hcju/stacksqa huggingface dataset:
Observe that $\Spec(F) : \Spec(A) \to \Spec(A)$ is the identity map. Being regular is defined in terms of the local rings and being flat is something about local rings, see Algebra, Lemma \ref{algebra-lemma-flat-localization}. Thus we may and do assume $A$ is a Noetherian local ring with maximal ideal $\mathfrak m$. \medskip\noindent Assume $A$ is regular. Let $x_1, \ldots, x_d$ be a system of parameters for $A$. Applying $F$ we find $F(x_1), \ldots, F(x_d) = x_1^p, \ldots, x_d^p$, which is a system of parameters for $A$. Hence $F$ is flat, see Algebra, Lemmas \ref{algebra-lemma-CM-over-regular-flat} and \ref{algebra-lemma-regular-ring-CM}. \medskip\noindent Conversely, assume $F$ is flat. Write $\mathfrak m = (x_1, \ldots, x_r)$ with $r$ minimal. Then $x_1, \ldots, x_r$ are independent in the sense defined above. Since $F$ is flat, we see that $x_1^p, \ldots, x_r^p$ are independent, see Lemma \ref{lemma-flat-extension-independent}. Hence $\text{length}_A(A/(x_1^p, \ldots, x_r^p)) = p^r$ by Lemma \ref{lemma-3}. Let $\chi(n) = \text{length}_A(A/\mathfrak m^n)$ and recall that this is a numerical polynomial of degree $\dim(A)$, see Algebra, Proposition \ref{algebra-proposition-dimension}. Choose $n \gg 0$. Observe that $$ \mathfrak m^{pn + pr} \subset F(\mathfrak m^n)A \subset \mathfrak m^{pn} $$ as can be seen by looking at monomials in $x_1, \ldots, x_r$. We have $$ A/F(\mathfrak m^n)A = A/\mathfrak m^n \otimes_{A, F} A $$ By flatness of $F$ this has length $\chi(n) \text{length}_A(A/F(\mathfrak m)A)$ (Algebra, Lemma \ref{algebra-lemma-pullback-module}) which is equal to $p^r\chi(n)$ by the above. We conclude $$ \chi(pn + pr) \geq p^r\chi(n) \geq \chi(pn) $$ Looking at the leading terms this implies $r = \dim(A)$, i.e., $A$ is regular.
Choose a set $J$ and an isomorphism $M[\mathfrak m] \to \bigoplus_{j \in J} k$. Since $\bigoplus_{j \in J} E$ is injective (Dualizing Complexes, Lemma \ref{dualizing-lemma-sum-injective-modules}) we can extend this isomorphism to an $A$-module homomorphism $\varphi : M \to \bigoplus_{j \in J} E$. We claim that $\varphi$ is an isomorphism, i.e., bijective. \medskip\noindent Injective. Let $z \in M$ be nonzero. Since $M$ is $\mathfrak m$-power torsion we can choose an element $f \in A$ such that $fz \in M[\mathfrak m]$ and $fz \not = 0$. Then $\varphi(fz) = f\varphi(z)$ is nonzero, hence $\varphi(z)$ is nonzero. \medskip\noindent Surjective. Let $z \in M$. Then $x_1^n z = 0$ for some $n \geq 0$. We will prove that $z \in x_1M$ by induction on $n$. If $n = 0$, then $z = 0$ and the result is true. If $n > 0$, then applying $D_1$ we find $0 = n x_1^{n - 1} z + x_1^nD_1(z)$. Hence $x_1^{n - 1}(nz + x_1D_1(z)) = 0$. By induction we get $nz + x_1D_1(z) \in x_1M$. Since $n$ is invertible, we conclude $z \in x_1M$. Thus we see that $M$ is $x_1$-divisible. If $\varphi$ is not surjective, then we can choose $e \in \bigoplus_{j \in J} E$ not in $M$. Arguing as above we may assume $\mathfrak m e \subset M$, in particular $x_1 e \in M$. There exists an element $z_1 \in M$ with $x_1 z_1 = x_1 e$. Hence $x_1(z_1 - e) = 0$. Replacing $e$ by $e - z_1$ we may assume $e$ is annihilated by $x_1$. Thus it suffices to prove that $$ \varphi[x_1] : M[x_1] \longrightarrow \left(\bigoplus\nolimits_{j \in J} E\right)[x_1] = \bigoplus\nolimits_{j \in J} E[x_1] $$ is surjective. If $d = 1$, this is true by construction of $\varphi$. If $d > 1$, then we observe that $E[x_1]$ is the injective hull of the residue field of $k[[x_2, \ldots, x_d]]$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-quotient}. Observe that $M[x_1]$ as a module over $k[[x_2, \ldots, x_d]]$ is $\mathfrak m/(x_1)$-power torsion and comes equipped with operators $D_2, \ldots, D_d$ satisfying the displayed Leibniz rule. Thus by induction on $d$ we conclude that $\varphi[x_1]$ is surjective as desired.
Let $f_1, \ldots, f_r$ be elements generating $I$. Recall that $R\Gamma_Z(A)$ is computed by the complex $$ A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r} $$ See Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-adjoint}. Since $\theta$ extends uniquely to an additive operator on any localization of $A$ satisfying the Leibniz rule with respect to $\theta$, the lemma is clear.
If $V$ is separated, then we can argue using an affine open covering $V = \bigcup_{j = 1, \ldots m} V_j$. Namely, because $V$ is separated we may write $V_{j_0 \ldots j_p} = \Spec(B_{j_0 \ldots j_p})$. See Schemes, Lemma \ref{schemes-lemma-characterize-separated}. Then we find that the $A$-module $H^i(V, \mathcal{O}_V)$ is the $i$th cohomology group of the {\v C}ech complex $$ \prod B_{j_0} \to \prod B_{j_0j_1} \to \prod B_{j_0j_1j_2} \to \ldots $$ See Cohomology of Schemes, Lemma \ref{coherent-lemma-cech-cohomology-quasi-coherent}. Each $B = B_{j_0 \ldots j_p}$ is an \'etale $A$-algebra. Hence $\Omega_B = \Omega_A \otimes_A B$ and we conclude $\theta$ extends uniquely to a derivation $\theta_B : B \to B$. These maps define an endomorphism of the {\v C}ech complex and define the desired operators on the cohomology groups. \medskip\noindent In the general case we use a hypercovering of $V$ by affine opens, exactly as in the first part of the proof of Cohomology of Schemes, Lemma \ref{coherent-lemma-hypercoverings}. We omit the details.
Observe that the relative Frobenius morphism $$ V \longrightarrow V \times_{\Spec(A), \Spec(F)} \Spec(A) $$ of $V$ over $A$ is an isomorphism, see \'Etale Morphisms, Lemma \ref{etale-lemma-relative-frobenius-etale}. Thus the lemma follows from cohomology and base change, see Derived Categories of Schemes, Lemma \ref{perfect-lemma-compare-base-change}. Observe that since $V$ is \'etale over $A$, it is flat over $A$.
We may choose a map of short exact sequences $$ \xymatrix{ 0 \ar[r] & K \ar[r] \ar[d] & R^{\oplus r} \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K' \ar[r] & \bigoplus_{i \in I} R \ar[r] & M' \ar[r] & 0 } $$ whose right vertical arrow is the given map. We can factor this map through the short exact sequence \begin{equation} \label{equation-pushout} 0 \to K' \to E \to M \to 0 \end{equation} which is the pushout of the first short exact sequence by $K \to K'$. By a diagram chase we see that the assumption in the lemma implies that the boundary map $\text{Tor}_1^R(M, N) \to K' \otimes_R N$ induced by (\ref{equation-pushout}) is zero, i.e., the sequence (\ref{equation-pushout}) is universally exact. This implies by Algebra, Lemma \ref{algebra-lemma-universally-exact-split} that (\ref{equation-pushout}) is split (this is where we use that $M$ is of finite presentation). Hence the map $M \to M'$ factors through $\bigoplus_{i \in I} R$ and we win.
It is clear that (2) implies (1). If $a_\bullet$ is as in (1), then we can consider the map of complexes $a'_\bullet : P_\bullet \to P'_\bullet$ with $a'_i = a_i$ for $i \leq e + 1$ and $a'_i = 0$ for $i \geq e + 1$ to get a map of complexes as in (2). Thus (1) is equivalent to (2). \medskip\noindent By the construction of the $\Ext$ and $\text{Tor}$ functors using resolutions (Algebra, Sections \ref{algebra-section-ext} and \ref{algebra-section-tor}) we see  that (1) and (2) imply all of the other conditions. \medskip\noindent It is clear that (3) implies (4) implies (5). Let $N$ be as in (5). The canonical map $\tilde \xi : P'_{e + 1} \to N$ precomposed with $P'_{e + 2} \to P'_{e + 1}$ is zero. Hence we may consider the class $\xi$ of $\tilde \xi$ in $$ \Ext^{e + 1}_R(M', N) = \frac{\Ker(\Hom(P'_{e + 1}, N \to \Hom(P'_{e + 2}, N)}{ \Im(\Hom(P'_e, N \to \Hom(P'_{e + 1}, N)} $$ Choose a map of complexes $a_\bullet : P_\bullet \to P'_\bullet$ lifting $\alpha$, see Derived Categories, Lemma \ref{derived-lemma-morphisms-lift-projective}. If $\xi$ maps to zero in $\Ext^{e + 1}_R(M', N)$, then we find a map $\varphi : P_e \to N$ such that $\tilde \xi  \circ a_{e + 1} = \varphi \circ d$. Thus we obtain a map of complexes $$ \xymatrix{ \ldots \ar[r] & P_{e + 1} \ar[r] \ar[d]^0 & P_e \ar[r] \ar[d]^{a_e - \varphi} & P_{e - 1} \ar[r] \ar[d]^{a_{e - 1}} & \ldots \\ \ldots \ar[r] & P'_{e + 1} \ar[r] & P'_e \ar[r] & P'_{e - 1} \ar[r] & \ldots } $$ as in (2). Hence (1) -- (5) are equivalent. \medskip\noindent The equivalence of (6) and (7) follows from dimension shifting; we omit the details. \medskip\noindent Assume $M$ is $(-e - 1)$-pseudo-coherent. (The parenthetical statement in the lemma follows from More on Algebra, Lemma \ref{more-algebra-lemma-Noetherian-pseudo-coherent}.) We will show that (7) implies (4) which finishes the proof. We will use induction on $e$. The base case is $e = 0$. Then $M$ is of finite presentation by More on Algebra, Lemma \ref{more-algebra-lemma-n-pseudo-module} and we conclude from Lemma \ref{lemma-map-tor-1-zero} that $M \to M'$ factors through a free module. Of course if $M \to M'$ factors through a free module, then $\Ext^i_R(M', N) \to \Ext^i_R(M, N)$ is zero for all $i > 0$ as desired. Assume $e > 0$. We may choose a map of short exact sequences $$ \xymatrix{ 0 \ar[r] & K \ar[r] \ar[d] & R^{\oplus r} \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K' \ar[r] & \bigoplus_{i \in I} R \ar[r] & M' \ar[r] & 0 } $$ whose right vertical arrow is the given map. We obtain $\text{Tor}_{i + 1}^R(M, N) = \text{Tor}^R_i(K, N)$ and $\Ext^{i + 1}_R(M, N) = \Ext^i_R(K, N)$ for $i \geq 1$ and all $R$-modules $N$ and similarly for $M', K'$. Hence we see that $\text{Tor}_e^R(K, N) \to \text{Tor}_e^R(K', N)$ is zero for all $R$-modules $N$. By More on Algebra, Lemma \ref{more-algebra-lemma-cone-pseudo-coherent} we see that $K$ is $(-e)$-pseudo-coherent. By induction we conclude that $\Ext^e(K', N) \to \Ext^e(K, N)$ is zero for all $R$-modules $N$, which gives what we want.
Let $\xi \in \Ext^{e + 1}_A(A/I^n, N)$ be the element constructed in Lemma \ref{lemma-characterize-vanishing-tor-ext-above-e} part (5). Since $e = \text{cd}(A, I)$ we have $0 = H^{e + 1}_Z(N) = H^{e + 1}_I(N) = \colim \Ext^{e + 1}(A/I^m, N)$ by Dualizing Complexes, Lemmas \ref{dualizing-lemma-local-cohomology-noetherian} and \ref{dualizing-lemma-local-cohomology-ext}. Thus we may pick $m \geq n$ such that $\xi$ maps to zero in $\Ext^{e + 1}_A(A/I^m, N)$ as desired.
By induction on $i$. Base case $i = 1$. The short exact sequence $0 \to I^n \to A \to A/I^n \to 0$ determines an injection $\text{Tor}_1^A(M, A/I^n) \subset I^n \otimes_A M$, see Algebra, Remark \ref{algebra-remark-Tor-ring-mod-ideal}. As $M$ is annihilated by $I^m$ we see that the map $I^n \otimes_A M \to I^{n - m} \otimes_A M$ is zero for $n \geq m$. Hence the result holds with $c = m$. \medskip\noindent Induction step. Let $i > 1$ and assume $c$ works for $i - 1$. By More on Algebra, Lemma \ref{more-algebra-lemma-tor-strictly-pro-zero} applied to $M = A/I^m$ we can choose $c' \geq 0$ such that $\text{Tor}_i(A/I^m, A/I^n) \to \text{Tor}_i(A/I^m, A/I^{n - c'})$ is zero for $n \geq c'$. Let $M$ be annihilated by $I^m$. Choose a short exact sequence $$ 0 \to S \to \bigoplus\nolimits_{i \in I} A/I^m \to M \to 0 $$ The corresponding long exact sequence of tors gives an exact sequence $$ \text{Tor}_i^A(\bigoplus\nolimits_{i \in I} A/I^m, A/I^n) \to \text{Tor}_i^A(M, A/I^n) \to \text{Tor}_{i - 1}^A(S, A/I^n) $$ for all integers $n \geq 0$. If $n \geq c + c'$, then the map $\text{Tor}_{i - 1}^A(S, A/I^n) \to \text{Tor}_{i - 1}^A(S, A/I^{n - c})$ is zero and the map $\text{Tor}_i^A(A/I^m, A/I^{n - c}) \to  \text{Tor}_i^A(A/I^m, A/I^{n - c - c'})$ is zero. Combined with the short exact sequences this implies the result holds for $i$ with constant $c + c'$.
Recall that $B$ is the quotient of $A[x_2, \ldots, x_t]/(a_1x_2 - a_2, \ldots, a_1x_t - a_t)$ by its $a_1$-torsion, see Algebra, Lemma \ref{algebra-lemma-affine-blowup-quotient-description}. Let $$ B_\bullet = \text{Koszul complex on }a_1x_2 - a_2, \ldots, a_1x_t - a_t \text{ over }A[x_2, \ldots, x_t] $$ viewed as a chain complex sitting in degrees $(t - 1), \ldots, 0$. The complex $B_\bullet[1/a_1]$ is isomorphic to the Koszul complex on $x_2 - a_2/a_1, \ldots, x_t - a_t/a_1$ which is a regular sequence in $A[1/a_1][x_2, \ldots, x_t]$. Since regular sequences are Koszul regular, we conclude that the augmentation $$ \epsilon : B_\bullet \longrightarrow B $$ is a quasi-isomorphism after inverting $a_1$. Since the homology modules of the cone $C_\bullet$ on $\epsilon$ are finite $A[x_2, \ldots, x_n]$-modules and since $C_\bullet$ is bounded, we conclude that there exists a $c \geq 0$ such that $a_1^c$ annihilates all of these. By Derived Categories, Lemma \ref{derived-lemma-trick-vanishing-composition} this implies that, after possibly replacing $c$ by a larger integer, that $a_1^c$ is zero on $C_\bullet$ in $D(A)$. The proof is finished once the reader contemplates the distinguished triangle $$ B_\bullet \otimes_A^\mathbf{L} M \to B \otimes_A^\mathbf{L} M \to C_\bullet \otimes_A^\mathbf{L} M $$ Namely, the first term is represented by $B_\bullet \otimes_A M$ which is sitting in homological degrees $(t - 1), \ldots, 0$ in view of the fact that the terms in the Koszul complex $B_\bullet$ are free (and hence flat) $A$-modules. Whence $\text{Tor}_i^A(B, M) = H_i(C_\bullet \otimes_A^\mathbf{L} M)$ for $i > t - 1$ and this is annihilated by $a_1^c$. Since $a_1^cB = I^cB$ and since the tor module is a module over $B$ we conclude.
Choose a free resolution $F_\bullet \to M$. Then $\widetilde{F}_\bullet$ is a flat resolution of $\widetilde{M}$. Hence $Lp^*\widetilde{M}$ is given by the complex $p^*\widetilde{F}_\bullet$. Thus $Lp^*\widetilde{M}(q)$ is given by the complex $p^*\widetilde{F}_\bullet(q)$. Since $p^*\widetilde{F}_i(q)$ are right acyclic for $\Gamma(X, -)$ by our choice of $q \geq q(A, I)$ and since we have $\Gamma(X, p^*\widetilde{F}_i(q)) = I^qF_i$ by our choice of $q \geq q(A, I)$, we get that $R\Gamma(X, Lp^*\widetilde{M}(q))$ is given by the complex with terms $I^qF_i$ by Derived Categories of Schemes, Lemma \ref{perfect-lemma-acyclicity-lemma-global}. The result follows as the complex $I^qF_\bullet$ computes $M \otimes_A^\mathbf{L} I^q$ by definition.
Say $I = (a_1, \ldots, a_t)$. The question is affine local on $X$. For $1 \leq i \leq t$ let $B_i = A[\frac{I}{a_i}]$ be the affine blowup algebra. Then $X$ has an affine open covering by the spectra of the rings $B_i$, see Divisors, Lemma \ref{divisors-lemma-blowing-up-affine}. By the description of derived pullback given in Derived Categories of Schemes, Lemma \ref{perfect-lemma-quasi-coherence-pullback} we conclude it suffices to prove that for each $i$ there exists a $c \geq 0$ such that $$ \text{Tor}_j^A(B_i, M) $$ is annihilated by $I^c$ for $j \geq t$. This is Lemma \ref{lemma-annihilates-affine}.
Let $q(A, I)$ be as above. For $q \geq q(A, I)$ we have $$ R\Gamma(X, Lp^*\widetilde{M}(q)) = M \otimes_A^\mathbf{L} I^q $$ by Lemma \ref{lemma-compute-tor-Iq}. We have a bounded and convergent spectral sequence $$ H^a(X, H^b(Lp^*\widetilde{M}(q))) \Rightarrow \text{Tor}_{-a - b}^A(M, I^q) $$ by Derived Categories of Schemes, Lemma \ref{perfect-lemma-spectral-sequence}. Let $d$ be an integer as in Cohomology of Schemes, Lemma \ref{coherent-lemma-vanishing-nr-affines-quasi-separated} (actually we can take $d = t$, see Cohomology of Schemes, Lemma \ref{coherent-lemma-vanishing-nr-affines}). Then we see that $H^{-i}(X, Lp^*\widetilde{M}(q)) = \text{Tor}_i^A(M, I^q)$ has a finite filtration with at most $d$ steps whose graded are subquotients of the modules $$ H^a(X, H^{- i - a}(Lp^*\widetilde{M})(q)),\quad a = 0, 1, \ldots, d - 1 $$ If $i \geq t$ then all of these modules are annihilated by $I^c$ where $c = c(A, I)$ is as in Lemma \ref{lemma-annihilates} because the cohomology sheaves $H^{- i - a}(Lp^*\widetilde{M})$ are all annihilated by $I^c$ by the lemma. Hence we see that $\text{Tor}_i^A(M, I^q)$ is annihilated by $I^{dc}$ for $q \geq q(A, I)$ and $i \geq t$. Using the short exact sequence $0 \to I^q \to A \to A/I^q \to 0$ we find that $\text{Tor}_i(M, A/I^q)$ is annihilated by $I^{dc}$ for $q \geq q(A, I)$ and $i > t$. We conclude that $I^m$ with $m = \max(dc, q(A, I) - 1)$ annihilates $\text{Tor}_i^A(M, A/I^q)$ for all $q \geq 0$ and $i > t$ as desired.
Let $c_1$ be the constant found in Lemma \ref{lemma-annihilates-tors}. Please keep in mind that this constant $c_1$ works for $\text{Tor}_i$ for all $i > t$ simultaneously. \medskip\noindent Say $I = (a_1, \ldots, a_t)$. For an $A$-module $M$ we set $$ \ell(M) = \#\{i \mid 1 \leq i \leq t,\ a_i^{c_1}\text{ is zero on }M\} $$ This is an element of $\{0, 1, \ldots, t\}$. We will prove by descending induction on $0 \leq s \leq t$ the following statement $H_s$: there exist $N, c \geq 0$ such that for every module $M$ with $\ell(M) \geq s$ the maps $$ \text{Tor}_{t + 1 + i}^A(M, A/I^n) \to \text{Tor}_{t + 1 + i}^A(M, A/I^{n - c}) $$ are zero for $i = 0, \ldots, s$ for all $n \geq N$. \medskip\noindent Base case: $s = t$. If $\ell(M) = t$, then $M$ is annihilated by $(a_1^{c_1}, \ldots, a_t^{c_1}\}$ and hence by $I^{t(c_1 - 1) + 1}$. We conclude from Lemma \ref{lemma-maps-zero-fixed-torsion} that $H_t$ holds by taking $c = N$ to be the maximum of the integers $c(A, I, t(c_1 - 1) + 1, t + 1), \ldots, c(A, I, t(c_1 - 1) + 1, 2t + 1)$ found in the lemma. \medskip\noindent Induction step. Say $0 \leq s < t$ we have $N, c$ as in $H_{s + 1}$. Consider a module $M$ with $\ell(M) = s$. Then we can choose an $i$ such that $a_i^{c_1}$ is nonzero on $M$. It follows that $\ell(M[a_i^c]) \geq s + 1$ and $\ell(M/a_i^{c_1}M) \geq s + 1$ and the induction hypothesis applies to them. Consider the exact sequence $$ 0 \to M[a_i^{c_1}] \to M \xrightarrow{a_i^{c_1}} M \to M/a_i^{c_1}M \to 0 $$ Denote $E \subset M$ the image of the middle arrow. Consider the corresponding diagram of Tor modules $$ \xymatrix{ & & \text{Tor}_{i + 1}(M/a_i^{c_1}M, A/I^q) \ar[d] \\ \text{Tor}_i(M[a_i^{c_1}], A/I^q) \ar[r] & \text{Tor}_i(M, A/I^q) \ar[r] \ar[rd]^0 & \text{Tor}_i(E, A/I^q) \ar[d] \\ & & \text{Tor}_i(M, A/I^q) } $$ with exact rows and columns (for every $q$). The south-east arrow is zero by our choice of $c_1$. We conclude that the module $\text{Tor}_i(M, A/I^q)$ is sandwiched between a quotient module of $\text{Tor}_i(M[a_i^{c_1}], A/I^q)$ and a submodule of $\text{Tor}_{i + 1}(M/a_i^{c_1}M, A/I^q)$. Hence we conclude $H_s$ holds with $N$ replaced by $N + c$ and $c$ replaced by $2c$. Some details omitted.
If (2) holds we get (1) by a combination of Lemmas \ref{lemma-check-finiteness-pushforward-on-associated-points}, Remark \ref{remark-closure}, and Lemma \ref{lemma-sharp-finiteness-pushforward}. If (2) does not hold, then $j_*i_{x, *}\mathcal{O}_{W_x}$ is not finite for some $x \in \text{Ass}(\mathcal{F})$ by the discussion in Remark \ref{remark-no-finiteness-pushforward} (and Remark \ref{remark-closure}). Thus $j_*\mathcal{F}$ is not coherent by Lemma \ref{lemma-check-finiteness-pushforward-on-associated-points}.
Since $A \to A^\wedge$ is faithfully flat we have $H^d_{V(I)}(A) \otimes_A A^\wedge = H^d_{V(IA^\wedge)}(A^\wedge)$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-torsion-change-rings}. Thus we may assume $A$ is complete. \medskip\noindent Assume $A$ is complete. Let $\mathfrak p_1, \ldots, \mathfrak p_n \subset A$ be the minimal primes of dimension $d$. Consider the complete local ring $A_i = A/\mathfrak p_i$. We have $H^d_{V(I)}(A_i) = H^d_{V(IA_i)}(A_i)$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-and-restriction}. By Lemma \ref{lemma-cd-top-vanishing-some-module} it suffices to prove the lemma for $(A_i, IA_i)$. Thus we may assume $A$ is a complete local domain. \medskip\noindent Assume $A$ is a complete local domain. We can choose a prime ideal $\mathfrak p \supset I$ with $\dim(A/\mathfrak p) = 1$. By Lemma \ref{lemma-cd-top-vanishing} it suffices to prove the lemma for $\mathfrak p$. \medskip\noindent By Lemma \ref{lemma-cd-bound-dualizing} it suffices to show that $H^0_{V(\mathfrak p)}(\omega_A^\bullet) = 0$. Recall that $$ H^0_{V(\mathfrak p)}(\omega_A^\bullet) = \colim \text{Ext}^0_A(A/\mathfrak p^n, \omega_A^\bullet) $$ By Lemma \ref{lemma-inverse-system-symbolic-powers} we see that the colimit is the same as $$ \colim \text{Ext}^0_A(A/\mathfrak p^{(n)}, \omega_A^\bullet) $$ Since $\text{depth}(A/\mathfrak p^{(n)}) = 1$ we see that these ext groups are zero by Lemma \ref{lemma-sitting-in-degrees} as desired.
If the characteristic of $k$ is zero, then the equivalence of (1) and (2) follows from Lemmas \ref{lemma-fs-implies-regular} and \ref{lemma-regular-implies-fs}. \medskip\noindent If the characteristic of $k$ is $p > 0$, then it follows from Proposition \ref{proposition-characterization-geometrically-regular} that (1), (3), (4), and (5) are equivalent. Assume (2) holds. By Lemma \ref{lemma-base-change-fs} we see that $k' \to A' = A \otimes_k k'$ is formally smooth for the $\mathfrak m' = \mathfrak mA$-adic topology. Hence if $k \subset k'$ is finite purely inseparable, then $A'$ is a regular local ring by Lemma \ref{lemma-fs-implies-regular}. Thus we see that (1) holds. \medskip\noindent Finally, we will prove that (5) implies (2). Choose a solid diagram $$ \xymatrix{ A \ar[r]_{\bar\psi} \ar@{-->}[rd] & B/J \\ k \ar[u]^i \ar[r]^\varphi & B \ar[u]_\pi } $$ as in Definition \ref{definition-formally-smooth}. As $J^2 = 0$ we see that $J$ has a canonical $B/J$ module structure and via $\bar\psi$ an $A$-module structure. As $\bar\psi$ is continuous for the $\mathfrak m$-adic topology we see that $\mathfrak m^nJ = 0$ for some $n$. Hence we can filter $J$ by $B/J$-submodules $0 \subset J_1 \subset J_2 \subset \ldots \subset J_n = J$ such that each quotient $J_{t + 1}/J_t$ is annihilated by $\mathfrak m$. Considering the sequence of ring maps $B \to B/J_1 \to B/J_2 \to \ldots \to B/J$ we see that it suffices to prove the existence of the dotted arrow when $J$ is annihilated by $\mathfrak m$, i.e., when $J$ is a $K$-vector space. \medskip\noindent Assume given a diagram as above such that $J$ is annihilated by $\mathfrak m$. By Lemma \ref{lemma-regular-implies-fs} we see that $\mathbf{F}_p \to A$ is formally smooth in the $\mathfrak m$-adic topology. Hence we can find a ring map $\psi : A \to B$ such that $\pi \circ \psi = \bar \psi$. Then $\psi \circ i, \varphi : k \to B$ are two maps whose compositions with $\pi$ are equal. Hence $D = \psi \circ i - \varphi : k \to J$ is a derivation. By Algebra, Lemma \ref{algebra-lemma-universal-omega} we can write $D = \xi \circ \text{d}$ for some $k$-linear map $\xi : \Omega_{k/\mathbf{F}_p} \to J$. Using the $K$-vector space structure on $J$ we extend $\xi$ to a $K$-linear map $\xi' : \Omega_{k/\mathbf{F}_p} \otimes_k K \to J$. Using (5) we can find a $K$-linear map $\xi'' : \Omega_{A/\mathbf{F}_p} \otimes_A K$ whose restriction to $\Omega_{k/\mathbf{F}_p} \otimes_k K$ is $\xi'$. Write $$ D' : A \xrightarrow{\text{d}} \Omega_{A/\mathbf{F}_p} \to \Omega_{A/\mathbf{F}_p} \otimes_A K \xrightarrow{\xi''} J. $$ Finally, set $\psi' = \psi - D' : A \to B$. The reader verifies that $\psi'$ is a ring map such that $\pi \circ \psi' = \bar \psi$ and such that $\psi' \circ i = \varphi$ as desired.
The category appearing on the right side of the arrow is the category of triples $(M', M_1, \alpha_1)$ where $M'$ is an $S$-module, $M_1$ is a $R_f$-module, and $\alpha_1 : M'_f \to M_1 \otimes_R S$ is a $S_f$-isomorphism, see Categories, Example \ref{categories-example-2-fibre-product-categories}. Hence this theorem is a special case of Proposition \ref{proposition-equivalence}.