hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-check-injective-on-ass} we see that $\varphi$ is injective. Let $N' \subset N$ be an finitely generated $R$-module containing the image of $M$. Then $\text{Ass}(N_\mathfrak p) = \emptyset$ implies $\text{Ass}(N'_\mathfrak p) = \emptyset$. Hence the assumptions of the lemma hold for $M \to N'$. In order to prove that $\varphi$ is an isomorphism, it suffices to prove the same thing for every such $N' \subset N$. Thus we may assume $N$ is a finite $R$-module. In this case, $\mathfrak p \not \in \text{Ass}(N) \Rightarrow \text{depth}(N_\mathfrak p) \geq 1$, see Algebra, Lemma \ref{algebra-lemma-ideal-nonzerodivisor}. Consider the short exact sequence $$ 0 \to M \to N \to Q \to 0 $$ defining $Q$. Looking at the conditions we see that either $Q_\mathfrak p = 0$ in case (1) or $\text{depth}(Q_\mathfrak p) \geq 1$ in case (2) by Algebra, Lemma \ref{algebra-lemma-depth-in-ses}. This implies that $Q$ does not have any associated primes, hence $Q = 0$ by Algebra, Lemma \ref{algebra-lemma-ass-zero}.
If (1) is true, then $M_\mathfrak p$ is a reflexive module for all primes of $\mathfrak p$ by Lemma \ref{lemma-check-reflexive}. Thus (1) $\Rightarrow$ (2). Assume (2). Set $N = \Hom_R(\Hom_R(M, R), R)$ so that $$ N_\mathfrak p = \Hom_{R_\mathfrak p}( \Hom_{R_\mathfrak p}(M_\mathfrak p, R_\mathfrak p), R_\mathfrak p) $$ for every prime $\mathfrak p$ of $R$. See Algebra, Lemma \ref{algebra-lemma-hom-from-finitely-presented}. We apply Lemma \ref{lemma-isom-depth-2-torsion-free} to the map $j : M \to N$. This is allowed because $M$ is finite and $N$ is torsion free by Lemma \ref{lemma-hom-into-torsion-free}. In case (2)(a) the map $M_\mathfrak p \to N_\mathfrak p$ is an isomorphism and in case (2)(b) we have $\text{depth}(M_\mathfrak p) \geq 2$.
Since formation of reflexive hull $\Hom_R(\Hom_R(M, R), R)$ commutes with localization (Algebra, Lemma \ref{algebra-lemma-hom-from-finitely-presented}) part (1) follows from Lemma \ref{lemma-hom-into-depth}. Part (2) is immediate from Lemma \ref{lemma-hom-into-S2}.
By Algebra, Lemma \ref{algebra-lemma-criterion-normal} we see that $R$ satisfies $(R_1)$ and $(S_2)$. \medskip\noindent Assume (1). Then $M$ is torsion free by Lemma \ref{lemma-reflexive-torsion-free} and satisfies $(S_2)$ by Lemma \ref{lemma-reflexive-S2}. Thus (2) holds. \medskip\noindent Assume (2). By definition $M' = \bigcap_{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ is the kernel of the map $$ M_K \longrightarrow \bigoplus\nolimits_{\text{height}(\mathfrak p) = 1} M_K/M_\mathfrak p \subset \prod\nolimits_{\text{height}(\mathfrak p) = 1} M_K/M_\mathfrak p $$ Observe that our map indeed factors through the direct sum as indicated since given $a/b \in K$ there are at most finitely many height $1$ primes $\mathfrak p$ with $b \in \mathfrak p$. Let $\mathfrak p_0$ be a prime of height $1$. Then $(M_K/M_\mathfrak p)_{\mathfrak p_0} = 0$ unless $\mathfrak p = \mathfrak p_0$ in which case we get $(M_K/M_\mathfrak p)_{\mathfrak p_0} = M_K/M_{\mathfrak p_0}$. Thus by exactness of localization and the fact that localization commutes with direct sums, we see that $M'_{\mathfrak p_0} = M_{\mathfrak p_0}$. Since $M$ has depth $\geq 2$ at primes of height $> 1$, we see that $M \to M'$ is an isomorphism by Lemma \ref{lemma-isom-depth-2-torsion-free}. Hence (3) holds. \medskip\noindent Assume (3). Let $\mathfrak p$ be a prime of height $1$. Then $R_\mathfrak p$ is a discrete valuation ring by $(R_1)$. By Lemma \ref{lemma-dedekind-torsion-free-flat} we see that $M_\mathfrak p$ is finite free, in particular reflexive. Hence the map $M \to M^{**}$ induces an isomorphism at all the primes $\mathfrak p$ of height $1$. Thus the condition $M = \bigcap_{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ implies that $M = M^{**}$ and (1) holds.
Let $\mathfrak p$ be a prime of height $1$. The kernel of $M_\mathfrak p \to M \otimes_R K$ is the torsion submodule $(M_\mathfrak p)_{tors}$ of $M_\mathfrak p$. Moreover, we have $(M/M_{tors})_\mathfrak p = M_\mathfrak p/(M_\mathfrak p)_{tors}$ and this is a finite free module over the discrete valuation ring $R_\mathfrak p$ (Lemma \ref{lemma-dedekind-torsion-free-flat}). Then $M_\mathfrak p/(M_\mathfrak p)_{tors} \to (M_\mathfrak p)^{**} = (M^{**})_\mathfrak p$ is an isomorphism, hence the lemma is a consequence of Lemma \ref{lemma-reflexive-over-normal}.
It suffices to show that $B = \bigcap B_\mathfrak p$ where the intersection is over height $1$ primes $\mathfrak p \subset A$, see Lemma \ref{lemma-reflexive-over-normal}. Let $b \in \bigcap B_\mathfrak p$. Let $x^d + a_1x^{d - 1} + \ldots + a_d$ be the minimal polynomial of $b$ over $K$. We want to show $a_i \in A$. By Algebra, Lemma \ref{algebra-lemma-minimal-polynomial-normal-domain} we see that $a_i \in A_\mathfrak p$ for all $i$ and all height one primes $\mathfrak p$. Hence we get what we want from Algebra, Lemma \ref{algebra-lemma-normal-domain-intersection-localizations-height-1} (or the lemma already cited as $A$ is a reflexive module over itself).
It is clear that $u(x) \in IN$. If $u(x) \in I'N$, then $u(x) \in (I' \cap I)N$, see discussion following Definition \ref{definition-content-ideal}. Hence it suffices to show: if $x \in I'N$ and $I' \subset I$, $I' \not =  I$, then $u(x) \not \in I'N$. Since $I/I'$ is a nonzero finite $A$-module (Lemma \ref{lemma-content-finitely-generated}) there is a nonzero map $\chi : I/I' \to A/\mathfrak m$ of $A$-modules by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}). Since $I$ is the content ideal of $x$ we see that $x \not \in I''M$ where $I'' = \Ker(\chi)$. Hence $x$ is not in the kernel of the map $$ IM = I \otimes_A M \xrightarrow{\chi \otimes 1} A/\mathfrak m \otimes M \cong M/\mathfrak m M $$ Applying our hypothesis on $\overline{u}$ we conclude that $u(x)$ does not map to zero under the map $$ IN = I \otimes_A N \xrightarrow{\chi \otimes 1} A/\mathfrak m \otimes N \cong N/\mathfrak m N $$ and we conclude.
Choose a presentation $$ 0 \to K \to S^{\oplus r} \to M \to 0 $$ of $M$ as an $S$-module. Let $\mathfrak q$ be a prime ideal of $S$ lying over a prime $\mathfrak p$ of $R$. By assumption there exist finitely many elements $k_1, \ldots, k_t \in K$ such that if we set $K' = \sum Sk_j \subset K$ then $K'_{\mathfrak p} = K_{\mathfrak p}$ and $K'_{\mathfrak p_j} = K_{\mathfrak p_j}$ for $j = 1, \ldots, m$. Setting $M' = S^{\oplus r}/K'$ we deduce that in particular $M'_{\mathfrak q} = M_{\mathfrak q}$. By openness of flatness, see Algebra, Theorem \ref{algebra-theorem-openness-flatness} we conclude that there exists a $g \in S$, $g \not \in \mathfrak q$ such that $M'_g$ is flat over $R$. Thus $M'_g \to M_g$ is a surjective map of flat $R$-modules. Consider the commutative diagram $$ \xymatrix{ M'_g \ar[r] \ar[d] & M_g \ar[d] \\ \prod (M'_g)_{\mathfrak p_j} \ar[r] & \prod (M_g)_{\mathfrak p_j} } $$ The bottom arrow is an isomorphism by choice of $k_1, \ldots, k_t$. The left vertical arrow is an injective map as $R \to \prod R_{\mathfrak p_j}$ is injective and $M'_g$ is flat over $R$. Hence the top horizontal arrow is injective, hence an isomorphism. This proves that $M_g$ is of finite presentation over $S_g$. We conclude by applying Algebra, Lemma \ref{algebra-lemma-cover}.
Let $M = \bigoplus M_d$ be the grading on $M$. Pick homogeneous generators $m_1, \ldots, m_r \in M$ of $M$. Say $\deg(m_i) = d_i \in \mathbf{Z}$. This gives us a presentation $$ 0 \to K \to \bigoplus\nolimits_{i = 1, \ldots, r} S(-d_i) \to M \to 0 $$ which in each degree $d$ leads to the short exact sequence $$ 0 \to K_d \to \bigoplus\nolimits_{i = 1, \ldots, r} S_{d - d_i} \to M_d \to 0. $$ By assumption each $M_d$ is a finite flat $R$-module. By Algebra, Lemma \ref{algebra-lemma-finite-flat-local} this implies each $M_d$ is a finite free $R$-module. Hence we see each $K_d$ is a finite $R$-module. Also each $K_d$ is flat over $R$ by Algebra, Lemma \ref{algebra-lemma-flat-ses}. Hence we conclude that each $K_d$ is finite free by Algebra, Lemma \ref{algebra-lemma-finite-flat-local} again. \medskip\noindent Let $\mathfrak m$ be the maximal ideal of $R$. By the flatness of $M$ over $R$ the short exact sequences above remain short exact after tensoring with $\kappa = \kappa(\mathfrak m)$. As the ring $S \otimes_R \kappa$ is Noetherian we see that there exist homogeneous elements $k_1, \ldots, k_t \in K$ such that the images $\overline{k}_j$ generate $K \otimes_R \kappa$ over $S \otimes_R \kappa$. Say $\deg(k_j) = e_j$. Thus for any $d$ the map $$ \bigoplus\nolimits_{j = 1, \ldots, t} S_{d - e_j} \longrightarrow K_d $$ becomes surjective after tensoring with $\kappa$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) this implies the map is surjective over $R$. Hence $K$ is generated by $k_1, \ldots, k_t$ over $S$ and we win.
As $S$ is finitely generated as an $R$-algebra, it is finitely generated as an $S_0$ algebra, say by homogeneous elements $t_1, \ldots, t_n \in S$ of degrees $d_1, \ldots, d_n > 0$. Set $P = R[x_1, \ldots, x_n]$ with $\deg(x_i) = d_i$. The ring map $P \to S$, $x_i \to t_i$ is finite as $S_0$ is a finite $R$-module. To prove (1) it suffices to prove that $S$ is a finitely presented $P$-module.  To prove (2) it suffices to prove that $M$ is a finitely presented $P$-module. Thus it suffices to prove that if $S = P$ is a graded polynomial ring and $M$ is a finite $S$-module flat over $R$, then $M$ is finitely presented as an $S$-module. By Lemma \ref{lemma-flat-graded-finite-type-finite-presentation-module} we see $M_{\mathfrak p}$ is a finitely presented $S_{\mathfrak p}$-module for every prime $\mathfrak p$ of $R$. Thus the result follows from Lemma \ref{lemma-flat-finite-type-finite-presentation-local-module}.
We are going to use that an $A$-module is flat if and only if it is torsion free, see Lemma \ref{lemma-valuation-ring-torsion-free-flat}. By Algebra, Lemma \ref{algebra-lemma-homogenize} we can find a graded $A$-algebra $S$ with $S_0 = A$ and generated by finitely many elements in degree $1$, an element $f \in S_1$ and a finite graded $S$-module $N$ such that $B \cong S_{(f)}$ and $M \cong N_{(f)}$. If $M$ is torsion free, then we can take $N$ torsion free by replacing it by $N/N_{tors}$, see Lemma \ref{lemma-torsion}. Similarly, if $B$ is torsion free, then we can take $S$ torsion free by replacing it by $S/S_{tors}$. Hence in case (1), we may apply Lemma \ref{lemma-flat-graded-finite-type-finite-presentation} to see that $S$ is a finitely presented $A$-algebra, which implies that $B = S_{(f)}$ is a finitely presented $A$-algebra. To see (2) we may first replace $S$ by a graded polynomial ring, and then we may apply Lemma \ref{lemma-flat-graded-finite-type-finite-presentation-module} to conclude.
First observe that $\text{Fit}_k(M \otimes_R R') = IR' = aR'$. The first equality by Lemma \ref{lemma-fitting-ideal-basics} part (3) and the second equality by Algebra, Lemma \ref{algebra-lemma-affine-blowup}. From Lemma \ref{lemma-principal-fitting-ideal} and exactness of localization we see that $M'_{\mathfrak p'}$ can be generated by $\leq k$ elements for every prime $\mathfrak p'$ of $R'$. Then $\text{Fit}_k(M') = R'$ for example by Lemma \ref{lemma-fitting-ideal-generate-locally}.
By Lemma \ref{lemma-blowup-fitting-ideal} we have $\text{Fit}_k(M') = R'$. By Lemma \ref{lemma-fitting-ideal-finite-locally-free} it suffices to show that $\text{Fit}_{k - 1}(M') = 0$. Recall that $R' \subset R'_a = R_a$, see Algebra, Lemma \ref{algebra-lemma-affine-blowup}. Hence it suffices to prove that $\text{Fit}_{k - 1}(M')$ maps to zero in $R'_a = R_a$. Since clearly $(M')_a = M_a$ this reduces us to showing that $\text{Fit}_{k - 1}(M_a) = 0$ because formation of Fitting ideals commutes with base change according to Lemma \ref{lemma-fitting-ideal-basics} part (3). This is true by our assumption that $M_a$ is finite locally free of rank $k$ (see Algebra, Lemma \ref{algebra-lemma-finite-projective}) and the already cited Lemma \ref{lemma-fitting-ideal-finite-locally-free}.
Choose a surjection $R^{\oplus n} \to M$. Choose a finite submodule $K \subset \Ker(R^{\oplus n} \to M)$ such that $R^{\oplus n}/K \to M$ becomes an isomorphism after inverting $f$. This is possible because $M_f$ is of finite presentation for example by Algebra, Lemma \ref{algebra-lemma-finite-projective}. Set $M_1 = R^{\oplus n}/K$ and suppose we can prove the lemma for $M_1$. Say $I \subset R$ is the corresponding ideal. Then for $a \in I$ the map $$ M_1' = (M_1 \otimes_R R')/a\text{-power torsion} \longrightarrow M' = (M \otimes_R R')/a\text{-power torsion} $$ is surjective. It is also an isomorphism after inverting $a$ in $R'$ as $R'_a = R_f$, see Algebra, Lemma \ref{algebra-lemma-blowup-in-principal}. But $a$ is a nonzerodivisor on $M'_1$, whence the displayed map is an isomorphism. Thus it suffices to prove the lemma in case $M$ is a finitely presented $R$-module. \medskip\noindent Assume $M$ is a finitely presented $R$-module. Then $J = \text{Fit}_r(M) \subset S$ is a finitely generated ideal. We claim that $I = fJ$ works. \medskip\noindent We first check that $V(f) = V(I)$. The inclusion $V(f) \subset V(I)$ is clear. Conversely, if $f \not \in \mathfrak p$, then $\mathfrak p$ is not an element of $V(J)$ by  Lemma \ref{lemma-fitting-ideal-generate-locally}. Thus $\mathfrak p \not \in V(fJ) = V(I)$. \medskip\noindent Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fb$ for some $b \in J$. By Algebra, Lemmas \ref{algebra-lemma-affine-blowup} and \ref{algebra-lemma-blowup-add-principal} we see that $J R' = b R'$ and $b$ is a nonzerodivisor in $R'$.  Let $\mathfrak p' \subset R' = R[\frac{I}{a}]$ be a prime ideal. Then $JR'_{\mathfrak p'}$ is generated by $b$. It follows from Lemma \ref{lemma-principal-fitting-ideal} that $M'_{\mathfrak p'}$ can be generated by $r$ elements. Since $M'$ is finite, there exist $m_1, \ldots, m_r \in M'$ and $g \in R'$, $g \not \in \mathfrak p'$ such that the corresponding map $(R')^{\oplus r} \to M'$ becomes surjective after inverting $g$. \medskip\noindent Finally, consider the ideal $J' = \text{Fit}_{k - 1}(M')$. Note that $J' R'_g$ is generated by the coefficients of relations between $m_1, \ldots, m_r$ (compatibility of Fitting ideal with base change). Thus it suffices to show that $J' = 0$, see Lemma \ref{lemma-fitting-ideal-finite-locally-free}. Since $R'_a = R_f$ (Algebra, Lemma \ref{algebra-lemma-blowup-in-principal}) and $M'_a = M_f$ is free of rank $r$ we see that $J'_a = 0$. Since $a$ is a nonzerodivisor in $R'$ we conclude that $J' = 0$ and we win.
Denote $R^\wedge$ the completion of $R$ with respect to $I$. As $R \to R^\wedge$ is flat by Algebra, Lemma \ref{algebra-lemma-completion-flat} it suffices to prove that $(\bigoplus\nolimits_{\alpha \in A} R)^\wedge$ is a flat $R^\wedge$-module (use Algebra, Lemma \ref{algebra-lemma-composition-flat}). Since $$ (\bigoplus\nolimits_{\alpha \in A} R)^\wedge = (\bigoplus\nolimits_{\alpha \in A} R^\wedge)^\wedge $$ we may replace $R$ by $R^\wedge$ and assume that $R$ is complete with respect to $I$ (see Algebra, Lemma \ref{algebra-lemma-completion-complete}). In this case Lemma \ref{lemma-ui-completion-direct-sum-into-product} tells us the map $(\bigoplus\nolimits_{\alpha \in A} R)^\wedge \to \prod_{\alpha \in A} R$ is universally injective. Thus, by Algebra, Lemma \ref{algebra-lemma-ui-flat-domain} it suffices to show that $\prod_{\alpha \in A} R$ is flat. By Algebra, Proposition \ref{algebra-proposition-characterize-coherent} (and Algebra, Lemma \ref{algebra-lemma-Noetherian-coherent}) we see that $\prod_{\alpha \in A} R$ is flat.
Proof for $p = 1$. Choose a short exact sequence $0 \to K \to R^{\oplus t} \to M \to 0$. Then $\text{Tor}_1^A(M, A/I^n) = K \cap (I^n)^{\oplus t}/I^nK$. By Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees}) there is a constant $c \geq 0$ such that $K \cap (I^n)^{\oplus t} \subset I^{n - c}K$ for $n \geq c$. Thus the result for $p = 1$. For $p > 1$ we have $\text{Tor}_p^A(M, A/I^n) = \text{Tor}^A_{p - 1}(K, A/I^n)$. Thus the lemma follows by induction.
We first show that $Q \otimes_A M = \lim Q \otimes_A M_n$ for every finite $A$-module $Q$. Choose a resolution $F_2 \to F_1 \to F_0 \to Q \to 0$ by finite free $A$-modules $F_i$. Then $$ F_2 \otimes_A M_n \to F_1 \otimes_A M_n \to F_0 \otimes_A M_n $$ is a chain complex whose homology in degree $0$ is $Q \otimes_A M_n$ and whose homology in degree $1$ is $$ \text{Tor}_1^A(Q, M_n) = \text{Tor}_1^A(Q, A/I^n) \otimes_{A/I^n} M_n $$ as $M_n$ is flat over $A/I^n$. By Lemma \ref{lemma-tor-strictly-pro-zero} we see that this system is essentially constant (with value $0$). It follows from Homology, Lemma \ref{homology-lemma-apply-Mittag-Leffler-again} that $\lim Q \otimes_A A/I^n = \Coker(\lim F_1 \otimes_A M_n \to \lim F_0 \otimes_A M_n)$. Since $F_i$ is finite free this equals $\Coker(F_1 \otimes_A M \to F_0 \otimes_A M) = Q \otimes_A M$. \medskip\noindent Next, let $Q \to Q'$ be an injective map of finite $A$-modules. We have to show that $Q \otimes_A M \to Q' \otimes_A M$ is injective (Algebra, Lemma \ref{algebra-lemma-flat}). By the above we see $$ \Ker(Q \otimes_A M \to Q' \otimes_A M) = \Ker(\lim Q \otimes_A M_n \to \lim Q' \otimes_A M_n). $$ For each $n$ we have an exact sequence $$ \text{Tor}_1^A(Q', M_n) \to \text{Tor}_1^A(Q'', M_n) \to Q \otimes_A M_n \to Q' \otimes_A M_n $$ where $Q'' = \Coker(Q \to Q')$. Above we have seen that the inverse systems of Tor's are essentially constant with value $0$. It follows from Homology, Lemma \ref{homology-lemma-apply-Mittag-Leffler-again} that the inverse limit of the right most maps is injective.
By Algebra, Lemma \ref{algebra-lemma-what-does-it-mean} the modules $M/I^nM$ are flat over $R/I^n$ for all $n$. By Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated} we have (a) $R^\wedge$ and $M^\wedge$ are $I$-adically complete and (b) $R/I^n = R^\wedge/I^nR^\wedge$ for all $n$. By Algebra, Lemma \ref{algebra-lemma-completion-Noetherian} the ring $R^\wedge$ is Noetherian. Applying Lemma \ref{lemma-limit-flat} we conclude that $M^\wedge = \lim M/I^nM$ is flat as an $R^\wedge$-module.
Set $g_i = \sum x_{ij}f_j$. The matrix $(x_{ji})$ gives an isomorphism $x : R^{\oplus r} \to R^{\oplus r}$ such that $(g_1, \ldots, g_r) = (f_1, \ldots, f_r) \circ x$. Hence this follows from the functoriality of the Koszul complex described in Lemma \ref{lemma-functorial}.
Part (1) follows from the fact that the extended alternating {\v C}ech complex is zero in degrees $< 0$ and $> r$. To prove (2) it suffices to show that for each $i$ there exists an $n \geq 1$ such that $f_i^n x = 0$. To see this it suffices to show that $(H^q)_{f_i} = 0$. Since localization is exact, $(H^q)_{f_i}$ is the $q$th cohomology module of the localization of the extended alternating complex of $M$ at $f_i$. By Lemma \ref{lemma-extended-alternating-base-change} this localization is the extended alternating {\v C}ech complex of $M_{f_i}$ over $R_{f_i}$ with respect to the images of $f_1, \ldots, f_r$ in $R_{f_i}$. Thus we reduce to showing that $H^q$ is zero if $f_i$ is invertible, which follows from Lemma \ref{lemma-extended-alternating-homotopy-zero}. Part (3) follows from the observation that $(H^q)_{f_i} = 0$ for all $i$ that we just proved. To see part (4) note that in this case $f$ acts invertibly on $H^q$ and $H^q$ is supported on $V(f)$ by (3). This forces $H^q$ to be zero (small detail omitted).
We urge the reader to prove this for themselves. Denote $K(R, f_1^n, \ldots, f_r^n)$ the Koszul complex of Definition \ref{definition-koszul-complex} viewed as a cochain complex sitting in degrees $0, \ldots, r$. Thus we have $$ K(R, f_1^n, \ldots, f_r^n) : 0 \to \wedge^r(R^{\oplus r}) \to \wedge^{r - 1}(R^{\oplus r}) \to \ldots \to R^{\oplus r} \to R \to 0 $$ with the term $\wedge^r(R^{\oplus r})$ sitting in degree $0$. Let $e^n_1, \ldots, e^n_r$ be the standard basis of $R^{\oplus r}$. Then the elements $e^n_{j_1} \wedge \ldots \wedge e^n_{j_{r - p}}$ for $1 \leq j_1 < \ldots < j_{r - p} \leq r$ form a basis for the term in degree $p$ of the Koszul complex. Further, observe that $$ d(e^n_{j_1} \wedge \ldots \wedge e^n_{j_{r - p}}) = \sum (-1)^{a + 1} f_{j_a}^n e^n_{j_1} \wedge \ldots \wedge \hat e^n_{j_a} \wedge \ldots \wedge e^n_{j_{r - p}} $$ by our construction of the Koszul complex in Section \ref{section-koszul}. The transition maps of our system $$ K(R, f_1^n, \ldots, f_r^n) \to K(R, f_1^{n + 1}, \ldots, f_r^{n + 1}) $$ are given by the rule $$ e^n_{j_1} \wedge \ldots \wedge e^n_{j_{r - p}} \longmapsto f_{i_0} \ldots f_{i_{p - 1}} e^{n + 1}_{j_1} \wedge \ldots \wedge e^{n + 1}_{j_{r - p}} $$ where the indices $1 \leq i_0 < \ldots < i_{p - 1} \leq r$ are such that $\{1, \ldots r\} = \{i_0, \ldots, i_{p - 1}\} \amalg \{j_1, \ldots, j_{r - p}\}$. We omit the short computation that shows this is compatible with differentials. Observe that the transition maps are always $1$ in degree $0$ and equal to $f_1 \ldots f_r$ in degree $r$. \medskip\noindent Denote $K^p(R, f_1^n, \ldots, f_r^n)$ the term of degree $p$ in the Koszul complex. Observe that for any $f \in R$ we have $$ R_f = \colim (R \xrightarrow{f} R \xrightarrow{f} R \to \ldots ) $$ Hence we see that in degree $p$ we obtain $$ \colim K^p(R, f_1^n, \ldots f_r^n) = \bigoplus\nolimits_{1 \leq i_0 < \ldots < i_{p - 1} \leq r} R_{f_{i_0} \ldots f_{i_{p - 1}}} $$ Here the element $e^n_{j_1} \wedge \ldots \wedge e^n_{j_{r - p}}$ of the Koszul complex above maps in the colimit to the element $(f_{i_0} \ldots f_{i_{p - 1}})^{-n}$ in the summand $R_{f_{i_0} \ldots f_{i_{p - 1}}}$ where the indices are chosen such that $\{1, \ldots r\} = \{i_0, \ldots, i_{p - 1}\} \amalg \{j_1, \ldots, j_{r - p - 2}\}$. Thus the differential on this complex is given by $$ d(1\text{ in }R_{f_{i_0} \ldots f_{i_{p - 1}}}) = \sum\nolimits_{i \not \in \{i_0, \ldots, i_{p - 1}\}} (-1)^{i - t}\text{ in } R_{f_{i_0} \ldots f_{i_t} f_i f_{i_{t + 1}} \ldots f_{i_{p - 1}}} $$ Thus if we consider the map of complexes given in degree $p$ by the map $$ \bigoplus\nolimits_{1 \leq i_0 < \ldots < i_{p - 1} \leq r} R_{f_{i_0} \ldots f_{i_{p - 1}}} \longrightarrow \bigoplus\nolimits_{1 \leq i_0 < \ldots < i_{p - 1} \leq r} R_{f_{i_0} \ldots f_{i_{p - 1}}} $$ determined by the rule $$ 1\text{ in }R_{f_{i_0} \ldots f_{i_{p - 1}}} \longmapsto (-1)^{i_0 + \ldots + i_{p - 1} + p}\text{ in }R_{f_{i_0} \ldots f_{i_{p - 1}}} $$ then we get an isomorphism of complexes from $\colim K(R, f_1^n, \ldots, f_r^n)$ to the extended alternating {\v C}ech complex defined in this section. We omit the verification that the signs work out.
Let $R$ be a ring and let $M$ be an $R$-module. It is immediate that an $M$-regular sequence of length $1$ is $M$-Koszul-regular. Let $f_1, \ldots, f_r$ be an $M$-regular sequence. Then $f_1$ is a nonzerodivisor on $M$. Hence $$ 0 \to K_\bullet(f_2, \ldots, f_r) \otimes M \xrightarrow{f_1} K_\bullet(f_2, \ldots, f_r) \otimes M \to K_\bullet(\overline{f}_2, \ldots, \overline{f}_r) \otimes M/f_1M \to 0 $$ is a short exact sequence of complexes where $\overline{f}_i$ is the image of $f_i$ in $R/(f_1)$. By Lemma \ref{lemma-cone-koszul} the complex $K_\bullet(R, f_1, \ldots, f_r)$ is isomorphic to the cone of multiplication by $f_1$ on $K_\bullet(f_2, \ldots, f_r)$. Thus $K_\bullet(R, f_1, \ldots, f_r) \otimes M$ is isomorphic to the cone on the first map. Hence $K_\bullet(\overline{f}_2, \ldots, \overline{f}_r) \otimes M/f_1M$ is quasi-isomorphic to $K_\bullet(f_1, \ldots, f_r) \otimes M$. As $\overline{f}_2, \ldots, \overline{f}_r$ is an $M/f_1M$-regular sequence in $R/(f_1)$ the result follows from the case $r = 1$ and induction.
By Lemma \ref{lemma-koszul-mult} we have exact sequences $$ H_i(K_\bullet(f_1, \ldots, f_{r - 1}, f) \otimes M) \to H_i(K_\bullet(f_1, \ldots, f_{r - 1}, fg) \otimes M) \to H_i(K_\bullet(f_1, \ldots, f_{r - 1}, g) \otimes M) $$ for all $i$.