\mathbf{Z})$ is separated. We will prove this using the refined Noetherian valuative criterion as stated in  More on Morphisms of Stacks, Lemma \ref{stacks-more-morphisms-lemma-refined-valuative-criterion-separated} \medskip\noindent Since $\overline{\mathcal{M}}_g$ is an open substack of $\Curvesstack$, we see $\overline{\mathcal{M}}_g \to \Spec(\mathbf{Z})$ is quasi-separated and locally of finite presentation by Lemma \ref{lemma-curves-qs-lfp}. In particular the stack $\overline{\mathcal{M}}_g$ is locally Noetherian (Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-locally-finite-type-locally-noetherian}). By Lemma \ref{lemma-smooth-dense-in-stable} the open immersion $\mathcal{M}_g \to \overline{\mathcal{M}}_g$ has dense image. Also, $\mathcal{M}_g \to \overline{\mathcal{M}}_g$ is quasi-compact (Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-locally-closed-in-noetherian}), hence of finite type. Thus all the preliminary assumptions of More on Morphisms of Stacks, Lemma \ref{stacks-more-morphisms-lemma-refined-valuative-criterion-separated} are satisfied for the morphisms $$ \mathcal{M}_g \to \overline{\mathcal{M}}_g \quad\text{and}\quad \overline{\mathcal{M}}_g \to \Spec(\mathbf{Z}) $$ and it suffices to check the following: given any $2$-commutative diagram $$ \xymatrix{ \Spec(K) \ar[r] \ar[d] & \mathcal{M}_g \ar[r] & \overline{\mathcal{M}}_g \ar[d] \\ \Spec(R) \ar[rr] \ar@{..>}[rru] & & \Spec(\mathbf{Z}) } $$ where $R$ is a discrete valuation ring with field of fractions $K$ the category of dotted arrows is either empty or a setoid with exactly one isomorphism class. (Observe that we don't need to worry about $2$-arrows too much, see Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-cat-dotted-arrows-independent}). Unwinding what this means using that $\mathcal{M}_g$, resp.\ $\overline{\mathcal{M}}_g$ are the algebraic stacks parametrizing smooth, resp.\ stable families of genus $g$ curves, we find that what we have to prove is exactly the uniqueness result stated and proved in Lemma \ref{lemma-unique-stable-model}.
We will use the notation from Section \ref{section-polarized-curves}. Consider the subset $$ T \subset |\textit{PolarizedCurves}| $$ of points $\xi$ such that there exists a field $k$ and a pair $(X, \mathcal{L})$ over $k$ representing $\xi$ with the following two properties \begin{enumerate} \item $X$ is a stable genus $g$ curve, and \item $\mathcal{L} = \omega_X^{\otimes 3}$. \end{enumerate} Clearly, under the continuous map $$ |\textit{PolarizedCurves}| \longrightarrow |\Curvesstack| $$ the image of the set $T$ is exactly the open subset $$ |\overline{\mathcal{M}}_g| \subset |\Curvesstack| $$ Thus it suffices to show that $T$ is quasi-compact. By Lemma \ref{lemma-polarized-curves-in-polarized} we see that $$ |\textit{PolarizedCurves}| \subset |\Polarizedstack| $$ is an open and closed immersion. Thus it suffices to prove quasi-compactness of $T$ as a subset of $|\Polarizedstack|$. For this we use the criterion of Moduli Stacks, Lemma \ref{moduli-lemma-bounded-polarized}. First, we observe that for $(X, \mathcal{L})$ as above the Hilbert polynomial $P$ is the function $P(t) = (6g - 6)t + (1 - g)$ by Riemann-Roch, see Algebraic Curves, Lemma \ref{curves-lemma-rr}. Next, we observe that $H^1(X, \mathcal{L}) = 0$ and $\mathcal{L}$ is very ample by Algebraic Curves, Lemma \ref{curves-lemma-tricanonical}. This means exactly that with $n = P(3) - 1$ there is a closed immersion $$ i : X \longrightarrow \mathbf{P}^n_k $$ such that $\mathcal{L} = i^*\mathcal{O}_{\mathbf{P}^1_k}(1)$ as desired.
Proof of (1). If $\gamma$ is a divided power structure, then condition (2) (applied to $1$ and $n-1$ instead of $n$ and $m$) implies that $n \gamma_n(x) = \gamma_1(x)\gamma_{n - 1}(x)$. Hence by induction and condition (1) we get $n! \gamma_n(x) = x^n$. \medskip\noindent Assume $A$ is torsion free as a $\mathbf{Z}$-module. Proof of (2). This is clear from (1). \medskip\noindent Proof of (3). Assume that $n! \gamma_n(x) = x^n$ for all $x \in I$ and $n \geq 1$. Since $A \subset A \otimes_{\mathbf{Z}} \mathbf{Q}$ it suffices to prove the axioms (1) -- (5) of Definition \ref{definition-divided-powers} in case $A$ is a $\mathbf{Q}$-algebra. In this case $\gamma_n(x) = x^n/n!$ and it is straightforward to verify (1) -- (5); for example, (4) corresponds to the binomial formula $$ (x + y)^n = \sum_{i = 0, \ldots, n} \frac{n!}{i!(n - i)!} x^iy^{n - i} $$ We encourage the reader to do the verifications to make sure that we have the coefficients correct. \medskip\noindent Proof of (4). Assume we have generators $x_i$ of $I$ as an ideal such that $x_i^n \in (n!)I$ for all $n \geq 1$. We claim that for all $x \in I$ we have $x^n \in (n!)I$. If the claim holds then we can set $\gamma_n(x) = x^n/n!$ which is a divided power structure by (3). To prove the claim we note that it holds for $x = ax_i$. Hence we see that the claim holds for a set of generators of $I$ as an abelian group. By induction on the length of an expression in terms of these, it suffices to prove the claim for $x + y$ if it holds for $x$ and $y$. This follows immediately from the binomial theorem.
The numbers (1), (2), (3), (4), (5) in this proof refer to the conditions listed in Definition \ref{definition-divided-powers}. Applying (3) we see that if (2) and (5) hold for $x$ then (2) and (5) hold for $ax$ for all $a \in A$. Hence we see (b) implies (2) and (5) hold for a set of generators of $I$ as an abelian group. Hence, by induction of the length of an expression in terms of these it suffices to prove that, given $x, y \in I$ such that (2) and (5) hold for $x$ and $y$, then (2) and (5) hold for $x + y$. \medskip\noindent Proof of (2) for $x + y$. By (4) we have $$ \gamma_n(x + y)\gamma_m(x + y) = \sum\nolimits_{i + j = n,\ k + l = m} \gamma_i(x)\gamma_k(x)\gamma_j(y)\gamma_l(y) $$ Using (2) for $x$ and $y$ this equals $$ \sum \frac{(i + k)!}{i!k!}\frac{(j + l)!}{j!l!} \gamma_{i + k}(x)\gamma_{j + l}(y) $$ Comparing this with the expansion $$ \gamma_{n + m}(x + y) = \sum \gamma_a(x)\gamma_b(y) $$ we see that we have to prove that given $a + b = n + m$ we have $$ \sum\nolimits_{i + k = a,\ j + l = b,\ i + j = n,\ k + l = m} \frac{(i + k)!}{i!k!}\frac{(j + l)!}{j!l!} = \frac{(n + m)!}{n!m!}. $$ Instead of arguing this directly, we note that the result is true for the ideal $I = (x, y)$ in the polynomial ring $\mathbf{Q}[x, y]$ because $\gamma_n(f) = f^n/n!$, $f \in I$ defines a divided power structure on $I$. Hence the equality of rational numbers above is true. \medskip\noindent Proof of (5) for $x + y$ given that (1) -- (4) hold and that (5) holds for $x$ and $y$. We will again reduce the proof to an equality of rational numbers. Namely, using (4) we can write $\gamma_n(\gamma_m(x + y)) = \gamma_n(\sum \gamma_i(x)\gamma_j(y))$. Using (4) we can write $\gamma_n(\gamma_m(x + y))$ as a sum of terms which are products of factors of the form $\gamma_k(\gamma_i(x)\gamma_j(y))$. If $i > 0$ then \begin{align*} \gamma_k(\gamma_i(x)\gamma_j(y)) & = \gamma_j(y)^k\gamma_k(\gamma_i(x)) \\ & = \frac{(ki)!}{k!(i!)^k} \gamma_j(y)^k \gamma_{ki}(x) \\ & = \frac{(ki)!}{k!(i!)^k} \frac{(kj)!}{(j!)^k} \gamma_{ki}(x) \gamma_{kj}(y) \end{align*} using (3) in the first equality, (5) for $x$ in the second, and (2) exactly $k$ times in the third. Using (5) for $y$ we see the same equality holds when $i = 0$. Continuing like this using all axioms but (5) we see that we can write $$ \gamma_n(\gamma_m(x + y)) = \sum\nolimits_{i + j = nm} c_{ij}\gamma_i(x)\gamma_j(y) $$ for certain universal constants $c_{ij} \in \mathbf{Z}$. Again the fact that the equality is valid in the polynomial ring $\mathbf{Q}[x, y]$ implies that the coefficients $c_{ij}$ are all equal to $(nm)!/n!(m!)^n$ as desired.
Let $x \in I$ and $y \in J$. Then $$ \gamma_n(xy) = y^n\gamma_n(x) = n! \delta_n(y) \gamma_n(x) = \delta_n(y) x^n = \delta_n(xy). $$ Hence $\gamma$ and $\delta$ agree on a set of (additive) generators of $IJ$. By property (4) of Definition \ref{definition-divided-powers} it follows that they agree on all of $IJ$. \medskip\noindent Assume $\gamma$ and $\delta$ agree on $I \cap J$. Let $z \in I + J$. Write $z = x + y$ with $x \in I$ and $y \in J$. Then we set $$ \epsilon_n(z) = \sum \gamma_i(x)\delta_{n - i}(y) $$ for all $n \geq 1$. To see that this is well defined, suppose that $z = x' + y'$ is another representation with $x' \in I$ and $y' \in J$. Then $w = x - x' = y' - y \in I \cap J$. Hence \begin{align*} \sum\nolimits_{i + j = n} \gamma_i(x)\delta_j(y) & = \sum\nolimits_{i + j = n} \gamma_i(x' + w)\delta_j(y) \\ & = \sum\nolimits_{i' + l + j = n} \gamma_{i'}(x')\gamma_l(w)\delta_j(y) \\ & = \sum\nolimits_{i' + l + j = n} \gamma_{i'}(x')\delta_l(w)\delta_j(y) \\ & = \sum\nolimits_{i' + j' = n} \gamma_{i'}(x')\delta_{j'}(y + w) \\ & = \sum\nolimits_{i' + j' = n} \gamma_{i'}(x')\delta_{j'}(y') \end{align*} as desired. Hence, we have defined maps $\epsilon_n : I + J \to I + J$ for all $n \geq 1$; it is easy to see that $\epsilon_n \mid_{I} = \gamma_n$ and $\epsilon_n \mid_{J} = \delta_n$. Next, we prove conditions (1) -- (5) of Definition \ref{definition-divided-powers} for the collection of maps $\epsilon_n$. Properties (1) and (3) are clear. To see (4), suppose that $z = x + y$ and $z' = x' + y'$ with $x, x' \in I$ and $y, y' \in J$ and compute \begin{align*} \epsilon_n(z + z') & = \sum\nolimits_{a + b = n} \gamma_a(x + x')\delta_b(y + y') \\ & = \sum\nolimits_{i + i' + j + j' = n} \gamma_i(x) \gamma_{i'}(x')\delta_j(y)\delta_{j'}(y') \\ & = \sum\nolimits_{k = 0, \ldots, n} \sum\nolimits_{i+j=k} \gamma_i(x)\delta_j(y) \sum\nolimits_{i'+j'=n-k} \gamma_{i'}(x')\delta_{j'}(y') \\ & = \sum\nolimits_{k = 0, \ldots, n}\epsilon_k(z)\epsilon_{n-k}(z') \end{align*} as desired. Now we see that it suffices to prove (2) and (5) for elements of $I$ or $J$, see Lemma \ref{lemma-check-on-generators}. This is clear because $\gamma$ and $\delta$ are divided power structures. \medskip\noindent The existence of a divided power structure $\epsilon$ on $I+J$ whose restrictions to $I$ and $J$ are $\gamma$ and $\delta$ is thus proven; its uniqueness is rather clear.
The empty limit is the zero ring (that's weird but we need it). The product of a collection of divided power rings $(A_t, I_t, \gamma_t)$, $t \in T$ is given by $(\prod A_t, \prod I_t, \gamma)$ where $\gamma_n((x_t)) = (\gamma_{t, n}(x_t))$. The equalizer of $\alpha, \beta : (A, I, \gamma) \to (B, J, \delta)$ is just $C = \{a \in A \mid \alpha(a) = \beta(a)\}$ with ideal $C \cap I$ and induced divided powers. It follows that all limits exist, see Categories, Lemma \ref{categories-lemma-limits-products-equalizers}.
The empty colimit is $\mathbf{Z}$ with divided power ideal $(0)$. Let's discuss general colimits. Let $\mathcal{C}$ be a category and let $c \mapsto (A_c, I_c, \gamma_c)$ be a diagram. Consider the functor $$ F(B, J, \delta) = \lim_{c \in \mathcal{C}} Hom((A_c, I_c, \gamma_c), (B, J, \delta)) $$ Note that any $f = (f_c)_{c \in C} \in F(B, J, \delta)$ has the property that all the images $f_c(A_c)$ generate a subring $B'$ of $B$ of bounded cardinality $\kappa$ and that all the images $f_c(I_c)$ generate a divided power sub ideal $J'$ of $B'$. And we get a factorization of $f$ as a $f'$ in $F(B')$ followed by the inclusion $B' \to B$. Also, $F$ commutes with limits. Hence we may apply Lemma \ref{lemma-a-version-of-brown} to see that $F$ is representable and we win.
Proof of (1). This is clear. Assume (2)(a). Define $\bar\gamma_n(x \bmod I') = \gamma_n(x) \bmod I'$ for $x \in I$. This is well defined since $\gamma_n(x + y) = \gamma_n(x) \bmod I'$ for $y \in I'$ by Definition \ref{definition-divided-powers} (4) and the fact that $\gamma_j(y) \in I'$ by assumption. It is clear that $\bar\gamma$ is a divided power structure as $\gamma$ is one. Hence (2)(b) holds. Also, (2)(b) implies (2)(a) by part (1). It is clear that (2)(a) implies (2)(c). Assume (2)(c). Note that $\gamma_n(x) = a^n\gamma_n(x_i) \in I'$ for $x = ax_i$. Hence we see that $\gamma_n(x) \in I'$ for a set of generators of $I'$ as an abelian group. By induction on the length of an expression in terms of these, it suffices to prove $\forall n : \gamma_n(x + y) \in I'$ if $\forall n : \gamma_n(x), \gamma_n(y) \in I'$. This follows immediately from the fourth axiom of a divided power structure.
Let $t \geq 1$ be an integer such that $p^tA/I = 0$, i.e., $p^tA \subset I$. The map $A^\wedge \to A/I$ is the composition $A^\wedge \to A/p^tA \to A/I$ which is surjective (for example by Algebra, Lemma \ref{algebra-lemma-completion-generalities}). As $p^eI \subset p^eA \cap I \subset p^{e - t}I$ for $e \geq t$ we see that the kernel of the composition $A^\wedge \to A/I$ is the $p$-adic completion of $I$. The map $\gamma_n$ is continuous because $$ \gamma_n(x + p^ey) = \sum\nolimits_{i + j = n} p^{je}\gamma_i(x)\gamma_j(y) = \gamma_n(x) \bmod p^eI $$ by the axioms of a divided power structure. It is clear that the axioms for divided power structures are inherited by the maps $\gamma_n^\wedge$ from the maps $\gamma_n$. Finally, to see the last statement say $e > t$. Then $p^eA \subset I$ and $\gamma_1(p^eA) \subset p^eA$ and for $n > 1$ we have $$ \gamma_n(p^ea) = p^n \gamma_n(p^{e - 1}a) = \frac{p^n}{n!} p^{n(e - 1)}a^n \in p^e A $$ as $p^n/n! \in \mathbf{Z}_{(p)}$ and as $n \geq 2$ and $e \geq 2$ so $n(e - 1) \geq e$. This proves that $\gamma$ extends to $A/p^eA$, see Lemma \ref{lemma-kernel}. The statement on limits is clear from the construction of limits in the proof of Lemma \ref{lemma-limits}.
We will prove the lemma in case of a divided power polynomial algebra in one variable. The result for the general case can be argued in exactly the same way, or by noting that $A\langle x_1, \ldots, x_t\rangle$ is isomorphic to the ring obtained by adjoining the divided power variables $x_1, \ldots, x_t$ one by one. \medskip\noindent Let $A\langle x \rangle_{+}$ be the ideal generated by $x, x^{[2]}, x^{[3]}, \ldots$. Note that $J = IA\langle x \rangle + A\langle x \rangle_{+}$ and that $$ IA\langle x \rangle \cap A\langle x \rangle_{+} = IA\langle x \rangle \cdot A\langle x \rangle_{+} $$ Hence by Lemma \ref{lemma-two-ideals} it suffices to show that there exist divided power structures on the ideals $IA\langle x \rangle$ and $A\langle x \rangle_{+}$. The existence of the first follows from Lemma \ref{lemma-gamma-extends} as $A \to A\langle x \rangle$ is flat. For the second, note that if $A$ is torsion free, then we can apply Lemma \ref{lemma-silly} (4) to see that $\delta$ exists. Namely, choosing as generators the elements $x^{[m]}$ we see that $(x^{[m]})^n = \frac{(nm)!}{(m!)^n} x^{[nm]}$ and $n!$ divides the integer $\frac{(nm)!}{(m!)^n}$. In general write $A = R/\mathfrak a$ for some torsion free ring $R$ (e.g., a polynomial ring over $\mathbf{Z}$). The kernel of $R\langle x \rangle \to A\langle x \rangle$ is $\bigoplus \mathfrak a x^{[m]}$. Applying criterion (2)(c) of Lemma \ref{lemma-kernel} we see that the divided power structure on $R\langle x \rangle_{+}$ extends to $A\langle x \rangle$ as desired. \medskip\noindent Proof of the universal property. Given a homomorphism $\varphi : A \to C$ of divided power rings and $k_1, \ldots, k_t \in K$ we consider $$ A\langle x_1, \ldots, x_t \rangle \to C,\quad x_1^{[n_1]} \ldots x_t^{[n_t]} \longmapsto \epsilon_{n_1}(k_1) \ldots \epsilon_{n_t}(k_t) $$ using $\varphi$ on coefficients. The only thing to check is that this is an $A$-algebra homomorphism (details omitted). The inverse construction is clear.
If $n$ is not divisible by $p$, then $\gamma_n(x) = c x \gamma_{n - 1}(x)$ where $c$ is a unit in $\mathbf{Z}_{(p)}$. Moreover, $$ \gamma_{pm}(x) = c \gamma_m(\gamma_p(x)) $$ where $c$ is a unit in $\mathbf{Z}_{(p)}$. Thus the first assertion is clear. For the second assertion, we can, working backwards, use these equalities to define all $\gamma_n$. More precisely, if $n = a_0 + a_1p + \ldots + a_e p^e$ with $a_i \in \{0, \ldots, p - 1\}$ then we set $$ \gamma_n(x) = c_n x^{a_0} \delta(x)^{a_1} \ldots \delta^e(x)^{a_e} $$ for $c_n \in \mathbf{Z}_{(p)}$ defined by $$ c_n = {(p!)^{a_1 + a_2(1 + p) + \ldots + a_e(1 + \ldots + p^{e - 1})}}/{n!}. $$ Now we have to show the axioms (1) -- (5) of a divided power structure, see Definition \ref{definition-divided-powers}. We observe that (1) and (3) are immediate. Verification of (2) and (5) is by a direct calculation which we omit. Let $x, y \in I$. We claim there is a ring map $$ \varphi : \mathbf{Z}_{(p)}\langle u, v \rangle \longrightarrow A $$ which maps $u^{[n]}$ to $\gamma_n(x)$ and $v^{[n]}$ to $\gamma_n(y)$. By construction of $\mathbf{Z}_{(p)}\langle u, v \rangle$ this means we have to check that $$ \gamma_n(x)\gamma_m(x) = \frac{(n + m)!}{n!m!} \gamma_{n + m}(x) $$ in $A$ and similarly for $y$. This is true because (2) holds for $\gamma$. Let $\epsilon$ denote the divided power structure on the ideal $\mathbf{Z}_{(p)}\langle u, v\rangle_{+}$ of $\mathbf{Z}_{(p)}\langle u, v\rangle$. Next, we claim that $\varphi(\epsilon_n(f)) = \gamma_n(\varphi(f))$ for $f \in \mathbf{Z}_{(p)}\langle u, v\rangle_{+}$ and all $n$. This is clear for $n = 0, 1, \ldots, p - 1$. For $n = p$ it suffices to prove it for a set of generators of the ideal $\mathbf{Z}_{(p)}\langle u, v\rangle_{+}$ because both $\epsilon_p$ and $\gamma_p = \delta$ satisfy properties (1) and (3) of the lemma. Hence it suffices to prove that $\gamma_p(\gamma_n(x)) = \frac{(pn)!}{p!(n!)^p}\gamma_{pn}(x)$ and similarly for $y$, which follows as (5) holds for $\gamma$. Now, if $n = a_0 + a_1p + \ldots + a_e p^e$ is an arbitrary integer written in $p$-adic expansion as above, then $$ \epsilon_n(f) = c_n f^{a_0} \gamma_p(f)^{a_1} \ldots \gamma_p^e(f)^{a_e} $$ because $\epsilon$ is a divided power structure. Hence we see that $\varphi(\epsilon_n(f)) = \gamma_n(\varphi(f))$ holds for all $n$. Applying this for $f = u + v$ we see that axiom (4) for $\gamma$ follows from the fact that $\epsilon$ is a divided power structure.
Observe that $A'_{even} = A_{even} \otimes_R R'$ and $A'_{even, +} = A_{even, +} \otimes_R R'$. Hence we are trying to show that the divided powers $\gamma$ extend to $A'_{even}$ (terminology as in Definition \ref{definition-extends}). Once we have shown $\gamma$ extends it follows easily that this extension has all the desired properties. \medskip\noindent Choose a polynomial $R$-algebra $P$ (on any set of generators) and a surjection of $R$-algebras $P \to R'$. The ring map $A_{even} \to A_{even} \otimes_R P$ is flat, hence the divided powers $\gamma$ extend to $A_{even} \otimes_R P$ uniquely by Lemma \ref{lemma-gamma-extends}. Let $J = \Ker(P \to R')$. To show that $\gamma$ extends to $A \otimes_R R'$ it suffices to show that $I' = \Ker(A_{even, +} \otimes_R P \to A_{even, +} \otimes_R R')$ is generated by elements $z$ such that $\gamma_n(z) \in I'$ for all $n > 0$. This is clear as $I'$ is generated by elements of the form $x \otimes f$ with $x \in A_{even, +}$ and $f \in \Ker(P \to R')$.
We write out the construction for the case that $R$ is Noetherian and $R\to S$ is of finite type. Without those assumptions, the proof is the same, except that we have to use some set (possibly infinite) of generators in each degree. \medskip\noindent Start of the construction: Let $A(0) = R[x_1, \ldots, x_n]$ be a (usual) polynomial ring and let $A(0) \to S$ be a surjection. As grading we take $A(0)_0 = A(0)$ and $A(0)_d = 0$ for $d \not = 0$. Thus $\text{d} = 0$ and $\gamma_n$, $n > 0$, is zero as well. \medskip\noindent Choose generators $f_1, \ldots, f_m \in R[x_1, \ldots, x_n]$ for the kernel of the given map $A(0) = R[x_1, \ldots, x_n] \to S$. We apply Example \ref{example-adjoining-odd} $m$ times to get $$ A(1) = A(0)\langle T_1, \ldots, T_m\rangle $$ with $\deg(T_i) = 1$ as a graded divided power polynomial algebra. We set $\text{d}(T_i) = f_i$. Since $A(1)$ is a divided power polynomial algebra over $A(0)$ and since $\text{d}(f_i) = 0$ this extends uniquely to a differential on $A(1)$ by Lemma \ref{lemma-extend-differential}. \medskip\noindent Induction hypothesis: Assume we are given factorizations $$ R \to A(0) \to A(1) \to \ldots \to A(m) \to S $$ where $A(0)$ and $A(1)$ are as above and each $R \to A(m') \to S$ for $2 \leq m' \leq m$ satisfies properties (1) and (4) of the statement of the lemma and (2) replaced by the condition that $H_i(A(m')) \to H_i(S)$ is an isomorphism for $m' > i \geq 0$. The base case is $m = 1$. \medskip\noindent Induction step: Assume we have $R \to A(m) \to S$ as in the induction hypothesis. Consider the group $H_m(A(m))$. This is a module over $H_0(A(m)) = S$. In fact, it is a subquotient of $A(m)_m$ which is a finite type module over $A(m)_0 = R[x_1, \ldots, x_n]$. Thus we can pick finitely many elements $$ e_1, \ldots, e_t \in \Ker(\text{d} : A(m)_m \to A(m)_{m - 1}) $$ which map to generators of this module. Applying Example \ref{example-adjoining-odd} or \ref{example-adjoining-even} $t$ times we get $$ A(m + 1) = A(m)\langle T_1, \ldots, T_t\rangle $$ with $\deg(T_i) = m + 1$ as a graded divided power algebra. We set $\text{d}(T_i) = e_i$. Since $A(m+1)$ is a divided power polynomial algebra over $A(m)$ and since $\text{d}(e_i) = 0$ this extends uniquely to a differential on $A(m + 1)$ compatible with the divided power structure. Since we've added only material in degree $m + 1$ and higher we see that $H_i(A(m + 1)) = H_i(A(m))$ for $i < m$. Moreover, it is clear that $H_m(A(m + 1)) = 0$ by construction. \medskip\noindent To finish the proof we observe that we have shown there exists a sequence of maps $$ R \to A(0) \to A(1) \to \ldots \to A(m) \to A(m + 1) \to \ldots \to S $$ and to finish the proof we set $A = \colim A(m)$.
This is proved in exactly the same way as Lemma \ref{lemma-tate-resolution}. The only additional twist is that, given $A(m) \to S$ we have to show that $H_m = H_m(A(m))$ is a finite $R[x_1, \ldots, x_m]$-module (so that in the next step we need only add finitely many variables). Consider the complex $$ \ldots \to A(m)_{m - 1} \to A(m)_m \to A(m)_{m - 1} \to \ldots \to A(m)_0 \to S \to 0 $$ Since $S$ is a pseudo-coherent $R[x_1, \ldots, x_n]$-module and since $A(m)_i$ is a finite free $R[x_1, \ldots, x_n]$-module we conclude that this is a pseudo-coherent complex, see More on Algebra, Lemma \ref{more-algebra-lemma-complex-pseudo-coherent-modules}. Since the complex is exact in (homological) degrees $> m$ we conclude that $H_m$ is a finite $R$-module by More on Algebra, Lemma \ref{more-algebra-lemma-finite-cohomology}.