hcju/stacksqa huggingface dataset:
Pick an integer $n$ and consider the distinguished triangle $$ \tau_{\leq n}K \to K \to \tau_{\geq n + 1}K \to \tau_{\leq n}K[1] $$ see Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}. By assumption (3) and Lemma \ref{lemma-internal-hom-evaluate-isomorphism} the map is an isomorphism for $\tau_{\leq n}K$. Hence it suffices to show that both $$ R\Hom_R(L, M) \otimes_R^\mathbf{L} \tau_{\geq n + 1}K \quad\text{and}\quad R\Hom_R(R\Hom_R(\tau_{\geq n + 1}K, L), M) $$ have vanishing cohomology in degrees $\leq n - c$ for some $c$. This follows immediately from assumptions (2) and (1).
Proof in case $M$ is perfect. Note that both sides of the arrow transform distinguished triangles in $M$ into distinguished triangles and commute with direct sums. Hence it suffices to check it holds when $M = R[n]$, see Derived Categories, Remark \ref{derived-remark-check-on-generator} and Lemma \ref{lemma-perfect-ring-classical-generator}. In this case the result is obvious. \medskip\noindent Proof in case $K$ is perfect. Same argument as in the previous case. \medskip\noindent Proof in case (3). We may represent $K$ and $L$ by bounded below complexes of $R$-modules $K^\bullet$ and $L^\bullet$. We may assume that $K^\bullet$ is a K-flat complex consisting of flat $R$-modules, see Lemma \ref{lemma-bounded-below-tor-amplitude}. We may represent $M$ by a bounded above complex $M^\bullet$ of finite free $R$-modules, see Definition \ref{definition-pseudo-coherent}. Then the object on the LHS is represented by $$ \text{Tot}(K^\bullet \otimes_R \Hom^\bullet(M^\bullet, L^\bullet)) $$ and the object on the RHS by $$ \Hom^\bullet(M^\bullet, \text{Tot}(K^\bullet \otimes_R L^\bullet)) $$ This uses Lemma \ref{lemma-RHom-out-of-projective}. Both complexes have in degree $n$ the module $$ \bigoplus\nolimits_{p + q + r = n} K^p \otimes \Hom_R(M^{-r}, L^q) = \bigoplus\nolimits_{p + q + r = n} \Hom_R(M^{-r}, K^p \otimes_R L^q) $$ because $M^{-r}$ is finite free (as well these are finite direct sums). The map defined in Lemma \ref{lemma-internal-hom-diagonal-better} comes from the map of complexes defined in Lemma \ref{lemma-diagonal-better} which uses the canonical isomorphisms between these modules.
The last statement is Lemma \ref{lemma-RHom-out-of-projective}. Since $P^\bullet$ represents a perfect object, there exists a finite complex of finite projective $R$-modules $F^\bullet$ such that $P^\bullet$ and $F^\bullet$ are isomorphic in $D(R)$, see Definition \ref{definition-perfect}. Then $P^\bullet$ and $F^\bullet$ are homotopy equivalent, see Derived Categories, Lemma \ref{derived-lemma-morphisms-from-projective-complex}. Then $\Hom^\bullet(P^\bullet, K^\bullet)$ and $\Hom^\bullet(F^\bullet, K^\bullet)$ are homotopy equivalent. Hence the first is K-flat if and only if the second is (follows from Definition \ref{definition-K-flat} and Lemma \ref{lemma-derived-tor-homotopy}). It is clear that $$ \Hom^\bullet(F^\bullet, K^\bullet) = \text{Tot}(E^\bullet \otimes_R K^\bullet) $$ where $E^\bullet$ is the dual complex to $F^\bullet$ with terms $E^n = \Hom_R(F^{-n}, R)$, see Lemma \ref{lemma-dual-perfect-complex} and its proof. Since $E^\bullet$ is a bounded complex of projectives we find that it is K-flat by Lemma \ref{lemma-derived-tor-quasi-isomorphism}. Then we conclude by Lemma \ref{lemma-tensor-product-K-flat}.
We may check the map is an isomorphism after applying the restriction functor $D(R') \to D(R)$. After applying this functor our map becomes the map $$ R\Hom_R(K, L) \otimes_R^\mathbf{L} R' \longrightarrow R\Hom_R(K, L \otimes_R^\mathbf{L} R') $$ of Lemma \ref{lemma-internal-hom-diagonal-better}. See discussion above the lemma to match the left and right hand sides; in particular, this uses Lemma \ref{lemma-upgrade-adjoint-tensor-RHom}. Thus we conclude by Lemma \ref{lemma-internal-hom-evaluate-tensor-isomorphism}.
Say $I = (f_1, \ldots, f_r)$. Let $K_n \in D(A)$ be the object represented by the Koszul complex on $f_1^n, \ldots, f_r^n$. Recall that we have maps $K_n \to A/I^n$ which induce a pro-isomorphism of inverse systems, see Lemma \ref{lemma-sequence-Koszul-complexes}. Hence it suffices to show that $$ M^\bullet \otimes_A^\mathbf{L} K_n \longrightarrow M^\bullet/I^nM^\bullet $$ defines an isomorphism of pro-objects of $D(A)$. Since $K_n$ is represented by a complex of finite free $A$-modules sitting in degrees $-r, \ldots, 0$ there exist $a, b \in \mathbf{Z}$ such that the source and target of the displayed arrow have vanishing cohomology in degrees outside $[a, b]$ for all $n$. Thus we may apply Derived Categories, Lemma \ref{derived-lemma-pro-isomorphism-bis} and we find that it suffices to show that the maps $$ H^i(M^\bullet \otimes_A^\mathbf{L} A/I^n) \to H^i(M^\bullet/I^nM^\bullet) $$ define isomorphisms of pro-systems of $A$-modules for any $i \in \mathbf{Z}$. To see this choose a quasi-isomorphism $P^\bullet \to M^\bullet$ where $P^\bullet$ is a bounded above complex of finite free $A$-modules. The arrows above are given by the maps $$ H^i(P^\bullet/I^nP^\bullet) \to H^i(M^\bullet/I^nM^\bullet) $$ These define an isomorphism of pro-systems by Lemma \ref{lemma-consequence-Artin-Rees}. Namely, the lemma shows both are isomorphic to the pro-system $H^i/I^nH^i$ with $H^i = H^i(M^\bullet) = H^i(P^\bullet)$.
Note that $\Hom_A(M_n, N_n) = \Hom_A(M, N_n)$. Choose a presentation $$ A^{\oplus t} \to A^{\oplus s} \to M \to 0 $$ Applying the right exact functor $\Hom_A(-, N)$ we obtain a complex $$ 0 \xrightarrow{\alpha} N^{\oplus s} \xrightarrow{\beta} N^{\oplus t} $$ whose cohomology in the middle is $\Hom_A(M, N)$ and such that for $n \geq 0$ the cohomology of $$ 0 \xrightarrow{\alpha_n} N_n^{\oplus s} \xrightarrow{\beta_n} N_n^{\oplus t} $$ is $\Hom_A(M_n, N_n)$. Let $c \geq 0$ be as in Lemma \ref{lemma-consequence-Artin-Rees} for this $A$, $I$, $\alpha$, and $\beta$. By part (3) of the lemma we deduce the Mittag-Leffler property for $(\Hom_A(M_n, N_n))$. The kernel and cokernel of the maps $\Hom_A(M, N)/I^n\Hom_A(M, N) \to \Hom_A(M_n, N_n)$ are killed by $I^c$ by [art part (6) of the lemma. We find that $\lim \Hom_A(M_n, N_n) = \Hom_A(M, N)^\wedge$ by part (2) of the lemma. The equality $$ \Hom_{A^\wedge}(M^\wedge, N^\wedge) = \lim \Hom_A(M_n, N_n) $$ follows formally from the fact that $M^\wedge = \lim M_n$ and $M_n = M^\wedge/I^nM^\wedge$ and the corresponding facts for $N$, see Algebra, Lemma \ref{algebra-lemma-completion-complete}. \medskip\noindent The result for isomorphisms follows from the case of homomorphisms applied to both $(\Hom(M_n, N_n))$ and $(\Hom(N_n, M_n))$ and the following fact: for $n > m > 0$, if we have maps $\alpha : M_n \to N_n$ and $\beta : N_n \to M_n$ which induce an isomorphisms $M_m \to N_m$ and $N_m \to M_m$, then $\alpha$ and $\beta$ are isomorphisms. Namely, then $\alpha \circ \beta$ is surjective by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) hence $\alpha \circ \beta$ is an isomorphism by Algebra, Lemma \ref{algebra-lemma-fun}.
By Lemma \ref{lemma-hom-systems-ML} the system $(\text{Isom}_A(M_n, N_n))$ is Mittag-Leffler. By assumption each of the sets $\text{Isom}_A(M_n, N_n)$ is nonempty. Hence $\lim \text{Isom}_A(M_n, N_n)$ is nonempty. Since $\lim \text{Isom}_A(M_n, N_n) = \text{Isom}_{A^\wedge}(M^\wedge, N^\wedge)$ we obtain an isomorphism.
Choose a short exact sequence $$ 0 \to K \to A^{\oplus r} \to M \to 0 $$ and set $K_n = K/I^nK$. For $n \geq 1$ define $K(n) = \Ker(A_n^{\oplus r} \to M_n)$ so that we have exact sequences $$ 0 \to K(n) \to A_n^{\oplus r} \to M_n \to 0 $$ and surjections $K_n \to K(n)$. In fact, by Lemma \ref{lemma-consequence-Artin-Rees} there is a $c \geq 0$ and maps $K(n) \to K_n/I^{n - c}K_n$ which are ``almost inverse''. Since $I^{n - c}K_n \subset K_n[I^c]$ these maps which witness the fact that the systems $\{K(n)\}_{n \geq 1}$ and $\{K_n\}_{n \geq 1}$ are isomorphic in $\mathcal{C}$. \medskip\noindent We claim the systems $$ \{\Ext^i_{A_n}(K(n), N_n)\}_{n \geq 1} \quad\text{and}\quad \{\Ext^i_{A_n}(K_n, N_n)\}_{n \geq 1} $$ are isomorphic in the category $\mathcal{C}$. Namely, the surjective maps $K_n \to K(n)$ have kernels annihilated by $I^c$ and therefore determine maps $$ \Ext^i_{A_n}(K(n), N_n) \to \Ext^i_{A_n}(K_n, N_n) $$ whose kernel and cokernel are annihilated by $I^c$. Hence the claim by Lemma \ref{lemma-iso}. \medskip\noindent For $i \geq 2$ we have isomorphisms $$ \Ext^{i - 1}_A(K, N) = \Ext^i_A(M, N) \quad\text{and}\quad \Ext^{i - 1}_{A_n}(K(n), N_n) = \Ext^i_{A_n}(M_n, N_n) $$ In this way we see that it suffices to prove the lemma for $i = 0, 1$. \medskip\noindent For $i = 0, 1$ we consider the commutative diagram $$ \xymatrix{ 0 \ar[r] & \Hom(M, N) \ar[r] \ar[dd] & N^{\oplus r} \ar[r]_-\varphi \ar[dd] & \Hom(K, N) \ar[r] \ar[d] & \Ext^1(M, N) \ar[r] & 0 \\ & & & \Hom(K_n, N_n) \\ 0 \ar[r] & \Hom(M_n, N_n) \ar[r] & N_n^{\oplus r} \ar[r] & \Hom(K(n), N_n) \ar[r] \ar[u] & \Ext^1(M_n, N_n) \ar[r] & 0 } $$ By Lemma \ref{lemma-hom-systems-ML} we see that the kernel and cokernel of $\Hom(M, N)/I^n \Hom(M, N) \to \Hom(M_n, N_n)$ and $\Hom(K, N)/I^n \Hom(K, N) \to \Hom(K_n, N_n)$ and are $I^c$-torsion for some $c \geq 0$ independent of $n$. Above we have seen the cokernel of the injective maps $\Hom(K(n), N_n) \to \Hom(K_n, N_n)$ are annihilated by $I^c$ after possibly increasing $c$. For such a $c$ we obtain maps $\delta_n : I^c\Hom(K, N)/I^n\Hom(K, N) \to \Hom(K(n), N_n)$ fitting into the diagram (precise formulation omitted). The kernel and cokernel of $\delta_n$ are annihilated by $I^c$ after possibly increasing $c$ since we know that the same thing is true for $\Hom(K, N)/I^n \Hom(K, N) \to \Hom(K_n, N_n)$ and $\Hom(K(n), N_n) \to \Hom(K_n, N_n)$. Then we can use commutativity of the solid diagram $$ \xymatrix{ \varphi^{-1}(I^c\Hom(K, N)) \ar[r]_-\varphi \ar[d] & I^c\Hom(K, N)/I^n\Hom(K, N) \ar[r] \ar[d]^{\delta_n} & I^c\Ext^1(M, N)/I^n\Ext^1(M, N) \ar[r] \ar@{..>}[d] & 0 \\ N_n^{\oplus r} \ar[r] & \Hom(K(n), N_n) \ar[r] & \Ext^1(M_n, N_n) \ar[r] & 0 } $$ to define the dotted arrow. A straightforward diagram chase (omitted) shows that the kernel and cokernel of the dotted arrow are annihilated buy $I^c$ after possibly increasing $c$ one final time.
Choose a resolution $$ \ldots \to P_2 \to P_1 \to P_0 \to M \to 0 $$ by finite free $B$-modues $P_i$. Set $P_{i, n} = P_i/I^nP_i$. Since $M$ and $B$ are flat over $A$, the sequence $$ \ldots \to P_{2, n} \to P_{1, n} \to P_{0, n} \to M_n \to 0 $$ is exact. We see that on the one hand the complex $$ \Hom_B(P_0, N) \to \Hom_B(P_1, N) \to \Hom_B(P_2, N) \to \ldots $$ computes the modules $\Ext^i_B(M, N)$ and on the other hand the complex $$ \Hom_{B_n}(P_{0, n}, N_n) \to \Hom_{B_n}(P_{1, n}, N_n) \to \Hom_{B_n}(P_{2, n}, N_n) \to \ldots $$ computes the modules $\Ext^i_{B_n}(M_n, N_n)$. Since $$ \Hom_{B_n}(P_{i, n}, N_n) = \Hom_B(P_i, N)/I^n \Hom_B(P_i, N) $$ we obtain the result from Lemma \ref{lemma-consequence-Artin-Rees} part (2).
We have $H_n = \Ker(\beta) \cap I^nL/\alpha(I^nK)$. Since $\Ker(\beta) \cap I^nL \subset \Ker(\beta) \cap I^{n - 1}L$ and $\alpha(I^nK) \subset \alpha(I^{n - 1}K)$ we get the maps $H_n \to H_{n - 1}$. Similarly for the map $H_1 \to H$. \medskip\noindent By Artin-Rees we may choose $c_1, c_2 \geq 0$ such that $\Im(\alpha) \cap I^nL \subset \alpha(I^{n - c_1}K)$ for $n \geq c_1$ and $\Ker(\beta) \cap I^nL \subset I^{n - c_2}\Ker(\beta)$ for $n \geq c_2$, see Algebra, Lemmas \ref{algebra-lemma-map-AR} and \ref{algebra-lemma-Artin-Rees}. Set $c = c_1 + c_2$. \medskip\noindent It follows immediately from our choice of $c \geq c_2$ that for $n \geq c$ the image of $H_n \to H$ is contained in $I^{n - c}H$. \medskip\noindent Let $n \geq c$. We define $\psi_n : I^nH \to H_{n - c}$ as follows. Say $x \in I^nH$. Choose $y \in I^n\Ker(\beta)$ representing $x$. We set $\psi_n(x)$ equal to the class of $y$ in $H_{n - c}$. To see this is well defined, suppose we have a second choice $y'$ as above for $x$. Then $y' - y \in \Im(\alpha)$. By our choice of $c \geq c_1$ we conclude that $y' - y \in \alpha(I^{n - c}K)$ which implies that $y$ and $y'$ represent the same element of $H_{n - c}$. Thus $\psi_n$ is well defined. \medskip\noindent The statements on the compositions $I^n H \to H_{n - c} \to  I^{n - 2c}H$ and $H_n \to I^{n - c}H \to H_{n - 2c}$ follow immediately from our definitions.
For $p = 0$, if $\varphi : M \to N$ is an $A$-linear map, then $\varphi(\sum f_i m_i) = \sum f_i \varphi(m_i)$ for $f_i \in A$ and $m_i \in M$. Hence $\varphi$ induces a map $I^nM \to I^nN$ for all $n$ and the result is true with $c = 0$. \medskip\noindent Choose a short exact sequence $0 \to K \to A^{\oplus t} \to M \to 0$. For each $n$ we pick a short exact sequence $0 \to L_n \to A^{\oplus s_n} \to I^nM \to 0$. It is clear that we can construct a map of short exact sequences $$ \xymatrix{ 0 \ar[r] & L_n \ar[r] \ar[d] & A^{\oplus s_n} \ar[r] \ar[d] & I^nM \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K \ar[r] & A^{\oplus t} \ar[r] & M \ar[r] & 0 } $$ such that $A^{\oplus s_n} \to A^{\oplus t}$ has image in $(I^n)^{\oplus t}$. By Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees}) there exists a $c \geq 0$ such that $L_n \to K$ factors through $I^{n - c}K$ if $n \geq c$. \medskip\noindent For $p = 1$ our choices above induce a solid commutative diagram $$ \xymatrix{ \Hom_A(A^{\oplus s_n}, N) \ar[r] & \Hom_A(L_n, N) \ar[r] & \Ext_A^1(I^nM, N) \ar[r] & 0 \\ \Hom_A((I^n)^{\oplus t}, I^{n - c}N) \ar[r] \ar[u] & \Hom_A(K \cap (I^n)^{\oplus t}, I^{n - c}N) \ar[r] \ar[u] & \Ext_A^1(I^nM, I^{n - c}N) \ar[u] \\ \Hom_A(A^{\oplus t}, N) \ar[r] \ar[u] & \Hom_A(K, N) \ar[r] \ar[u] & \Ext_A^1(M, N) \ar@{..>}[u] \ar[r] & 0 } $$ whose horizontal arrows are exact. The lower middle vertical arrow arises because $K \cap (I^n)^{\oplus t} \subset I^{n - c}K$ and hence any $A$-linear map $K \to N$ induces an $A$-linear map $(I^n)^{\oplus t} \to I^{n - c}N$ by the argument of the first paragraph. Thus we obtain the dotted arrow as desired. \medskip\noindent For $p > 1$ we obtain a commutative diagram $$ \xymatrix{ \Ext^{p - 1}_A(I^{n - c}K, N) \ar[r] & \Ext^{p - 1}_A(L_n, N) \ar[r] & \Ext_A^p(I^nM, N) \\ \Ext^{p - 1}_A(K, N) \ar[rr] \ar[u] & & \Ext_A^p(M, N) \ar[u] } $$ whose bottom horizontal arrow is an isomorphism. By induction on $p$ the left vertical map factors through $\Ext^{p - 1}_A(I^{n - c}K, I^{n - c - c'}N)$ for some $c' \geq 0$ and all $n \geq c + c'$. Using the composition $\Ext^{p - 1}_A(I^{n - c}K, I^{n - c - c'}N) \to \Ext^{p - 1}_A(L_n, I^{n - c - c'}N) \to \Ext^p_A(I^nM, I^{n - c - c'}N)$ we obtain the desired factorization (for $n \geq c + c'$ and with $c$ replaced by $c + c'$).
Choose a bounded above complex $P^\bullet$ of finite free $A$-modules representing $K$. Then $\Ext_A^p(K, M)$ is the cohomology of $$ \Hom_A(F^{-p + 1}, M) \xrightarrow{a} \Hom_A(F^{-p}, M) \xrightarrow{b} \Hom_A(F^{-p - 1}, M) $$ and $\Ext_A^p(K, I^nM)$ is computed by replacing these finite $A$-modules by $I^n$ times themselves. Thus the result by Lemma \ref{lemma-consequence-Artin-Rees-bis} (and much more is true).
It is elementary to show that the inverse system $I^n$ is pro-isomorphic to the inverse system $(f_1^n, \ldots, f_r^n)$ in the category of $A$-modules. Consider the inverse system of distinguished triangles $$ I_n^\bullet \to (f_1^n, \ldots, f_r^n) \to C_n^\bullet \to I_n^\bullet[1] $$ where $C_n^\bullet$ is the cone of the first arrow. By Derived Categories, Lemma \ref{derived-lemma-pro-isomorphism} it suffices to show that the inverse system $C_n^\bullet$ is pro-zero. The complex $I_n^\bullet$ has nonzero terms only in degrees $i$ with $-r + 1 \leq i \leq 0$ hence $C_n^\bullet$ is bounded similarly. Thus by Derived Categories, Lemma \ref{derived-lemma-essentially-constant-cohomology} it suffices to show that $H^p(C_n^\bullet)$ is pro-zero. By the discussion above we have $H^p(C_n^\bullet) = H^p(K_n^\bullet)$ for $p \leq -1$ and $H^p(C_n^\bullet) = 0$ for $p \geq 0$. The fact that the inverse systems $H^p(K_n^\bullet)$ are pro-zero was shown in the proof of Lemma \ref{lemma-sequence-Koszul-complexes} (and this is where the assumption that $A$ is Noetherian is used).
Choose generators $f_1, \ldots, f_r \in I$ of $I$. The inverse system $I^n$ is pro-isomorphic to the inverse system $(f_1^n, \ldots, f_r^n)$ in the category of $A$-modules. With notation as in Lemma \ref{lemma-sequence-powers-pro-bounded} we find that it suffices to prove the inverse system of maps $$ I_n^\bullet \otimes_A^\mathbf{L} M^\bullet \longrightarrow (f_1^n, \ldots, f_r^n)M^\bullet $$ defines an isomorphism of pro-objects of $D(A)$. Say we have $a \leq b$ such that $M^i = 0$ if $i \not \in [a, b]$. Then source and target of the arrows above have cohomology only in degrees $[-r + a, b]$. Thus it suffices to show that for any $p \in \mathbf{Z}$ the inverse system of maps $$ H^p(I_n^\bullet \otimes_A^\mathbf{L} M^\bullet) \longrightarrow H^p((f_1^n, \ldots, f_r^n)M^\bullet) $$ defines an isomorphism of pro-objects of $A$-modules, see Derived Categories, Lemma \ref{derived-lemma-pro-isomorphism-bis}. Using the pro-isomorphism between $I_n^\bullet \otimes_A^\mathbf{L} M^\bullet$ and $I^n \otimes_A^\mathbf{L} M^\bullet$ and the pro-isomorphism between $(f_1^n, \ldots, f_r^n)M^\bullet$ and $I^nM^\bullet$ this is equivalent to showing that the inverse system of maps $$ H^p(I^n \otimes_A^\mathbf{L} M^\bullet) \longrightarrow H^p(I^nM^\bullet) $$ defines an isomorphism of pro-objects of $A$-modules Choose a bounded above complex of finite free $A$-modules $P^\bullet$ and a quasi-isomorphism $P^\bullet \to M^\bullet$. Then it suffices to show that the inverse system of maps $$ H^p(I^nP^\bullet) \longrightarrow H^p(I^nM^\bullet) $$ is a pro-isomorphism. This follows from Lemma \ref{lemma-consequence-Artin-Rees-bis} as $H^p(P^\bullet) = H^p(M^\bullet)$.
This follows from Lemma \ref{lemma-tensoring-Deligne-system}. It can also been seen directly as follows. Consider the distinguished triangle $$ I \otimes_A^\mathbf{L} M \to M \to A/I \otimes_A^\mathbf{L} M \to I \otimes_A^\mathbf{L} M[1] $$ By the axioms of a triangulated category it suffices to prove that $I^nM \to A/I \otimes_A^\mathbf{L} M$ is zero in $D(A)$ for some $n$. Choose generators $f_1, \ldots, f_r$ of $I$ and let $K = K_\bullet(A, f_1, \ldots, f_r)$ be the Koszul complex and consider the factorization $A \to K \to A/I$ of the quotient map. Then we see that it suffices to show that $I^nM \to K \otimes_A M$ is zero in $D(A)$ for some $n > 0$. Suppose that we have found an $n > 0$ such that $I^nM \to K \otimes_A M$ factors through $\tau_{\geq t}(K \otimes_A M)$ in $D(A)$. Then the obstruction to factoring through $\tau_{\geq t + 1}(K \otimes_A M)$ is an element in $\Ext^t(I^nM, H_t(K \otimes_A M))$. The finite $A$-module $H_t(K \otimes_A M)$ is annihilated by $I$. Then by Lemma \ref{lemma-ext-annihilated} we can after increasing $n$ assume this obstruction element is zero. Repeating this a finite number of times we find $n$ such that $I^nM \to K \otimes_A M$ factors through $0 = \tau_{\geq r + 1}(K \otimes_A M)$ in $D(A)$ and we win.
Let $M$ be a finite $A$-module. Since $K$ is pseudo-coherent we see that $\Ext^i_A(K, M)$ is a finite $A$-module. Thus for $i \geq a$ it is annihilated by $I^t$ for some $t \geq 0$. By Lemma \ref{lemma-ext-induced-toplogy} we see that the image of $\Ext^i_A(K, I^nM) \to \Ext^i_A(K, M)$ is zero for some $n > 0$. The short exact sequence $0 \to I^nM \to M \to M/I^n M \to 0$ gives a long exact sequence $$ \Ext^i_A(K, I^nM) \to \Ext^i_A(K, M) \to \Ext^i_A(K, M/I^nM) \to \Ext^{i + 1}_A(K, I^nM) $$ The systems $\Ext^i_A(K, I^nM)$ and $\Ext^{i + 1}_A(K, I^nM)$ are essentially constant with value $0$ by what we just said (applied to the finite $A$-modules $I^mM$). A diagram chase shows $\Ext^i_A(K, M/I^nM)$ is essentially constant with value $\Ext^i_A(K, M)$.
By Lemma \ref{lemma-factor-through-derived-tensor-product} we can factor $I^nM \to M$ as $I^nM \to M \otimes_A^\mathbf{L} I \to M$. We claim the composition $$ I^nM \otimes_A^\mathbf{L} N \to (M \otimes_A^\mathbf{L} I) \otimes_A^\mathbf{L} N \to M \otimes_A^\mathbf{L} N $$ is zero. Namely, the diagram $$ \xymatrix{ (M \otimes_A^\mathbf{L} I) \otimes_A^\mathbf{L} N \ar[rr] \ar[rd] & & M \otimes_A^\mathbf{L} (I \otimes_A^\mathbf{L} N) \ar[ld] \\ & M \otimes_A^\mathbf{L} N } $$ commutes (details omitted) and the map $I \otimes_A^\mathbf{L} N \to N$ is zero as $N$ is annihilated by $I$.
We will use that, for a finite module, having finite projective dimension over $R$, resp.\ $R/I$ is the same as being a perfect module, see discussion following Definition \ref{definition-perfect}. We see that $E$ has finite projective dimension over $R$ by Lemma \ref{lemma-cohomology-perfect}. Thus we can apply Auslander-Buchsbaum (Algebra, Proposition \ref{algebra-proposition-Auslander-Buchsbaum}) to see that $$ \text{pd}_R(E) + \text{depth}(E) = \text{depth}(R),\quad \text{pd}_{R/I}(E) + \text{depth}(E) = \text{depth}(R/I), $$ and $$ \text{pd}_R(R/I) + \text{depth}(R/I) = \text{depth}(R) $$ Note that in the first equation we take the depth of $E$ as an $R$-module and in the second as an $R/I$-module. However these depths are the same (this is trivial but also follows from Algebra, Lemma \ref{algebra-lemma-depth-goes-down-finite}). This concludes the proof.
Let $N$ be a $B$-module. If $d = 0$, then $N$ is flat as an $A$-module, hence flat as a $B$-module by Lemma \ref{lemma-key}. Assume $d > 0$. Choose a resolution $F_\bullet \to N$ by free $B$-modules. Our assumption implies that $K = \Im(F_d \to F_{d - 1})$ is $A$-flat, see Lemma \ref{lemma-last-one-flat}. Hence it is $B$-flat by Lemma \ref{lemma-key}. Thus $0 \to K \to F_{d - 1} \to \ldots \to F_0 \to N \to 0$ is a flat resolution of length $d$ and we see that $N$ has tor dimension at most $d$.
The equivalence of (1) and (2) is immediate. Assume $A$ is absolutely flat. This implies every ideal of $A$ is pure, see Algebra, Definition \ref{algebra-definition-pure-ideal}. Hence every finitely generated ideal is generated by an idempotent by Algebra, Lemma \ref{algebra-lemma-finitely-generated-pure-ideal}. If $f \in A$, then $(f) = (e)$ for some idempotent $e \in A$ and $D(f) = D(e)$ is open and closed (Algebra, Lemma \ref{algebra-lemma-idempotent-spec}). This already implies every ideal of $A$ is maximal for example by Algebra, Lemma \ref{algebra-lemma-ring-with-only-minimal-primes}. Moreover, if $f$ is nilpotent, then $e = 0$ hence $f = 0$. Thus $A$ is reduced. \medskip\noindent Assume $A$ is reduced and every prime of $A$ is maximal. Let $M$ be an $A$-module. Our goal is to show that $M$ is flat. We may write $M$ as a filtered colimit of finite $A$-modules, hence we may assume $M$ is finite (Algebra, Lemma \ref{algebra-lemma-colimit-flat}). There is a finite filtration of $M$ by modules of the form $A/I$ (Algebra, Lemma \ref{algebra-lemma-trivial-filter-finite-module}), hence we may assume that $M = A/I$ (Algebra, Lemma \ref{algebra-lemma-flat-ses}). Thus it suffices to show every ideal of $A$ is pure. Since every local ring of $A$ is a field (by Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring} and the fact that every prime of $A$ is minimal), we see that every ideal $I \subset A$ is radical. Note that every closed subset of $\Spec(A)$ is closed under generalization. Thus every (radical) ideal of $A$ is pure by Algebra, Lemma \ref{algebra-lemma-pure-open-closed-specializations}.
Let $K_i$ be a family of fields. If $f = (f_i) \in \prod K_i$, then the ideal generated by $f$ is the same as the ideal generated by the idempotent $e = (e_i)$ with $e_i = 0, 1$ according to whether $f_i$ is $0$ or not. Thus $D(f) = D(e)$ is open and closed and we conclude by Lemma \ref{lemma-absolutely-flat} and Algebra, Lemma \ref{algebra-lemma-ring-with-only-minimal-primes}.
Assume $B \otimes_A B \to B$ is flat. The ring map $B' \otimes_{A'} B' \to B'$ is the base change of $B \otimes_A B \to B$ by $A \to A'$. Hence it is flat by Algebra, Lemma \ref{algebra-lemma-flat-base-change}. This proves (1). Part (2) follows from (1) and the fact (just used) that the base change of a flat ring map is flat.