hcju/stacksqa huggingface dataset:
The assumption means that $A$ is obtained from $R$ by successively adjoining some set of polynomial generators in degree zero, exterior generators in positive odd degrees, and divided power generators in positive even degrees. So we have a filtration $R \subset A(0) \subset A(1) \subset \ldots$ of $A$ such that $A(m + 1)$ is obtained from $A(m)$ by adjoining generators of the appropriate type (which we simply call ``divided power generators'') in degree $m + 1$. In particular, $A(0) \to H_0(A)$ is a surjection from a (usual) polynomial algebra over $R$ onto $H_0(A)$. Thus we can lift $\overline{\varphi}$ to an $R$-algebra map $\varphi(0) : A(0) \to B_0$. \medskip\noindent Write $A(1) = A(0)\langle T_j:j\in J\rangle$ for some set $J$ of divided power variables $T_j$ of degree $1$. Let $f_j \in B_0$ be $f_j = \varphi(0)(\text{d}(T_j))$. Observe that $f_j$ maps to zero in $H_0(B)$ as $\text{d}T_j$ maps to zero in $H_0(A)$. Thus we can find $b_j \in B_1$ with $\text{d}(b_j) = f_j$. By the universal property of divided power polynomial algebras from Lemma \ref{lemma-divided-power-polynomial-algebra}, we find a lift $\varphi(1) : A(1) \to B$ of $\varphi(0)$ mapping $T_j$ to $f_j$. \medskip\noindent Having constructed $\varphi(m)$ for some $m \geq 1$ we can construct $\varphi(m + 1) : A(m + 1) \to B$ in exactly the same manner. We omit the details.
Choose a factorization $R \to A \to S$ as above. Since $A \to S$ is a quasi-isomorphism and since $A_d$ is a free $R$-module, we see that the differential graded algebra $B = A \otimes_R T$ computes the Tor groups displayed in the lemma. Choose a surjection $R[y_j:j\in J] \to T$. Then we see that $B$ is a quotient of the differential graded algebra $A[y_j:j\in J]$ whose homology sits in degree $0$ (it is equal to $S[y_j:j\in J]$). By Lemma \ref{lemma-base-change-div} the differential graded algebras $B$ and $A[y_j:j\in J]$ have divided power structures compatible with the differentials. Hence we obtain our divided power structure on $H(B)$ by Lemma \ref{lemma-dpdga-good}. \medskip\noindent The divided power algebra structure constructed in this way is independent of the choice of $A$. Namely, if $A'$ is a second choice, then Lemma \ref{lemma-uniqueness-tate-resolution} implies there is a map $A \to A'$ preserving all structure and the augmentations towards $S$. Then the induced map $B = A \otimes_R T \to A' \otimes_R T' = B'$ also preserves all structure and is a quasi-isomorphism. The induced isomorphism of Tor algebras is therefore compatible with products and divided powers.
Let $A(0) \subset A(1) \subset A(2) \subset \ldots$ be the filtration of $A$ such that $A(m)$ is gotten from $A(m - 1)$ by adjoining divided power variables of degree $m$. Then $A(0)$ is a polynomial algebra over $R$ equipped with an $R$-algebra surjection $A(0) \to S$. Thus we can choose a map $$ \varphi : R[x_1, \ldots, x_n] \to A(0) $$ lifting the augmentations to $S$. Next, $A(1) = A(0)\langle T_1, \ldots, T_t \rangle$ for some divided power variables $T_i$ of degree $1$. Since $H_0(A) = S$ we can pick $\xi_j \in \sum A(0)T_i$ with $\text{d}(\xi_j) = \varphi(f_j)$. Then $$ \text{d}\left(\sum \varphi(r_j) \xi_j\right) = \sum  \varphi(r_j) \varphi(f_j) =  \sum \varphi(r_jf_j) = 0 $$ Since $H_1(A) = 0$ we can pick $\xi \in A_2$ with $\text{d}(\xi) = \sum \varphi(r_j) \xi_j$. If the coefficients of $r_j$ are in $I$, then the same is true for $\varphi(r_j)$. In this case $\text{d}(\xi)$ dies in $A_1/IA_1$ and hence $\xi$ defines a class in $H_2(A/IA)$. \medskip\noindent The construction of $\theta$ in the proof of Lemma \ref{lemma-get-derivation} proceeds by successively lifting $A(i)$ to $A'(i)$ and lifting the differential $\text{d}$. We lift $\varphi$ to $\varphi' : R'[x_1, \ldots, x_n] \to A'(0)$. Next, we have $A'(1) = A'(0)\langle T_1, \ldots, T_t\rangle$. Moreover, we can lift $\xi_j$ to $\xi'_j \in \sum A'(0)T_i$. Then $\text{d}(\xi'_j) = \varphi'(f'_j) + f a_j$ for some $a_j \in A'(0)$. Consider a lift $\xi' \in A'_2$ of $\xi$. Then we know that $$ \text{d}(\xi') = \sum \varphi'(r'_j)\xi'_j + \sum fb_iT_i $$ for some $b_i \in A(0)$. Applying $\text{d}$ again we find $$ \theta(\xi) = \sum \varphi'(r'_j)\varphi'(f'_j) + \sum f \varphi'(r'_j) a_j + \sum fb_i \text{d}(T_i) $$ The first term gives us what we want. The second term is zero because the coefficients of $r_j$ are in $I$ and hence are annihilated by $f$. The third term maps to zero in $H_0$ because $\text{d}(T_i)$ maps to zero.
Choose a Tate resolution $R \to A \to S$ as in Lemma \ref{lemma-tate-resolution}. Let $\xi \in H_2(A/IA)$ and $\theta : A/IA \to A/IA$ be the element and derivation found in Lemmas \ref{lemma-get-derivation} and \ref{lemma-compute-theta}. Observe that $$ \theta^n(\gamma_n(\xi)) = g^n $$ Hence if $g$ is not nilpotent, then $\xi^n$ is nonzero in $H_{2n}(A/IA)$ for all $n > 0$. Since $H_{2n}(A/IA) = \text{Tor}^R_{2n}(S, R/I)$ we conclude.
Let $f \in I$ be an element mapping to a nonzero element of $I/\mathfrak m I$ which is mapped to zero in $J/\mathfrak mJ$. We can choose an ideal $I'$ with $\mathfrak mI \subset I' \subset I$ such that $I/I'$ is generated by the image of $f$. Set $R = A/I$ and $R' = A/I'$. Let $J = (a_1, \ldots, a_m)$ for some $a_j \in A$. Then $f = \sum b_j a_j$ for some $b_j \in \mathfrak m$. Let $r_j, f_j \in R$ resp.\ $r'_j, f'_j \in R'$ be the image of $b_j, a_j$. Then we see we are in the situation of Lemma \ref{lemma-not-finite-pd} (with the ideal $I$ of that lemma equal to $\mathfrak m_R$) and the lemma is proved.
By Lemma \ref{lemma-injective} we see that $I/\mathfrak m I \to J/\mathfrak m J$ is injective. Thus we can find $s \leq r$ and a minimal system of generators $f_1, \ldots, f_r$ of $J$ such that $f_1, \ldots, f_s$ are in $I$ and form a minimal system of generators of $I$. The lemma follows as any minimal system of generators of $J$ is a regular sequence by More on Algebra, Lemmas \ref{more-algebra-lemma-independence-of-generators} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}.
If (3) holds, then $J$ is generated by a regular sequence, see for example More on Algebra, Lemmas \ref{more-algebra-lemma-join-koszul-regular-sequences} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. Moreover, if (3) holds, then $S/J = (S/I)/(J/I)$ has finite projective dimension over $S/IS$ because the Koszul complex will be a finite free resolution of $S/J$ over $S/IS$. Since $R/I \to S/IS$ is flat, it then follows that $S/J$ has finite tor dimension over $R/I$ by More on Algebra, Lemma \ref{more-algebra-lemma-flat-push-tor-amplitude}. Thus (3) implies (1). \medskip\noindent The implication (1) $\Rightarrow$ (2) is trivial. Assume (2). By More on Algebra, Lemma \ref{more-algebra-lemma-perfect-over-regular-local-ring} we find that $S/J$ has finite tor dimension over $S/IS$. Thus we can apply Lemma \ref{lemma-regular-sequence} to conclude that $IS$ and $J/IS$ are generated by regular sequences. Let $f_1, \ldots, f_r \in I$ be a minimal system of generators of $I$. Since $R \to S$ is flat, we see that $f_1, \ldots, f_r$ form a minimal system of generators for $IS$ in $S$. Thus $f_1, \ldots, f_r \in R$ is a sequence of elements whose images in $S$ form a regular sequence by More on Algebra, Lemmas \ref{more-algebra-lemma-independence-of-generators} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. Thus $f_1, \ldots, f_r$ is a regular sequence in $R$ by Algebra, Lemma \ref{algebra-lemma-flat-increases-depth}.
Let $k$ be the residue field of $A$. If the characteristic of $k$ is $p > 0$, then we denote $\Lambda$ a Cohen ring (Algebra, Definition \ref{algebra-definition-cohen-ring}) with residue field $k$ (Algebra, Lemma \ref{algebra-lemma-cohen-rings-exist}). If the characteristic of $k$ is $0$ we set $\Lambda = k$. Recall that $\Lambda[[x_1, \ldots, x_n]]$ for any $n$ is formally smooth over $\mathbf{Z}$, resp.\ $\mathbf{Q}$ in the $\mathfrak m$-adic topology, see More on Algebra, Lemma \ref{more-algebra-lemma-power-series-ring-over-Cohen-fs}. Fix a surjection $\Lambda[[x_1, \ldots, x_n]] \to A$ as in the Cohen structure theorem (Algebra, Theorem \ref{algebra-theorem-cohen-structure-theorem}). \medskip\noindent Let $R \to A$ be a surjection from a regular local ring $R$. Let $f_1, \ldots, f_r$ be a minimal sequence of generators of $\Ker(R \to A)$. We will use without further mention that an ideal in a Noetherian local ring is generated by a regular sequence if and only if any minimal set of generators is a regular sequence. Observe that $f_1, \ldots, f_r$ is a regular sequence in $R$ if and only if $f_1, \ldots, f_r$ is a regular sequence in the completion $R^\wedge$ by Algebra, Lemmas \ref{algebra-lemma-flat-increases-depth} and \ref{algebra-lemma-completion-flat}. Moreover, we have $$ R^\wedge/(f_1, \ldots, f_r)R^\wedge = (R/(f_1, \ldots, f_n))^\wedge = A^\wedge = A $$ because $A$ is $\mathfrak m_A$-adically complete (first equality by Algebra, Lemma \ref{algebra-lemma-completion-tensor}). Finally, the ring $R^\wedge$ is regular since $R$ is regular (More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}). Hence we may assume $R$ is complete. \medskip\noindent If $R$ is complete we can choose a map $\Lambda[[x_1, \ldots, x_n]] \to R$ lifting the given map $\Lambda[[x_1, \ldots, x_n]] \to A$, see More on Algebra, Lemma \ref{more-algebra-lemma-lift-continuous}. By adding some more variables $y_1, \ldots, y_m$ mapping to generators of the kernel of $R \to A$ we may assume that $\Lambda[[x_1, \ldots, x_n, y_1, \ldots, y_m]] \to R$ is surjective (some details omitted). Then we can consider the commutative diagram $$ \xymatrix{ \Lambda[[x_1, \ldots, x_n, y_1, \ldots, y_m]] \ar[r] \ar[d] & R \ar[d] \\ \Lambda[[x_1, \ldots, x_n]] \ar[r] & A } $$ By Algebra, Lemma \ref{algebra-lemma-ci-well-defined} we see that the condition for $R \to A$ is equivalent to the condition for the fixed chosen map $\Lambda[[x_1, \ldots, x_n]] \to A$. This finishes the proof of the lemma.
The completion of $A$ is equal to $A^\wedge = R_\mathfrak p^\wedge/(f_1, \ldots, f_r)R_\mathfrak p^\wedge$ because completion for finite modules over the Noetherian ring $R_\mathfrak p$ is exact (Algebra, Lemma \ref{algebra-lemma-completion-tensor}). The image of the sequence $f_1, \ldots, f_r$ in $R_\mathfrak p$ is a regular sequence by Algebra, Lemmas \ref{algebra-lemma-completion-flat} and \ref{algebra-lemma-flat-increases-depth}. Moreover, $R_\mathfrak p^\wedge$ is a regular local ring by More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}. Hence the result holds by our definition of complete intersection for complete local rings.
We may and do replace $R$ by its localization at $\mathfrak p$. Then $\mathfrak p = \mathfrak m$ is the maximal ideal of $R$ and $A = R/I$. Let $f_1, \ldots, f_r \in I$ be a minimal sequence of generators.  The completion of $A$ is equal to $A^\wedge = R^\wedge/(f_1, \ldots, f_r)R^\wedge$ because completion for finite modules over the Noetherian ring $R_\mathfrak p$ is exact (Algebra, Lemma \ref{algebra-lemma-completion-tensor}). \medskip\noindent If (1) holds, then the image of the sequence $f_1, \ldots, f_r$ in $R^\wedge$ is a regular sequence by assumption. Hence it is a regular sequence in $R$ by Algebra, Lemmas \ref{algebra-lemma-completion-flat} and \ref{algebra-lemma-flat-increases-depth}. Thus (1) implies (2). \medskip\noindent Assume (3) holds. Set $c = \dim(R) - \dim(A)$ and let $f_1, \ldots, f_c \in I$ map to generators of $I/I^2$. by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we see that $I = (f_1, \ldots, f_c)$. Since $R$ is regular and hence Cohen-Macaulay (Algebra, Proposition \ref{algebra-proposition-CM-module}) we see that $f_1, \ldots, f_c$ is a regular sequence by Algebra, Proposition \ref{algebra-proposition-CM-module}. Thus (3) implies (2). Finally, (2) implies (1) by Lemma \ref{lemma-quotient-regular-ring-by-regular-sequence}.
Choose a prime $\mathfrak q$ of $A^\wedge$ lying over $\mathfrak p$ (this is possible as $A \to A^\wedge$ is faithfully flat by Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}). Then $A_\mathfrak p \to (A^\wedge)_\mathfrak q$ is a flat local ring homomorphism. Thus by Proposition \ref{proposition-avramov} we see that $A_\mathfrak p$ is a complete intersection if and only if $(A^\wedge)_\mathfrak q$ is a complete intersection. Thus it suffices to prove the lemma in case $A$ is complete (this is the key step of the proof). \medskip\noindent Assume $A$ is complete. By definition we may write $A = R/(f_1, \ldots, f_r)$ for some regular sequence $f_1, \ldots, f_r$ in a regular local ring $R$. Let $\mathfrak q \subset R$ be the prime corresponding to $\mathfrak p$. Observe that $f_1, \ldots, f_r \in \mathfrak q$ and that $A_\mathfrak p = R_\mathfrak q/(f_1, \ldots, f_r)R_\mathfrak q$. Hence $A_\mathfrak p$ is a complete intersection by Lemma \ref{lemma-quotient-regular-ring-by-regular-sequence}.
Proof of (1). Let $k[x_1, \ldots, x_n] \to S$ be a surjection. Let $\mathfrak p \subset k[x_1, \ldots, x_n]$ be the prime ideal corresponding to $\mathfrak q$. Let $I \subset k[x_1, \ldots, x_n]$ be the kernel of our surjection. Note that $k[x_1, \ldots, x_n]_\mathfrak p \to S_\mathfrak q$ is surjective with kernel $I_\mathfrak p$. Observe that $k[x_1, \ldots, x_n]$ is a regular ring by Algebra, Proposition \ref{algebra-proposition-finite-gl-dim-polynomial-ring}. Hence the equivalence of the two notions in (1) follows by combining Lemma \ref{lemma-quotient-regular-ring} with Algebra, Lemma \ref{algebra-lemma-lci-local}. \medskip\noindent Having proved (1) the equivalence in (2) follows from the definition and Algebra, Lemma \ref{algebra-lemma-lci-global}.
Choose a surjection $R = A[x_1, \ldots, x_n] \to B$. Observe that $A \to R$ is flat with regular fibres. Let $I$ be the kernel of $R \to B$. Assume (2). Then we see that $I$ is locally generated by a regular sequence by Lemma \ref{lemma-well-defined-if-you-can-find-good-factorization} and Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood}. In other words, (1) holds. Conversely, assume (1). Then after localizing on $R$ and $B$ we can assume that $I$ is generated by a Koszul regular sequence. By More on Algebra, Lemma \ref{more-algebra-lemma-noetherian-finite-all-equivalent} we find that $I$ is locally generated by a regular sequence. Hence (2) hold by Lemma \ref{lemma-well-defined-if-you-can-find-good-factorization}. Some details omitted.
By assumption we may choose a finite free resolution $$ 0 \to R^{\oplus n_e} \to R^{\oplus n_{e-1}} \to \ldots \to R^{\oplus n_1} \to R \to R/I \to 0 $$ Then $\varphi_1 : R^{\oplus n_1} \to R$ has rank $1$ and we see that $I$ contains a nonzerodivisor $y$ by Algebra, Proposition \ref{algebra-proposition-what-exact}. Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be the associated primes of $R$, see Algebra, Lemma \ref{algebra-lemma-finite-ass}. Let $I^2 \subset J \subset I$ be an ideal such that $J/I^2 = F$. Then $J \not \subset \mathfrak p_i$ for all $i$ as $y^2 \in J$ and $y^2 \not \in \mathfrak p_i$, see Algebra, Lemma \ref{algebra-lemma-ass-zero-divisors}. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we have $J \not \subset \mathfrak m J + I^2$. By Algebra, Lemma \ref{algebra-lemma-silly} we can pick $x \in J$, $x \not \in \mathfrak m J + I^2$ and $x \not \in \mathfrak p_i$ for $i = 1, \ldots, n$. Then $x$ is a nonzerodivisor and the image of $x$ in $I/I^2$ generates (by Nakayama's lemma) the summand $J/I^2 \cong R/I$.
If $r = 0$ there is nothing to prove. Assume $r > 0$. We may pick $x \in I$ such that $x$ is a nonzerodivisor and $x \bmod I^2$ generates a summand of $F$ isomorphic to $R/I$, see Lemma \ref{lemma-free-summand-in-ideal-finite-proj-dim}. Consider the ring $R' = R/(x)$ and the ideal $I' = I/(x)$. Of course $R'/I' = R/I$. The short exact sequence $$ 0 \to R/I \xrightarrow{x} I/xI \to I' \to 0 $$ splits because the map $I/xI \to I/I^2$ sends $xR/xI$ to a direct summand. Now $I/xI = I \otimes_R^\mathbf{L} R'$ has finite projective dimension over $R'$, see More on Algebra, Lemmas \ref{more-algebra-lemma-perfect-module} and \ref{more-algebra-lemma-pull-perfect}. Hence the summand $I'$ has finite projective dimension over $R'$. On the other hand, we have the short exact sequence $0 \to xR/xI \to I/I^2 \to I'/(I')^2 \to 0$ and we conclude $I'/(I')^2$ has the free direct summand $F' = F/(R/I \cdot x)$ of rank $r - 1$. By induction on $r$ we may we pick a regular sequence $x'_2, \ldots, x'_r \in I'$ such that there congruence classes freely generate $F'$. If $x_1 = x$ and $x_2, \ldots, x_r$ are any elements lifting $x'_1, \ldots, x'_r$ in $R$, then we see that the lemma holds.
Write $B = A[x_1, \ldots, x_n]/I$. Then $\NL_{B/A}$ is represented by the complex $$ I/I^2 \longrightarrow \bigoplus B \text{d}x_i $$ of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is finite free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see More on Algebra, Lemma \ref{more-algebra-lemma-last-one-flat}. Since $I/I^2$ is a finite $B$-module and $B$ is Noetherian, this is true if and only if $I/I^2$ is a finite locally free $B$-module (Algebra, Lemma \ref{algebra-lemma-finite-projective}). Thus the equivalence of (1) and (2) is clear. Moreover, the equivalence of (1) and (3) also follows if we apply Proposition \ref{proposition-regular-ideal} (and the observation that a regular ideal is a Koszul regular ideal as well as a quasi-regular ideal, see More on Algebra, Section \ref{more-algebra-section-ideals}).
The equivalence of (3) and (4) is More on Algebra, Lemma \ref{more-algebra-lemma-syntomic-lci}. \medskip\noindent If $A \to B$ is syntomic, then we can find a cocartesian diagram $$ \xymatrix{ B_0 \ar[r] & B \\ A_0 \ar[r] \ar[u] & A \ar[u] } $$ such that $A_0 \to B_0$ is syntomic and $A_0$ is Noetherian, see Algebra, Lemmas \ref{algebra-lemma-limit-module-finite-presentation} and \ref{algebra-lemma-colimit-lci}. By Lemma \ref{lemma-perfect-NL-lci} we see that $\NL_{B_0/A_0}$ is perfect of tor-amplitude in $[-1, 0]$. By More on Algebra, Lemma \ref{more-algebra-lemma-base-change-NL-flat} we conclude the same thing is true for $\NL_{B/A} = \NL_{B_0/A_0} \otimes_{B_0}^\mathbf{L} B$ (see also More on Algebra, Lemmas \ref{more-algebra-lemma-pull-tor-amplitude} and \ref{more-algebra-lemma-pull-perfect}). This proves that (3) implies (2). \medskip\noindent Assume (1). By More on Algebra, Lemma \ref{more-algebra-lemma-base-change-NL-flat} for every ring map $A \to k$ where $k$ is a field, we see that $\NL_{B \otimes_A k/k}$ has tor-amplitude in $[-1, 0]$ (see More on Algebra, Lemma \ref{more-algebra-lemma-pull-tor-amplitude}). Hence by Lemma \ref{lemma-perfect-NL-lci} we see that $k \to B \otimes_A k$ is a local complete intersection homomorphism. Thus $A \to B$ is syntomic by definition. This proves (1) implies (3) and finishes the proof.
Consider the diagram $$ \xymatrix{ B \ar[r] & B^\wedge \\ A \ar[u] \ar[r] & A^\wedge \ar[u] } $$ Since the horizontal maps are faithfully flat (Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}) we conclude that the right vertical arrow is flat (for example by Algebra, Lemma \ref{algebra-lemma-criterion-flatness-fibre-Noetherian}). Moreover, we have $(B/\mathfrak m_A B)^\wedge = B^\wedge/\mathfrak m_{A^\wedge} B^\wedge$ by Algebra, Lemma \ref{algebra-lemma-completion-tensor}. Thus we may assume $A$ and $B$ are complete local Noetherian rings. \medskip\noindent Assume $A$ and $B$ are complete local Noetherian rings. Choose a diagram $$ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } $$ as in More on Algebra, Lemma \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings}. Let $I = \Ker(R \to A)$ and $J = \Ker(S \to B)$. Note that since $R/I = A \to B = S/J$ is flat the map $J/IS \otimes_R R/\mathfrak m_R \to J/J \cap \mathfrak m_R S$ is an isomorphism. Hence a minimal system of generators of $J/IS$ maps to a minimal system of generators of $\Ker(S/\mathfrak m_R S \to B/\mathfrak m_A B)$. Finally, $S/\mathfrak m_R S$ is a regular local ring. \medskip\noindent Assume (1) holds, i.e., $J$ is generated by a regular sequence. Since $A = R/I \to B = S/J$ is flat we see Lemma \ref{lemma-perfect-map-ci} applies and we deduce that $I$ and $J/IS$ are generated by regular sequences. We have $\dim(B) = \dim(A) + \dim(B/\mathfrak m_A B)$ and $\dim(S/IS) = \dim(A) + \dim(S/\mathfrak m_R S)$ (Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}). Thus $J/IS$ is generated by $$ \dim(S/J) - \dim(S/IS) = \dim(S/\mathfrak m_R S) - \dim(B/\mathfrak m_A B) $$ elements (Algebra, Lemma \ref{algebra-lemma-one-equation}). It follows that $\Ker(S/\mathfrak m_R S \to B/\mathfrak m_A B)$ is generated by the same number of elements (see above). Hence $\Ker(S/\mathfrak m_R S \to B/\mathfrak m_A B)$ is generated by a regular sequence, see for example Lemma \ref{lemma-quotient-regular-ring}. In this way we see that (2) holds. \medskip\noindent If (2) holds, then $I$ and $J/J \cap \mathfrak m_RS$ are generated by regular sequences. Lifting these generators (see above), using flatness of $R/I \to S/IS$, and using Grothendieck's lemma (Algebra, Lemma \ref{algebra-lemma-grothendieck-regular-sequence}) we find that $J/IS$ is generated by a regular sequence in $S/IS$. Thus Lemma \ref{lemma-perfect-map-ci} tells us that $J$ is generated by a regular sequence, whence (1) holds.
Let $F_\bullet \to A/\mathfrak m_A$ be a resolution by finite free $A$-modules. Observe that $\text{Tor}^A_p(B, A/\mathfrak m_A)$ is the $p$th homology of the complex $F_\bullet \otimes_A B$. Let $F_\bullet^\wedge = F_\bullet \otimes_A A^\wedge$ be the completion. Then $F_\bullet^\wedge$ is a resolution of $A^\wedge/\mathfrak m_{A^\wedge}$ by finite free $A^\wedge$-modules (as $A \to A^\wedge$ is flat and completion on finite modules is exact, see Algebra, Lemmas \ref{algebra-lemma-completion-tensor} and \ref{algebra-lemma-completion-flat}). It follows that $$ F_\bullet^\wedge \otimes_{A^\wedge} B^\wedge = F_\bullet \otimes_A B \otimes_B B^\wedge $$ By flatness of $B \to B^\wedge$ we conclude that $$ \text{Tor}^{A^\wedge}_p(B^\wedge, A^\wedge/\mathfrak m_{A^\wedge}) = \text{Tor}^A_p(B, A/\mathfrak m_A) \otimes_B B^\wedge $$ In this way we see that the condition in (1) on the local ring map $A \to B$ is equivalent to the same condition for the local ring map $A^\wedge \to B^\wedge$. Thus we may assume $A$ and $B$ are complete local Noetherian rings (since the other conditions are formulated in terms of the completions in any case). \medskip\noindent Assume $A$ and $B$ are complete local Noetherian rings. Choose a diagram $$ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } $$ as in More on Algebra, Lemma \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings}. Let $I = \Ker(R \to A)$ and $J = \Ker(S \to B)$. The proposition now follows from Lemma \ref{lemma-perfect-map-ci}.
By Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood} it suffices to show that $I_\mathfrak p \subset R_\mathfrak p$ is generated by a regular sequence for every $\mathfrak p \supset I$. Thus we may assume $R$ is local. If $I/I^2$ has rank $r$, then by Lemma \ref{lemma-vasconcelos} we find a regular sequence $x_1, \ldots, x_r \in I$ generating $I/I^2$. By Nakayama (Algebra, Lemma \ref{algebra-lemma-NAK}) we conclude that $I$ is generated by $x_1, \ldots, x_r$.
The representability of the diagonal (by algebraic spaces) was shown in Quot, Lemma \ref{quot-lemma-coherent-diagonal}. From the proof we find that we have to show $\mathit{Isom}(\mathcal{F}, \mathcal{G}) \to T$ is affine and of finite presentation for a pair of finitely presented $\mathcal{O}_{X_T}$-modules $\mathcal{F}$, $\mathcal{G}$ flat over $T$ with support proper over $T$. This was discussed in Section \ref{section-hom-isom}.
To check $\Cohstack_{X/B} \to B$ is quasi-separated we have to show that its diagonal is quasi-compact and quasi-separated. This is immediate from Lemma \ref{lemma-coherent-diagonal-affine-fp}. To prove that $\Cohstack_{X/B} \to B$ is locally of finite presentation, we have to show that $\Cohstack_{X/B} \to B$ is limit preserving, see Limits of Stacks, Proposition \ref{stacks-limits-proposition-characterize-locally-finite-presentation}. This follows from Quot, Lemma \ref{quot-lemma-coherent-limits} (small detail omitted).