hcju/stacksqa huggingface dataset:
Choose a surjection $R[x_1, \ldots, x_n] \to A$. Choose a surjection $A[y_1, \ldots, y_m] \to B$. Then we get a surjection $$ R[x_1, \ldots, x_n, y_1, \ldots, y_m] \to A[y_1, \ldots, y_m] $$ which is a flat base change of $R[x_1, \ldots, x_n] \to A$. By assumption $A$ is a pseudo-coherent module over $R[x_1, \ldots, x_n]$ hence by Lemma \ref{lemma-flat-base-change-pseudo-coherent} we see that $A[y_1, \ldots, y_m]$ is pseudo-coherent over $R[x_1, \ldots, x_n, y_1, \ldots, y_m]$. Thus the lemma follows from Lemma \ref{lemma-finite-push-pseudo-coherent} and the definitions.
The implication (2) $\Rightarrow$ (1) is in Lemma \ref{lemma-localize-relative-pseudo-coherent}. Assume (1). Write $1 = \sum f_ig_i$ in $A$. Choose a surjection $R[x_1, \ldots, x_n, y_1, \ldots, y_r, z_1, \ldots, z_r] \to A$. such that $y_i$ maps to $f_i$ and $z_i$ maps to $g_i$. Then we see that there exists a surjection $$ P = R[x_1, \ldots, x_n, y_1, \ldots, y_r, z_1, \ldots, z_r]/(\sum y_iz_i - 1) \longrightarrow A. $$ Note that $P$ is pseudo-coherent as an $R[x_1, \ldots, x_n, y_1, \ldots, y_r, z_1, \ldots, z_r]$-module and that $P[1/y_i]$ is pseudo-coherent as an $R[x_1, \ldots, x_n, y_1, \ldots, y_r, z_1, \ldots, z_r, 1/y_i]$-module. Hence by Lemma \ref{lemma-finite-push-pseudo-coherent} we see that $K^\bullet \otimes_A A_{f_i}$ is an $m$-pseudo-coherent complex of $P[1/y_i]$-modules for each $i$. Thus by Lemma \ref{lemma-glue-pseudo-coherent} we see that $K^\bullet$ is pseudo-coherent as a complex of $P$-modules, and Lemma \ref{lemma-finite-push-pseudo-coherent} shows that  $K^\bullet$ is pseudo-coherent as a complex of $R[x_1, \ldots, x_n, y_1, \ldots, y_r, z_1, \ldots, z_r]$-modules.
If $A \to B$ is perfect, then $B = A[x_1, \ldots, x_n]/I$ and $B$ is pseudo-coherent as an $A[x_1, \ldots, x_n]$-module and has finite tor dimension as an $A$-module. Hence Lemma \ref{lemma-perfect-over-polynomial-ring} implies that $B$ is perfect as a $A[x_1, \ldots, x_n]$-module, i.e., it has a finite resolution by finite projective $A[x_1, \ldots, x_n]$-modules (Lemma \ref{lemma-perfect-module}). Conversely, if $B = A[x_1, \ldots, x_n]/I$ and $B$ as an $A[x_1, \ldots, x_n]$-module has a finite resolution by finite projective $A[x_1, \ldots, x_n]$-modules then $B$ is pseudo-coherent as an $A[x_1, \ldots, x_n]$-module, hence $A \to B$ is pseudo-coherent. Moreover, the given resolution over $A[x_1, \ldots, x_n]$ is a finite resolution by flat $A$-modules and hence $B$ has finite tor dimension as an $A$-module.
Let $A \to B$ be a ring map which is flat and of finite presentation. It is clear that $B$ has finite tor dimension. By Algebra, Lemma \ref{algebra-lemma-flat-finite-presentation-limit-flat} there exists a finite type $\mathbf{Z}$-algebra $A_0 \subset A$ and a flat finite type ring map $A_0 \to B_0$ such that $B = B_0 \otimes_{A_0} A$. By Lemma \ref{lemma-Noetherian-relative-pseudo-coherent} we see that $A_0 \to B_0$ is pseudo-coherent. As $A_0 \to B_0$ is flat we see that $B_0$ and $A$ are tor independent over $A_0$, hence we may use Lemma \ref{lemma-base-change-relative-pseudo-coherent} to conclude that $A \to B$ is pseudo-coherent.
By Algebra, Lemma \ref{algebra-lemma-finite-gl-dim-finite-dim-regular} the assumption on $A$ means that $A$ has finite global dimension. Hence every module has finite tor dimension, see Lemma \ref{lemma-finite-gl-dim-tor-dimension}, in particular $B$ does. By Lemma \ref{lemma-Noetherian-pseudo-coherent-ring-map} the map is pseudo-coherent.
Let $A \to B$ be a local complete intersection homomorphism. By Definition \ref{definition-local-complete-intersection} this means that $B = A[x_1, \ldots, x_n]/I$ where $I$ is a Koszul ideal in $A[x_1, \ldots, x_n]$.  By Lemmas \ref{lemma-perfect-ring-map} and \ref{lemma-perfect-module} it suffices to show that $I$ is a perfect module over $A[x_1, \ldots, x_n]$. By Lemma \ref{lemma-glue-perfect} this is a local question. Hence we may assume that $I$ is generated by a Koszul-regular sequence (by Definition \ref{definition-regular-ideal}). Of course this means that $I$ has a finite free resolution and we win.
Observe that $R \to A$ and $R \to B$ are perfect ring maps (Lemma \ref{lemma-flat-finite-presentation-perfect}) hence a fortiori pseudo-coherent ring maps. Thus (1) $\Leftrightarrow$ (2) and (4) $\Leftrightarrow$ (5) by Lemma \ref{lemma-relative-pseudo-coherent-is-moot}. \medskip\noindent Using that $A$ is pseudo-coherent relative to $R$ we use Lemma \ref{lemma-composition-relative-pseudo-coherent} to see that (2) $\Leftrightarrow$ (3). However, since $A \to B$ is surjective, we see directly from Definition \ref{definition-relatively-pseudo-coherent} that (3) is equivalent with (4).
To prove (2) implies (1) it suffices by Lemma \ref{lemma-cone-relatively-perfect} to show that an $R$-flat, finitely presented $A$-module $M$ defines an $R$-perfect object of $D(A)$. Since $M$ has finite tor dimension over $R$, it suffices to show that $M$ is pseudo-coherent. By Algebra, Lemma \ref{algebra-lemma-flat-finite-presentation-limit-flat} there exists a finite type $\mathbf{Z}$-algebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ and a finite $A_0$-module $M_0$ flat over $R_0$ such that $A = A_0 \otimes_{R_0} R$ and $M = M_0 \otimes_{R_0} R$. By Lemma \ref{lemma-Noetherian-pseudo-coherent} we see that $M_0$ is pseudo-coherent $A_0$-module. Choose a resolution $P_0^\bullet \to M_0$ by finite free $A_0$-modules $P_0^n$. Since $A_0$ is flat over $R_0$, this is a flat resolution. Since $M_0$ is flat over $R_0$ we find that $P^\bullet = P_0^\bullet \otimes_{R_0} R$ still resolves $M = M_0 \otimes_{R_0} R$. (You can use Lemma \ref{lemma-base-change-comparison} to see this.) Hence $P^\bullet$ is a finite free resolution of $M$ over $A$ and we conclude that $M$ is pseudo-coherent. \medskip\noindent Assume (1). We can represent $K$ by a bounded above complex $P^\bullet$ of finite free $A$-modules. Assume that $K$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. By Lemma \ref{lemma-last-one-flat} we see that $\tau_{\geq a}P^\bullet$ is a complex of $R$-flat, finitely presented $A$-modules representing $K$.
By Lemma \ref{lemma-pull-pseudo-coherent} we see that $K \otimes_A^\mathbf{L} A'$ is pseudo-coherent. By Lemma \ref{lemma-base-change-comparison} we see that $K \otimes_A^\mathbf{L} A'$ is equal to $K \otimes_R^\mathbf{L} R'$ in $D(R')$. Then we can apply Lemma \ref{lemma-pull-tor-amplitude} to see that $K \otimes_R^\mathbf{L} R'$ in $D(R')$ has finite tor dimension.
The existence of $P^\bullet$ is the definition of a pseudo-coherent complex. We first represent $L$ by a bounded above complex $F^\bullet$ of free $A$-modules (this is possible because bounded tor dimension in particular implies bounded). Next, say $L$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. Then, after replacing $F^\bullet$ by $\tau_{\geq a}F^\bullet$, we get a complex as in (2). This follows from Lemma \ref{lemma-last-one-flat}. \medskip\noindent Proof of (a). Since $F^\bullet$ is bounded an since $P^\bullet$ is bounded above, we see that $E^n = 0$ for $n \ll 0$ and that $E^n$ is a finite (!) direct sum $$ E^n = \bigoplus\nolimits_{p + q = n} \Hom_A(P^{-q}, F^p) $$ and since $P^{-q}$ is finite free, this is indeed an $R$-flat $A$-module. The fact that $E^\bullet$ represents $R\Hom_A(K, L)$ follows from Lemma \ref{lemma-RHom-out-of-projective}. \medskip\noindent Proof of (b). Let $R \to R'$ be a ring map and $A' = A \otimes_R R'$. By Lemma \ref{lemma-base-change-comparison} the object $L \otimes_A^\mathbf{L} A'$ is represented by $F^\bullet \otimes_R R'$ viewed as a complex of $A'$-modules (by flatness of $F^p$ over $R$). Similarly for $P^\bullet \otimes_R R'$. As above $R\Hom_{A'}(K \otimes_A^\mathbf{L} A', L \otimes_A^\mathbf{L} A')$ is represented by $$ \Hom^\bullet(P^\bullet \otimes_R R', F^\bullet \otimes_R R') = E^\bullet \otimes_R R' $$ The equality holds by looking at the terms of the complex individually and using that $\Hom_{A'}(P^{-q} \otimes_R R', F^p \otimes_R R') = \Hom_A(P^{-q}, F^p) \otimes_R R'$.
By Algebra, Lemma \ref{algebra-lemma-colimit-category-fp-modules} the category of finitely presented $A$-modules is the colimit of the categories of finitely presented $A_i$-modules. Given this, Algebra, Lemma \ref{algebra-lemma-flat-finite-presentation-limit-flat} tells us that category of $R$-flat, finitely presented $A$-modules is the colimit of the categories of $R_i$-flat, finitely presented $A_i$-modules. Thus the characterization in Lemma \ref{lemma-structure-relatively-perfect} proves that (1) is true. \medskip\noindent To prove (2) we choose $P_0^\bullet$ representing $K_0$ and $F_0^\bullet$ representing $L_0$ as in Lemma \ref{lemma-compute-RHom-relatively-perfect}. Then $E_0^\bullet = \Hom^\bullet(P_0^\bullet, F_0^\bullet)$ satisfies $$ H^0(E_0^\bullet \otimes_{R_0} R_i) = \Hom_{D(A_i)}(K_0 \otimes_{A_0}^\mathbf{L} A_i, L_0 \otimes_{A_0}^\mathbf{L} A_i) $$ and $$ H^0(E_0^\bullet \otimes_{R_0} R) = \Hom_{D(A)}(K_0 \otimes_{A_0}^\mathbf{L} A, L_0 \otimes_{A_0}^\mathbf{L} A) $$ by the lemma. Thus the result because tensor product commutes with colimits and filtered colimits are exact (Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}).
We can represent $K$ by a bounded above complex of finite free $A$-modules $E^\bullet$, see Lemma \ref{lemma-pseudo-coherent}. By Lemma \ref{lemma-lift-complex-projectives} we conclude that $K'$ is pseudo-coherent because it can be represented by a bounded above complex $P^\bullet$ of finite free $A'$-modules with $P^\bullet \otimes_{A'} A = E^\bullet$. Observe that this also means $P^\bullet \otimes_{R'} R = E^\bullet$ (since $A = A' \otimes_{R'} R$). \medskip\noindent Let $I = \Ker(R' \to R)$. Then $I^n = 0$ for some $n$. Choose $[a, b]$ such that $K$ has tor amplitude in $[a, b]$ as a complex of $R$-modules. We will show $K'$ has tor amplitude in $[a, b]$. To do this, let $M'$ be an $R'$-module. If $IM' = 0$, then $$ K' \otimes_{R'}^\mathbf{L} M' = P^\bullet \otimes_{R'} M' = E^\bullet \otimes_R M' = K \otimes_R^\mathbf{L} M' $$ (because $A'$ is flat over $R'$ and $A$ is flat over $R$) which has nonzero cohomology only for degrees in $[a, b]$ by choice of $a, b$. If $I^{t + 1}M' = 0$, then we consider the short exact sequence $$ 0 \to IM' \to M' \to M'/IM' \to 0 $$ with $M = M'/IM'$. By induction on $t$ we have that both $K' \otimes_{R'}^\mathbf{L} IM'$ and $K' \otimes_{R'}^\mathbf{L} M'/IM'$ have nonzero cohomology only for degrees in $[a, b]$. Then the distinguished triangle $$ K' \otimes_{R'}^\mathbf{L} IM' \to K' \otimes_{R'}^\mathbf{L} M' \to K' \otimes_{R'}^\mathbf{L} M'/IM' \to (K' \otimes_{R'}^\mathbf{L} IM')[1] $$ proves the same is true for $K' \otimes_{R'}^\mathbf{L} M'$. This proves the desired bound for all $M'$ and hence the desired bound on the tor amplitude of $K'$.
By Lemma \ref{lemma-more-relative-pseudo-coherent-is-moot} $K$ is pseudo-coherent as a complex of $R[x_1, \ldots, x_d]$-modules. Therefore we may assume $A = R[x_1, \ldots, x_d]$. Applying Lemma \ref{lemma-perfect-over-regular-local-ring} to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$ with tor amplitude in $[a - d, b]$. Since $R_\mathfrak p \to A_\mathfrak q$ is flat, we conclude by Lemma \ref{lemma-flat-push-tor-amplitude}.
Observe that (1) implies (2) as an $R$-perfect complex has bounded tor dimension as a complex of $R$-modules by definition. Let us prove the other implication. \medskip\noindent Write $A = R[x_1, \ldots, x_d]/I$. Denote $L$ in $D(R[x_1, \ldots, x_d])$ the restriction of $K$. By Lemma \ref{lemma-more-relative-pseudo-coherent-is-moot} we see that $L$ is pseudo-coherent. Since $L$ and $K$ have the same image in $D(R)$ we see that $L$ is $R$-perfect if and only if $K$ is $R$-perfect. Also $L \otimes_R^\mathbf{L} \kappa(\mathfrak p)$ and $K \otimes_R^\mathbf{L} \kappa(\mathfrak p)$ are the same objects of $D(\kappa(\mathfrak p))$. This reduces us to the case $A = R[x_1, \ldots, x_d]$. \medskip\noindent Say $A = R[x_1, \ldots, x_d]$ and $K$ satisfies (2). Let $\mathfrak q \subset A$ be a prime lying over a prime $\mathfrak p \subset R$. By Lemma \ref{lemma-perfect-over-regular-local-ring} applied to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$. Since $K$ is bounded below, we see that $K$ is perfect in $D(A)$ by  Lemma \ref{lemma-check-perfect-stalks}. This implies that $K$ is $R$-perfect by Lemma \ref{lemma-perfect-relatively-perfect} and the proof is complete.
The equivalence of (1) and (2) follows from Lemma \ref{lemma-projective-amplitude}. If $R$ is Noetherian and $H^i(K)$ is a finite $R$-module for $i = -1, 0$, then $K$ is pseudo-coherent, see Lemma \ref{lemma-Noetherian-pseudo-coherent}. Thus the equivalence of (1) and (3) follows from Lemma \ref{lemma-projective-amplitude-pseudo-coherent}.
Proof of (1). Suppose $K$ is given by the complex of modules $M^\bullet$. We may first replace $M^\bullet$ by $\tau_{\leq 0}M^\bullet$. Thus we may assume $M^i = 0$ for $i > 0$, Next, we may choose a free resolution $P^\bullet \to M^\bullet$ with $P^i = 0$ for $i > 0$, see Derived Categories, Lemma \ref{derived-lemma-subcategory-left-resolution}. Finally, we can set $K^\bullet = \tau_{\geq -1}P^\bullet$. \medskip\noindent Proof of (2). Assume $R$ is Noetherian and $H^i(K)$ is a finite $R$-module for $i = -1, 0$. By Lemma \ref{lemma-pseudo-coherent} we can choose a quasi-isomorphism $F^\bullet \to M^\bullet$ with $F^i = 0$ for $i > 0$ and $F^i$ finite free. Then we can set $K^\bullet = \tau_{\geq -1}F^\bullet$.
Set $F^0 = M^0$. Choose a free $R$-module $F^{-1}$ and a surjection $F^{-1} \to M^{-1}$. Choose a free $R$-module $F^{-2}$ and a surjection $F^{-2} \to M^{-2} \times_{M^{-1}} F^{-1}$. Continuing in this way we obtain a quasi-isomorphism $p^\bullet : F^\bullet \to M^\bullet$ which is termwise surjective and with $F^i$ projective for all $i$. \medskip\noindent Proof of (1). By Derived Categories, Lemma \ref{derived-lemma-morphisms-from-projective-complex} we have $$ \Hom_{D(R)}(M^\bullet, K^\bullet) = \Hom_{K(R)}(F^\bullet, K^\bullet) $$ If $K^i = 0$ for $i \leq -2$, then any morphism of complexes $F^\bullet \to K^\bullet$ factors through $p^\bullet$. Similarly, any homotopy $\{h^i : F^i \to K^{i - 1}\}$ factors through $p^\bullet$. Thus (1) holds. \medskip\noindent Proof of (2). If (2)(b) holds, then $a^\bullet$ is homotopic to a map of complexes $(a')^\bullet : M^\bullet \to K^\bullet$ which is zero in degree $-1$. On the other hand, let $N \to I^\bullet$ be an injective resolution. We have $$ \Ext^1_R(K^\bullet, N) = \Hom_{D(R)}(K^\bullet, I^\bullet[1]) = \Hom_{K(R)}(K^\bullet, I^\bullet[1]) $$ by Derived Categories, Lemma \ref{derived-lemma-morphisms-into-injective-complex}. Let $b^\bullet : K^\bullet \to I^\bullet[1]$ be a map of complexes. Since $K^1 = 0$ the map $b^0 : K^0 \to I^1$ maps into the kernel of $I^1 \to I^2$ which is the image of $I^0 \to I^1$. Since $K^0$ is projective we can lift $b^0$ to a map $h : K^0 \to I^0$. Thus we see that $b^\bullet$ is homotopic to a map of complexes $(b')^\bullet$ with $(b')^0 = 0$. Since $K^i = 0$ for $i \not \in [-1, 0]$ it follows that $(b')^\bullet \circ (a')^\bullet = 0$ as a map of complexes. Hence the map $\Ext^1_R(K^\bullet, N) \to \Ext^1_R(M^\bullet, N)$ is zero. In this way we see that (2)(b) implies (2)(a). Conversely, assume (2)(a). We see that the canonical element in $\Ext^1_R(K^\bullet, K^{-1})$ maps to zero in $\Ext^1_R(M^\bullet, K^{-1})$. Using (1) we see immediately that we get a map $h^0$ as in (2)(b). \medskip\noindent Proof of (3). Choose $b^\bullet : F^\bullet \to K^\bullet$ representing $\alpha$. The composition of $\alpha$ with $K^\bullet \to K^{-2}[2]$ is represented by $b^{-2} : F^{-2} \to K^{-2}$. As this is homotopic to $a \circ p^{-2} : F^{-2} \to M^{-2} \to K^{-2}$, there is a map $h : F^{-1} \to K^{-2}$ such that $b^{-2} = a \circ p^{-2} + h \circ d_F^{-2}$. Adjusting $b^\bullet$ by $h$ viewed as a homotopy from $F^\bullet$ to $K^\bullet$, we find that $b^{-2} = a \circ p^{-2}$. Hence $b^{-2}$ factors through $p^{-2}$. Since $F^0 = M^0$ the kernel of $p^{-2}$ surjects onto the kernel of $p^{-1}$ (for example because the kernel of $p^\bullet$ is an acyclic complex or by a diagram chase). Hence $b^{-1}$ necessarily factors through $p^{-1}$ as well and we see that (3) holds for these factorizations and $a^0 = b^0$. \medskip\noindent Proof of (4) is omitted. Hint: There is a homotopy between $a^\bullet \circ p^\bullet$ and $(a')^\bullet \circ p^\bullet$ and we argue as before that this homotopy factors through $p^\bullet$.
\medskip\noindent Assume (1) and let $K^{-1} \to K^0$ be a two term complex representing $K$ with $K^0$ projective. We will use the description of maps in $D(R)$ out of $K^\bullet$ given in Lemma \ref{lemma-map-out-of-almost-free} without further mention. Choosing $N = K^{-1}$ consider the element $\xi$ of $\Ext^1_R(K, N)$ given by $\text{id}_{K^{-1}} : K^{-1} \to K^{-1}$. Since is annihilated by $a \in I$ we see that we get the dotted arrow fitting into the following commutative diagram $$ \xymatrix{ K^{-1} \ar[d]_a \ar[r]_{d_K^{-1}} & K^0 \ar@{..>}[ld]^h \\ K^{-1} } $$ This proves that (3) holds. Part (3) implies (2) in view of Lemma \ref{lemma-represent-two-term-complex} part (1). Assume $K^\bullet$ is as in (2) and $N$ is an arbitrary $R$-module. Any element $\xi$ of $\Ext^1_R(K, N)$ is given as the class of a map $\varphi : K^{-1} \to N$. Then for $a \in I$ by assumption we may choose a map $h$ as in the diagram above and we see that $a\varphi = \varphi \circ a = \varphi \circ h \circ \text{d}_K^{-1}$ which proves that $a \xi$ is zero in $\Ext^1_R(K, N)$. Thus (1), (2), and (3) are equivalent. \medskip\noindent Assume $R$ is Noetherian and $H^i(K)$ is a finite $R$-module for $i = -1, 0$. Part (3) implies (5) in view of Lemma \ref{lemma-represent-two-term-complex} part (2). It is clear that (5) implies (2). Trivially (1) implies (4). Thus to finish the proof it suffices to show that (4) implies any of the other conditions. Let $K^{-1} \to K^0$ be a complex representing $K$ with $K^0$ finite free and $K^{-1}$ finite as in Lemma \ref{lemma-represent-two-term-complex} part (2). The argument given in the proof of (2) $\Rightarrow$ (1) shows that if $\Ext^1_R(K, K^{-1})$ is annihilated by $I$, then (1) holds. In this way we see that (4) implies (1) and the proof is complete.
We may assume $K$ is represented by a two term complex $K^{-1} \to K^0$ with $K^0$ free such that for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ is equal to $h_a \circ d_K^{-1}$ for some map $h_a : K^0 \to K^{-1}$. By Lemma \ref{lemma-two-term-base-change} we see that $\tau_{\geq -1}(K \otimes_R^\mathbf{L} R')$ is represented by $K^\bullet \otimes_R R'$. Then of course for every $a \in I$ we see that $a \otimes 1 : K^{-1} \otimes_R R' \to K^{-1} \otimes_R R'$ is equal to $(h_a \otimes 1) \circ (d_K^{-1} \otimes 1)$. Since the collection of maps $K^{-1} \otimes_R R' \to K^{-1} \otimes_R R'$ which factor through $\text{d}_K^{-1} \otimes 1$ forms an $R'$-module we conclude.
Set $M = \Ker(H^{-1}(\alpha))$. Then $\alpha$ fits into a distinguished triangle $$ M[1] \to K \to K' \to M[2] $$ Since $K \to K' \xrightarrow{f} K'$ is zero by our assumption, we see that $f : K' \to K'$ factors over a map $M[2] \to K'$. However $\Hom(M[2], K') = 0$ for example by Derived Categories, Lemma \ref{derived-lemma-negative-exts}.
The distinguished triangle $H^{-1}(K)[1] \to K \to H^0(K)[0] \to H^{-1}(K)[2]$ determines an exact sequence $$ 0 \to \Ext^1_R(H^0(K), N) \to \Ext^1_R(K, N) \to \Hom_R(H^{-1}(K), N) \to \Ext^2_R(H^0(K), N) $$ Thus (2) implies that $I^{2c}$ annihilates $\Ext^1_R(K, N)$ for every $R$-module $N$. Assuming (1) we immediately see that $H^0(K)$ is $I^c$-projective. On the other hand, we may choose an injective map $H^{-1}(K) \to N$ for some injective $R$-module $N$. Then this map is the image of an element of $\Ext^1_R(K, N)$ by the vanishing of the $\Ext^2$ in the sequence and we conclude $H^{-1}(K)$ is annihilated by $I^c$. \medskip\noindent Assume $R$ is Noetherian and $H^i(K)$ is a finite $R$-module for $i = -1, 0$. By Lemma \ref{lemma-ext-1-annihilated-definite} we see that (3) is equivalent to (1) and (2). Also, if (3) holds then for $f \in I$ the multiplication by $f$ on $H^0(K)$ factors through a projective module, which implies that $H^0(K)_f$ is a summand of a projective $R_f$-module and hence itself a projective $R_f$-module. Choosing $f_1, \ldots, f_s$ to be generators of $I$ we find the equivalent conditions (1), (2), and (3) imply (5). Of course (5) trivially implies (4). \medskip\noindent Assume (4). Since $H^{-1}(K)$ is a finite $R$-module and $I$-power torsion we see that $I^{c_1}$ annihilates $H^{-1}(K)$ for some $c_1 \geq 0$. Choose a short exact sequence $$ 0 \to M \to R^{\oplus r} \to H^0(K) \to 0 $$ which determines an element $\xi \in \Ext^1_R(H^0(K), M)$. For any $f \in I$ we have $\Ext^1_R(H^0(K), M)_f = \Ext^1_{R_f}(H^0(K)_f, M_f)$ by Lemma \ref{lemma-pseudo-coherence-and-base-change-ext}. Hence if $H^0(K)_f$ is projective, then a power of $f$ annihilates $\xi$. We conclude that $\xi$ is annihilated by $(f_1, \ldots, f_s)^{c_2}$ for some $c_2 \geq 0$. Since $V(f_1, \ldots, f_s) \subset V(I)$ we have $\sqrt{I} \subset (f_1, \ldots, f_s)$ (Algebra, Lemma \ref{algebra-lemma-Zariski-topology}). Since $R$ is Noetherian we find $I^{c_3} \subset (f_1, \ldots, f_s)$ for some $c_3 \geq 0$ (Algebra, Lemma \ref{algebra-lemma-Noetherian-power}). Hence $I^{c2c3}$ annihilates $\xi$. This in turn says that $H^0(K)$ is $I^{c_2c_3}$-projective (as multiplication by $a \in I$ which annihilate $\xi$ factor through $R^{\oplus r}$). Hence taking $c = \max(c_1, c_2c_3)$ we see that (2) holds.
Choose a presentation $\alpha : P \to B$ as in Algebra, Section \ref{algebra-section-netherlander}. Let $I = \Ker(\alpha)$. Set $P' = P \otimes_A A'$ and denote $\alpha' : P' \to B'$ the corresponding presentation of $B'$ over $A'$. As $B$ is flat over $A$ we see that $I' = \Ker(\alpha')$ is equal to $I \otimes_A A'$. Hence $$ I'/(I')^2 = \Coker(I^2 \otimes_A A' \to I \otimes_A A') = I/I^2 \otimes_A A' = I/I^2 \otimes_B B' $$ We have $\Omega_{P'/A'} = \Omega_{P/A} \otimes_A A'$ because both sides have the same basis. It follows that $\Omega_{P'/A'} \otimes_{P'} B' = \Omega_{P/A} \otimes_P B \otimes_B B'$. This proves that $\NL(\alpha) \otimes_B B' \to \NL(\alpha')$ is an isomorphism of complexes and hence the first statement holds. \medskip\noindent We have $$ \NL(\alpha) = I/I^2 \longrightarrow \Omega_{P/A} \otimes_P B $$ as a complex of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see Lemma \ref{lemma-last-one-flat}. If this holds, then $\NL(\alpha) \otimes_B^\mathbf{L} B' = \NL(\alpha) \otimes_B B'$ and we get the second statement.
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 Lemma \ref{lemma-last-one-flat}. By definition $I$ is a Koszul regular ideal and hence a quasi-regular ideal, see Section \ref{section-ideals}. Thus $I/I^2$ is a finite projective $B$-module (Lemma \ref{lemma-quasi-regular-ideal-finite-projective}) and we conclude both that $\NL_{B/A}$ is perfect and that it has tor amplitude in $[-1, 0]$.