hcju/stacksqa huggingface dataset:
Assume the formal fibres of $R_\mathfrak m$ have $P$ for all maximal ideals $\mathfrak m$ of $R$. Let $\mathfrak p$ be a prime of $R$ and choose a maximal ideal $\mathfrak p \subset \mathfrak m$. Since $R_\mathfrak m \to R_\mathfrak m^\wedge$ is faithfully flat we can choose a prime $\mathfrak p'$ if $R_\mathfrak m^\wedge$ lying over $\mathfrak pR_\mathfrak m$. Consider the commutative diagram $$ \xymatrix{ R_\mathfrak m^\wedge \ar[r] & (R_\mathfrak m^\wedge)_{\mathfrak p'} \ar[r] & (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge \\ R_\mathfrak m \ar[u] \ar[r] & R_\mathfrak p \ar[u] \ar[r] & R_\mathfrak p^\wedge \ar[u] } $$ By assumption the fibres of the ring map $R_\mathfrak m \to R_\mathfrak m^\wedge$ have $P$. By Proposition \ref{proposition-Noetherian-complete-G-ring} $(R_\mathfrak m^\wedge)_{\mathfrak p'} \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ is regular. The localization $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge)_{\mathfrak p'}$ is regular. Hence $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ is regular by Lemma \ref{lemma-regular-composition}. Hence the fibres of $R_\mathfrak m \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ have $P$ by (C). Since $R_\mathfrak m \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ factors through the localization $R_\mathfrak p$, also the fibres of $R_\mathfrak p \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ have $P$. Thus we may apply (D) to see that the fibres of $R_\mathfrak p \to R_\mathfrak p^\wedge$ have $P$.
The ring map $A \to A^\wedge$ is flat by Algebra, Lemma \ref{algebra-lemma-completion-flat}. The ring $A^\wedge$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian}. Thus it suffices to check the third condition of Lemma \ref{lemma-P-local}. Let $\mathfrak m' \subset A^\wedge$ be a maximal ideal lying over $\mathfrak m \subset A$. By Algebra, Lemma \ref{algebra-lemma-radical-completion} we have $IA^\wedge \subset \mathfrak m'$. Since $A^\wedge/IA^\wedge = A/I$ we see that $I \subset \mathfrak m$, $\mathfrak m/I = \mathfrak m'/IA^\wedge$, and $A/\mathfrak m = A^\wedge/\mathfrak m'$. Since $A^\wedge/\mathfrak m'$ is a field, we conclude that $\mathfrak m$ is a maximal ideal as well. Then $A_\mathfrak m \to A^\wedge_{\mathfrak m'}$ is a flat local ring homomorphism of Noetherian local rings which identifies residue fields and such that $\mathfrak m A^\wedge_{\mathfrak m'} = \mathfrak m'A^\wedge_{\mathfrak m'}$. Thus it induces an isomorphism on complete local rings, see Lemma \ref{lemma-flat-unramified}. Let $(A_\mathfrak m)^\wedge$ be the completion of $A_\mathfrak m$ with respect to its maximal ideal. The ring map $$ (A^\wedge)_{\mathfrak m'} \to ((A^\wedge)_{\mathfrak m'})^\wedge = (A_\mathfrak m)^\wedge $$ is faithfully flat (Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}). Thus we can apply (D) to the ring maps $$ A_\mathfrak m \to (A^\wedge)_{\mathfrak m'} \to (A_\mathfrak m)^\wedge $$ to conclude because the fibres of $A_\mathfrak m \to (A_\mathfrak m)^\wedge$ have $P$ as $A$ is a $P$-ring.
Let $\mathfrak m^h \subset A^h$ be a maximal ideal. We have to show that the fibres of $A^h_{\mathfrak m^h} \to (A^h_{\mathfrak m^h})^\wedge$ have $P$, see Lemma \ref{lemma-check-P-ring-maximal-ideals}. Let $\mathfrak m$ be the inverse image of $\mathfrak m^h$ in $A$. Note that $I^h \subset \mathfrak m^h$ and hence $I \subset \mathfrak m$ as $(A^h, I^h)$ is a henselian pair. Recall that $A^h$ is Noetherian, $I^h = IA^h$, and that $A \to A^h$ induces an isomorphism on $I$-adic completions, see Lemma \ref{lemma-henselization-Noetherian-pair}. Then the local homomorphism of Noetherian local rings $$ A_\mathfrak m \to A^h_{\mathfrak m^h} $$ induces an isomorphism on completions at maximal ideals by Lemma \ref{lemma-flat-unramified} (details omitted). Let $\mathfrak q^h$ be a prime of $A^h_{\mathfrak m^h}$ lying over $\mathfrak q \subset A_\mathfrak m$. Set $\mathfrak q_1 = \mathfrak q^h$ and let $\mathfrak q_2, \ldots, \mathfrak q_t$ be the other primes of $A^h$ lying over $\mathfrak q$, so that $A^h \otimes_A \kappa(\mathfrak q) = \prod\nolimits_{i = 1, \ldots, t} \kappa(\mathfrak q_i)$, see Lemma \ref{lemma-filtered-colimit-etale-noetherian-fibres}. Using that $(A^h)_{\mathfrak m^h}^\wedge = (A_\mathfrak m)^\wedge$ as discussed above we see $$ \prod\nolimits_{i = 1, \ldots, t} (A^h_{\mathfrak m^h})^\wedge \otimes_{A^h_{\mathfrak m^h}} \kappa(\mathfrak q_i) = (A^h_{\mathfrak m^h})^\wedge \otimes_{A^h_{\mathfrak m^h}} (A^h_{\mathfrak m^h} \otimes_{A_{\mathfrak m}} \kappa(\mathfrak q)) = (A_{\mathfrak m})^\wedge \otimes_{A_{\mathfrak m}} \kappa(\mathfrak q) $$ Hence, looking at local rings and using (B), we see that $$ \kappa(\mathfrak q) \longrightarrow (A^h_{\mathfrak m^h})^\wedge \otimes_{A^h_{\mathfrak m^h}} \kappa(\mathfrak q^h) $$ has $P$ as $\kappa(\mathfrak q) \to (A_\mathfrak m)^\wedge \otimes_{A_\mathfrak m} \kappa(\mathfrak q)$ does by assumption on $A$. Since $\kappa(\mathfrak q^h)/\kappa(\mathfrak q)$ is separable algebraic, by (E) we find that $\kappa(\mathfrak q^h) \to (A^h_{\mathfrak m^h})^\wedge \otimes_{A^h_{\mathfrak m^h}} \kappa(\mathfrak q^h)$ has $P$ as desired.
We have seen this for the henselization in Lemma \ref{lemma-henselization-pair-P-ring}. To prove it for the strict henselization, it suffices to show that the formal fibres of $R^{sh}$ have $P$, see Lemma \ref{lemma-check-P-ring-maximal-ideals}. Let $\mathfrak r \subset R^{sh}$ be a prime and set $\mathfrak p = R \cap \mathfrak r$. Set $\mathfrak r_1 = \mathfrak r$ and let $\mathfrak r_2, \ldots, \mathfrak r_s$ be the other primes of $R^{sh}$ lying over $\mathfrak p$, so that $R^{sh} \otimes_R \kappa(\mathfrak p) = \prod\nolimits_{i = 1, \ldots, s} \kappa(\mathfrak r_i)$, see Lemma \ref{lemma-fibres-henselization}. Then we see that $$ \prod\nolimits_{i = 1, \ldots, t} (R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r_i) = (R^{sh})^\wedge \otimes_{R^{sh}} (R^{sh} \otimes_R \kappa(\mathfrak p)) = (R^{sh})^\wedge \otimes_R \kappa(\mathfrak p) $$ Note that $R^\wedge \to (R^{sh})^\wedge$ is formally smooth in the $\mathfrak m_{(R^{sh})^\wedge}$-adic topology, see Lemma \ref{lemma-henselization-noetherian}. Hence $R^\wedge \to (R^{sh})^\wedge$ is regular by Proposition \ref{proposition-fs-regular}. We conclude that property $P$ holds for $\kappa(\mathfrak p) \to (R^{sh})^\wedge \otimes_R \kappa(\mathfrak p)$ by (C) and our assumption on $R$. Using property (B), using the decomposition above, and looking at local rings we conclude that property $P$ holds for $\kappa(\mathfrak p) \to (R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r)$. Since $\kappa(\mathfrak r)/\kappa(\mathfrak p)$ is separable algebraic, it follows from (E) that $P$ holds for $\kappa(\mathfrak r) \to (R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r)$.
Part (A) follows from the definition of geometrically reduced algebras (Algebra, Definition \ref{algebra-definition-geometrically-reduced}). Part (B) follows too: a ring is reduced if and only if all local rings are reduced. Part (C). This follows from Lemma \ref{lemma-reduced-goes-up}. Part (D). This follows from Algebra, Lemma \ref{algebra-lemma-descent-reduced}. Part (E). This follows from Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-over-separable-algebraic}.
Part (A) follows from the definition of geometrically normal algebras (Algebra, Definition \ref{algebra-definition-geometrically-normal}). Part (B) follows too: a ring is normal if and only if all of its local rings are normal. Part (C). This follows from Lemma \ref{lemma-normal-goes-up}. Part (D). This follows from Algebra, Lemma \ref{algebra-lemma-descent-normal}. Part (E). This follows from Algebra, Lemma \ref{algebra-lemma-geometrically-normal-over-separable-algebraic}.
Let $k \to R$ be a ring map where $k$ is a field and $R$ a Noetherian ring. Let $k \subset k'$ be a finitely generated field extension. Then the fibres of the ring map $R \to R \otimes_k k'$ are Cohen-Macaulay by Algebra, Lemma \ref{algebra-lemma-tensor-fields-CM}. Hence we may apply Algebra, Lemma \ref{algebra-lemma-Sk-goes-up} to the ring map $R \to R \otimes_k k'$ to see that if $R$ has $(S_n)$ so does $R \otimes_k k'$. This proves (A). Part (B) follows too: a Noetherian rings has $(S_n)$ if and only if all of its local rings have $(S_n)$. Part (C). This follows from Algebra, Lemma \ref{algebra-lemma-Sk-goes-up} as the fibres of a regular homomorphism are regular and in particular Cohen-Macaulay. Part (D). This follows from Algebra, Lemma \ref{algebra-lemma-descent-Sk}. Part (E). This is immediate as the condition does not refer to the ground field.
Let $k \to R$ be a ring map where $k$ is a field and $R$ a Noetherian ring. Assume $P(k \to R)$ is true. Let $k \subset K$ be a finitely generated field extension. By Algebra, Lemma \ref{algebra-lemma-make-separable} we can find a diagram $$ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } $$ where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is separable. By Algebra, Lemma \ref{algebra-lemma-localization-smooth-separable} there exists a smooth $k'$-algebra $B$ such that $K'$ is the fraction field of $B$. Now we can argue as follows: Step 1: $R \otimes_k k'$ satisfies $(S_n)$ because we assumed $P$ for $k \to R$. Step 2: $R \otimes_k k' \to R \otimes_k k' \otimes_{k'} B$ is a smooth ring map (Algebra, Lemma \ref{algebra-lemma-base-change-smooth}) and we conclude $R \otimes_k k' \otimes_{k'} B$ satisfies $(S_n)$ by Algebra, Lemma \ref{algebra-lemma-Rk-goes-up} (and using Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field} to see that the hypotheses are satisfied). Step 3. $R \otimes_k k' \otimes_{k'} K' = R \otimes_k K'$ satisfies $(R_n)$ as it is a localization of a ring having $(R_n)$. Step 4. Finally $R \otimes_k K$ satisfies $(R_n)$ by descent of $(R_n)$ along the faithfully flat ring map $K \otimes_k A \to K' \otimes_k A$ (Algebra, Lemma \ref{algebra-lemma-descent-Rk}). This proves (A). Part (B) follows too: a Noetherian ring has $(R_n)$ if and only if all of its local rings have $(R_n)$. Part (C). This follows from Algebra, Lemma \ref{algebra-lemma-Rk-goes-up} as the fibres of a regular homomorphism are regular (small detail omitted). Part (D). This follows from Algebra, Lemma \ref{algebra-lemma-descent-Rk} (small detail omitted). \medskip\noindent Part (E). Let $l/k$ be a separable algebraic extension of fields and let $l \to R$ be a ring map with $R$ Noetherian. Assume that $k \to R$ has $P$. We have to show that $l \to R$ has $P$. Let $l'/l$ be a finite extension. First observe that there exists a finite subextension $l/m/k$ and a finite extension $m'/m$ such that $l' = l \otimes_m m'$. Then $R \otimes_l l' = R \otimes_m m'$. Hence it suffices to prove that $m \to R$ has property $P$, i.e., we may assume that $l/k$ is finite. If $l/k$ is finite, then $l'/k$ is finite and we see that $$ l' \otimes_l R = (l' \otimes_k R) \otimes_{l \otimes_k l} l $$ is a localization (by Algebra, Lemma \ref{algebra-lemma-separable-algebraic-diagonal}) of the Noetherian ring $l' \otimes_k R$ which has property $(R_n)$ by assumption $P$ for $k \to R$. This proves that $l' \otimes_l R$ has property $(R_n)$ as desired.
Assume (2). By Algebra, Lemma \ref{algebra-lemma-local-nagata-and-analytically-unramified} we have to show that if $A \to B$ is finite, $B$ is a domain, and $\mathfrak m' \subset B$ is a maximal ideal, then $B_{\mathfrak m'}$ is analytically unramified. Combining Lemmas \ref{lemma-formal-fibres-reduced} and \ref{lemma-check-P-ring-maximal-ideals} and Proposition \ref{proposition-finite-type-over-P-ring} we see that the formal fibres of $B_{\mathfrak m'}$ are geometrically reduced. In particular $B_{\mathfrak m'}^\wedge \otimes_B L$ is reduced where $L$ is the fraction field of $B$. It follows that $B_{\mathfrak m'}^\wedge$ is reduced, i.e., $B_{\mathfrak m'}$ is analytically unramified. \medskip\noindent Assume (1). Let $\mathfrak q \subset A$ be a prime ideal and let $\kappa(\mathfrak q) \subset K$ be a finite extension. We have to show that $A^\wedge \otimes_A K$ is reduced. Let $A/\mathfrak q \subset B \subset K$ be a local subring finite over $A$ whose fraction field is $K$. To construct $B$ choose $x_1, \ldots, x_n \in K$ which generate $K$ over $\kappa(\mathfrak q)$ and which satisfy monic polynomials $P_i(T) = T^{d_i} + a_{i, 1} T^{d_i - 1} + \ldots + a_{i, d_i} = 0$ with $a_{i, j} \in \mathfrak m$. Then let $B$ be the $A$-subalgebra of $K$ generated by $x_1, \ldots, x_n$. (For more details see the proof of Algebra, Lemma \ref{algebra-lemma-local-nagata-and-analytically-unramified}.) Then $$ A^\wedge \otimes_A K = (A^\wedge \otimes_A B)_\mathfrak q = B^\wedge_\mathfrak q $$ Since $B^\wedge$ is reduced by  Algebra, Lemma \ref{algebra-lemma-local-nagata-and-analytically-unramified} the proof is complete.
Let $R$ be quasi-excellent. Using that a finite type algebra over $R$ is quasi-excellent (Lemma \ref{lemma-finite-type-over-excellent}) we see that it suffices to show that any quasi-excellent domain is N-1, see Algebra, Lemma \ref{algebra-lemma-check-universally-japanese}. Applying Algebra, Lemma \ref{algebra-lemma-characterize-N-1} (and using that a quasi-excellent ring is J-2) we reduce to showing that a quasi-excellent local domain $R$ is N-1. As $R \to R^\wedge$ is regular we see that $R^\wedge$ is reduced by Lemma \ref{lemma-reduced-goes-up}. In other words, $R$ is analytically unramified. Hence $R$ is N-1 by Algebra, Lemma \ref{algebra-lemma-analytically-unramified-easy}.
Let $0 \to M'' \to M' \to M \to 0$ be a short exact sequence of $R$-modules. Consider the long exact sequence $$ \begin{matrix} 0 \to \Hom_R(M, J) \to \Hom_R(M', J) \to \Hom_R(M'', J) \\ \phantom{0\ } \to \Ext^1_R(M, J) \to \Ext^1_R(M', J) \to \Ext^1_R(M'', J) \to \ldots \end{matrix} $$ of Algebra, Lemma \ref{algebra-lemma-reverse-long-exact-seq-ext}. Thus we see that (2) implies (1). Conversely, if $J$ is injective then the $\Ext$-group is zero by Homology, Lemma \ref{homology-lemma-characterize-injectives} and Lemma \ref{lemma-relation-ext-ext}.
If $I \subset R$ is an ideal, then the short exact sequence $0 \to I \to R \to R/I \to 0$ gives an exact sequence $$ \Hom_R(R, J) \to \Hom_R(I, J) \to \Ext^1_R(R/I, J) \to 0 $$ by Algebra, Lemma \ref{algebra-lemma-reverse-long-exact-seq-ext} and the fact that $\Ext^1_R(R, J) = 0$ as $R$ is projective (Algebra, Lemma \ref{algebra-lemma-characterize-projective}). Thus (2) and (3) are equivalent. In this proof we will show that (1) $\Leftrightarrow$ (3) which is known as Baer's criterion. \medskip\noindent Assume (1). Given a module map $I \to J$ as in (3) we find the extension $R \to J$ because the map $\Hom_R(R, J) \to \Hom_R(I, J)$ is surjective by definition. \medskip\noindent Assume (3). Let $M \subset N$ be an inclusion of $R$-modules. Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$ extends to $N$ which finishes the proof of the lemma. Consider the set of homomorphisms $\varphi' : M' \to J$ with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$. Define $(M', \varphi') \geq (M'', \varphi'')$ if and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$. If $(M_i, \varphi_i)_{i \in I}$ is a totally ordered collection of such pairs, then we obtain a map $\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$ maps to $\varphi_i(a)$. Thus Zorn's lemma applies. To conclude we have to show that if the pair $(M', \varphi')$ is maximal then $M' = N$. In other words, it suffices to show, given any subgroup $M \subset N$, $M \not = N$ and any $\varphi : M \to J$, then we can find $\varphi' : M' \to J$ with $M \subset M' \subset N$ such that (a) the inclusion $M \subset M'$ is strict, and (b) the morphism $\varphi'$ extends $\varphi$. \medskip\noindent To prove this, pick $x \in N$, $x \not \in M$. Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$. Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$. Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3). By our choice of $I$ the kernel of $M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$ contains the kernel of the map $M \oplus R \to N$, $(y, f) \mapsto y + fx$. Hence this homomorphism factors through the image $M' = M + Rx$ and this extends the given homomorphism as desired.
Note that $J(M) \cong \prod_{\varphi \in M^\vee} R^\vee$ as an $R$-module. As the product of injective modules is injective, it suffices to show that $R^\vee$ is injective. For this we use that $$ \Hom_R(N, R^\vee) = \Hom_R(N, \Hom_{\mathbf{Z}}(R, \mathbf{Q}/\mathbf{Z})) = N^\vee $$ and the fact that $(-)^\vee$ is an exact functor by Lemma \ref{lemma-vee-exact}.
Let $L^\bullet$ be a complex of $R$-modules. Then $$ 0 \to \text{Tot}(L^\bullet \otimes_R K_1^\bullet) \to \text{Tot}(L^\bullet \otimes_R K_2^\bullet) \to \text{Tot}(L^\bullet \otimes_R K_3^\bullet) \to 0 $$ is a short exact sequence of complexes. Namely, for each $n, m$ the sequence of modules $0 \to L^n \otimes_R K_1^m \to L^n \otimes_R K_2^m \to L^n \otimes_R K_3^m \to 0$ is exact by Algebra, Lemma \ref{algebra-lemma-flat-tor-zero} and the sequence of complexes is a direct sum of these. Thus the lemma follows from this and the fact that in a short exact sequence of complexes if two out of three are acyclic, so is the third.
Let $L^\bullet$ be an acyclic complex of $R$-modules. Let $\xi \in H^n(\text{Tot}(L^\bullet \otimes_R P^\bullet))$. We have to show that $\xi = 0$. Since $\text{Tot}^n(L^\bullet \otimes_R P^\bullet)$ is a direct sum with terms $L^a \otimes_R P^b$ we see that $\xi$ comes from an element in $H^n(\text{Tot}(\tau_{\leq m}L^\bullet \otimes_R P^\bullet))$ for some $m \in \mathbf{Z}$. Since $\tau_{\leq m}L^\bullet$ is also acyclic we may replace $L^\bullet$ by $\tau_{\leq m}L^\bullet$. Hence we may assume that $L^\bullet$ is bounded above. In this case the spectral sequence of Homology, Lemma \ref{homology-lemma-first-quadrant-ss} has $$ {}'E_1^{p, q} = H^p(L^\bullet \otimes_R P^q) $$ which is zero as $P^q$ is flat and $L^\bullet$ acyclic. Hence $H^*(\text{Tot}(L^\bullet \otimes_R P^\bullet)) = 0$.
Because we are taking termwise colimits we have $$ \colim_i \text{Tot}(M^\bullet \otimes_R K_i^\bullet) = \text{Tot}(M^\bullet \otimes_R \colim_i K_i^\bullet) $$ by Algebra, Lemma \ref{algebra-lemma-tensor-products-commute-with-limits}. Hence the lemma follows from the fact that filtered colimits are exact, see Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}.
We will use repeatedly that tensor product commute with colimits (Algebra, Lemma \ref{algebra-lemma-tensor-products-commute-with-limits}). Thus we see that $K^\bullet \otimes_R M$ is acyclic for any $R$-module $M$, because any $R$-module is a filtered colimit of finitely presented $R$-modules $M$, see Algebra, Lemma \ref{algebra-lemma-module-colimit-fp}. Let $M^\bullet$ be an acyclic complex of $R$-modules. We have to show that $\text{Tot}(M^\bullet \otimes_R K^\bullet)$ is acyclic. Since $M^\bullet = \colim \tau_{\leq n} M^\bullet$ (termwise colimit) we have $$ \text{Tot}(M^\bullet \otimes_R K^\bullet) = \colim \text{Tot}(\tau_{\leq n} M^\bullet \otimes_R K^\bullet) $$ with truncations as in Homology, Section \ref{homology-section-truncations}. As filtered colimits are exact (Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}) we may replace $M^\bullet$ by $\tau_{\leq n}M^\bullet$ and assume that $M^\bullet$ is bounded above. In the bounded above case, we can write $M^\bullet = \colim \sigma_{\geq -n} M^\bullet$ where the complexes $\sigma_{\geq -n} M^\bullet$ are bounded but possibly no longer acyclic. Arguing as above we reduce to the case where $M^\bullet$ is a bounded complex. Finally, for a bounded complex $M^a \to \ldots \to M^b$ we can argue by induction on the length $b - a$ of the complex. The case $b - a = 1$ we have seen above. For $b - a > 1$ we consider the split short exact sequence of complexes $$ 0 \to \sigma_{\geq a + 1}M^\bullet \to M^\bullet \to M^a[-a] \to 0 $$ and we apply Lemma \ref{lemma-derived-tor-exact} to do the induction step. Some details omitted.
Let $\mathcal{P} \subset \Ob(\text{Mod}_R)$ be the class of flat $R$-modules. By Derived Categories, Lemma \ref{derived-lemma-special-direct-system} there exists a system $K_1^\bullet \to K_2^\bullet \to \ldots$ and a diagram $$ \xymatrix{ K_1^\bullet \ar[d] \ar[r] & K_2^\bullet \ar[d] \ar[r] & \ldots \\ \tau_{\leq 1}M^\bullet \ar[r] & \tau_{\leq 2}M^\bullet \ar[r] & \ldots } $$ with the properties (1), (2), (3) listed in that lemma. These properties imply each complex $K_i^\bullet$ is a bounded above complex of flat modules. Hence $K_i^\bullet$ is K-flat by Lemma \ref{lemma-derived-tor-quasi-isomorphism}. The induced map $\colim_i K_i^\bullet \to M^\bullet$ is a quasi-isomorphism and termwise surjective by construction. The complex $\colim_i K_i^\bullet$ is K-flat by Lemma \ref{lemma-colimit-K-flat}. The terms $\colim K_i^n$ are flat because filtered colimits of flat modules are flat, see Algebra, Lemma \ref{algebra-lemma-colimit-flat}.
Choose a quasi-isomorphism $K^\bullet \to L^\bullet$ with $K^\bullet$ a K-flat complex, see Lemma \ref{lemma-K-flat-resolution}. Consider the commutative diagram $$ \xymatrix{ \text{Tot}(K^\bullet \otimes_R P^\bullet) \ar[r] \ar[d] & \text{Tot}(K^\bullet \otimes_R Q^\bullet) \ar[d] \\ \text{Tot}(L^\bullet \otimes_R P^\bullet) \ar[r] & \text{Tot}(L^\bullet \otimes_R Q^\bullet) } $$ The result follows as by Lemma \ref{lemma-K-flat-quasi-isomorphism} the vertical arrows and the top horizontal arrow are quasi-isomorphisms.
We will use that the homotopy category $K(\text{Mod}_R)$ is a triangulated category, see Derived Categories, Proposition \ref{derived-proposition-homotopy-category-triangulated}. Choose a distinguished triangle $K^\bullet \to L^\bullet \to C^\bullet \to K^\bullet[1]$. Choose a quasi-isomorphism $M^\bullet \to C^\bullet$ with $M^\bullet$ K-flat with flat terms, see Lemma \ref{lemma-K-flat-resolution}. By the axioms of triangulated categories, we may fit the composition $M^\bullet \to C^\bullet \to K^\bullet[1]$ into a distinguished triangle $K^\bullet \to N^\bullet \to M^\bullet \to K^\bullet[1]$. By Lemma \ref{lemma-K-flat-two-out-of-three} we see that $N^\bullet$ is K-flat. Again using the axioms of triangulated categories, we can choose a map $N^\bullet \to L^\bullet$ fitting into the following morphism of distinghuised triangles $$ \xymatrix{ K^\bullet \ar[r] \ar[d] & N^\bullet \ar[r] \ar[d] & M^\bullet \ar[r] \ar[d] & K^\bullet[1] \ar[d] \\ K^\bullet \ar[r] & L^\bullet \ar[r] & C^\bullet \ar[r] & K^\bullet[1] } $$ Since two out of three of the arrows are quasi-isomorphisms, so is the third arrow $N^\bullet \to L^\bullet$ by the long exact sequences of cohomology associated to these distinguished triangles (or you can look at the image of this diagram in $D(R)$ and use Derived Categories, Lemma \ref{derived-lemma-third-isomorphism-triangle} if you like). This finishes the proof of (1), (2), and (3). To prove the final assertion, we may choose $N^\bullet$ such that $N^n \cong M^n \oplus K^n$, see Derived Categories, Lemma \ref{derived-lemma-improve-distinguished-triangle-homotopy}. Hence we get the desired flatness if the terms of $K^\bullet$ are flat.