hcju/stacksqa huggingface dataset:
The first statement follows because $\text{Supp}(M) = \bigcup V(\mathfrak p_i)$, see Lemma \ref{lemma-filter-primes-in-support}. Let $\mathfrak p \in \text{Supp}(M)$ be minimal. The support of $M_{\mathfrak p}$ is the set consisting of the maximal ideal $\mathfrak p R_{\mathfrak p}$. Hence by Lemma \ref{lemma-support-point} the length of $M_{\mathfrak p}$ is finite and $> 0$. Next we note that $M_{\mathfrak p}$ has a filtration with subquotients $ (R/\mathfrak p_i)_{\mathfrak p} = R_{\mathfrak p}/{\mathfrak p_i}R_{\mathfrak p} $. These are zero if $\mathfrak p_i \not \subset \mathfrak p$ and equal to $\kappa(\mathfrak p)$ if $\mathfrak p_i \subset \mathfrak p$ because by minimality of $\mathfrak p$ we have $\mathfrak p_i = \mathfrak p$ in this case. The result follows since $\kappa(\mathfrak p)$ has length $1$.
Let $M_i, \mathfrak p_i$ be as in Lemma \ref{lemma-filter-Noetherian-module}. By Lemma \ref{lemma-hilbert-ses-chi} we obtain the equality $d(M) = \max \{ d(R/\mathfrak p_i) \}$. By Proposition \ref{proposition-dimension} we have $d(R/\mathfrak p_i) = \dim(R/\mathfrak p_i)$. Trivially $\dim(R/\mathfrak p_i) = \dim V(\mathfrak p_i)$. Since all minimal primes of $\text{Supp}(M)$ occur among the $\mathfrak p_i$ (Lemma \ref{lemma-filter-minimal-primes-in-support}) we win.
If $R$ is local, this follows immediately from Lemmas \ref{lemma-support-dimension-d} and \ref{lemma-hilbert-ses-chi}. A more elementary argument, which works also if $R$ is not local, is to use that $\text{Supp}(M')$, $\text{Supp}(M'')$, and $\text{Supp}(M)$ are closed (Lemma \ref{lemma-support-closed}) and that $\text{Supp}(M) = \text{Supp}(M') \cup \text{Supp}(M'')$ (Lemma \ref{lemma-support-quotient}).
If $M = (0)$, then $\text{Ass}(M) = \emptyset$ by definition. If $M \not = 0$, pick any nonzero finitely generated submodule $M' \subset M$, for example a submodule generated by a single nonzero element. By Lemma \ref{lemma-support-zero} we see that $\text{Supp}(M')$ is nonempty. By Proposition \ref{proposition-minimal-primes-associated-primes} this implies that $\text{Ass}(M')$ is nonempty. By Lemma \ref{lemma-ass} this implies $\text{Ass}(M) \not = \emptyset$.
If $M$ is a finite $R$-module, then this is a consequence of Proposition \ref{proposition-minimal-primes-associated-primes}. In general write $M = \bigcup M_\lambda$ as the union of its finite submodules, and use that $\text{Supp}(M) = \bigcup \text{Supp}(M_\lambda)$ and $\text{Ass}(M) = \bigcup \text{Ass}(M_\lambda)$.
Any element in any associated prime clearly is a zerodivisor on $M$. Conversely, suppose $x \in R$ is a zerodivisor on $M$. Consider the submodule $N = \{m \in M \mid xm = 0\}$. Since $N$ is not zero it has an associated prime $\mathfrak q$ by Lemma \ref{lemma-ass-zero}. Then $x \in \mathfrak q$ and $\mathfrak q$ is an associated prime of $M$ by Lemma \ref{lemma-ass}.
(The parenthetical statement follows from Lemma \ref{lemma-ass-zero-divisors}.) The first inequality follows from $\text{Supp}(M/fM) \subset \text{Supp}(M)$, see Lemma \ref{lemma-support-quotient}. For the second inequality, note that $\text{Supp}(M/fM) = \text{Supp}(M) \cap V(f)$, see Lemma \ref{lemma-support-quotient}. It follows, for example by Lemma \ref{lemma-filter-primes-in-support} and elementary properties of dimension, that it suffices to show $\dim V(\mathfrak p) \leq \dim (V(\mathfrak p) \cap V(f)) + 1$ for primes $\mathfrak p$ of $R$. This is a consequence of Lemma \ref{lemma-one-equation}. Finally, if $f$ is not contained in any minimal prime of the support of $M$, then the chains of primes in $\text{Supp}(M/fM)$ all give rise to chains in $\text{Supp}(M)$ which are at least one step away from being maximal.
We have already seen in Lemma \ref{lemma-ass-functorial} that $\Spec(\varphi)(\text{Ass}_S(M)) \subset \text{Ass}_R(M)$. For the converse, choose a prime $\mathfrak p \in \text{Ass}_R(M)$. Let $m \in M$ be an element such that the annihilator of $m$ in $R$ is $\mathfrak p$. Let $I = \{g \in S \mid gm = 0\}$ be the annihilator of $m$ in $S$. Then $R/\mathfrak p \subset S/I$ is injective. Combining Lemmas \ref{lemma-injective-minimal-primes-in-image} and \ref{lemma-minimal-prime-image-minimal-prime} we see that there is a prime $\mathfrak q \subset S$ minimal over $I$ mapping to $\mathfrak p$. By Proposition \ref{proposition-minimal-primes-associated-primes} we see that $\mathfrak q$ is an associated prime of $S/I$, hence $\mathfrak q$ is an associated prime of $M$ by Lemma \ref{lemma-ass} and we win.
If $\mathfrak p \in \text{Ass}(M)$ there exists an element $m \in M$ whose annihilator is $\mathfrak p$. As localization is exact (Proposition \ref{proposition-localization-exact}) we see that the annihilator of $m/1$ in $M_{\mathfrak p}$ is $\mathfrak pR_{\mathfrak p}$ hence (1) holds. Assume $\mathfrak pR_{\mathfrak p} \in \text{Ass}(M_{\mathfrak p})$ and $\mathfrak p = (f_1, \ldots, f_n)$. Let $m/g$ be an element of $M_{\mathfrak p}$ whose annihilator is $\mathfrak pR_{\mathfrak p}$. This implies that the annihilator of $m$ is contained in $\mathfrak p$. As $f_i m/g = 0$ in $M_{\mathfrak p}$ we see there exists a $g_i \in R$, $g_i \not \in \mathfrak p$ such that $g_i f_i m = 0$ in $M$. Combined we see the annihilator of $g_1\ldots g_nm$ is $\mathfrak p$. Hence $\mathfrak p \in \text{Ass}(M)$.
The first equality follows, since if $m \in S^{-1}M$, then the annihilator of $m$ in $R$ is the intersection of the annihilator of $m$ in $S^{-1}R$ with $R$. The displayed inclusion and equality in the Noetherian case follows from Lemma \ref{lemma-associated-primes-localize} since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset$ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$.
As $M \subset S^{-1}M$ by assumption we get the inclusion $\text{Ass}(M) \subset \text{Ass}(S^{-1}M)$ from Lemma \ref{lemma-ass}. Conversely, suppose that $n/s \in S^{-1}M$ is an element whose annihilator is a prime ideal $\mathfrak p$. Then the annihilator of $n \in M$ is also $\mathfrak p$.
If there exists a nonzerodivisor $x$ in $I$, then $x$ clearly cannot be in any associated prime of $M$. Conversely, suppose $I \not \subset \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$. In this case we can choose $x \in I$, $x \not \in \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$ by Lemmas \ref{lemma-finite-ass} and \ref{lemma-silly}. By Lemma \ref{lemma-ass-zero-divisors} the element $x$ is not a zerodivisor on $M$.
Let $x \in M$ be an element of the kernel of the map. Then if $\mathfrak p$ is an associated prime of $Rx \subset M$ we see on the one hand that $\mathfrak p \in \text{Ass}(M)$ (Lemma \ref{lemma-ass}) and on the other hand that $(Rx)_{\mathfrak p} \subset M_{\mathfrak p}$ is not zero. This contradiction shows that $\text{Ass}(Rx) = \emptyset$. Hence $Rx = 0$ by Lemma \ref{lemma-ass-zero}.
If $\mathfrak p \in \text{Ass}_R(M)$ then there exists an injection $R/\mathfrak p \to M$. As $N$ is flat over $R$ we obtain an injection $R/\mathfrak p \otimes_R N \to M \otimes_R N$. Since $R/\mathfrak p \otimes_R N = N/\mathfrak pN$ we conclude that $\text{Ass}_S(N/\mathfrak pN) \subset \text{Ass}_S(M \otimes_R N)$, see Lemma \ref{lemma-ass}. Hence the right hand side is contained in the left hand side. \medskip\noindent Write $M = \bigcup M_\lambda$ as the union of its finitely generated $R$-submodules. Then also $N \otimes_R M = \bigcup N \otimes_R M_\lambda$ (as $N$ is $R$-flat). By definition of associated primes we see that $\text{Ass}_S(N \otimes_R M) = \bigcup \text{Ass}_S(N \otimes_R M_\lambda)$ and $\text{Ass}_R(M) = \bigcup \text{Ass}(M_\lambda)$. Hence we may assume $M$ is finitely generated. \medskip\noindent Let $\mathfrak q \in \text{Ass}_S(M \otimes_R N)$, and assume $R$ is Noetherian and $M$ is a finite $R$-module. To finish the proof we have to show that $\mathfrak q$ is an element of the right hand side. First we observe that $\mathfrak qS_{\mathfrak q} \in \text{Ass}_{S_{\mathfrak q}}((M \otimes_R N)_{\mathfrak q})$, see Lemma \ref{lemma-associated-primes-localize}. Let $\mathfrak p$ be the corresponding prime of $R$. Note that $$ (M \otimes_R N)_{\mathfrak q} = M \otimes_R N_{\mathfrak q} = M_{\mathfrak p} \otimes_{R_{\mathfrak p}} N_{\mathfrak q} $$ If $\mathfrak pR_{\mathfrak p} \not \in \text{Ass}_{R_{\mathfrak p}}(M_{\mathfrak p})$ then there exists an element $x \in \mathfrak pR_{\mathfrak p}$ which is a nonzerodivisor in $M_{\mathfrak p}$ (see Lemma \ref{lemma-ideal-nonzerodivisor}). Since $N_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ we see that the image of $x$ in $\mathfrak qS_{\mathfrak q}$ is a nonzerodivisor on $(M \otimes_R N)_{\mathfrak q}$. This is a contradiction with the assumption that $\mathfrak qS_{\mathfrak q} \in \text{Ass}_S((M \otimes_R N)_{\mathfrak q})$. Hence we conclude that $\mathfrak p$ is one of the associated primes of $M$. \medskip\noindent Continuing the argument we choose a filtration $$ 0 = M_0 \subset M_1 \subset \ldots \subset M_n = M $$ such that each quotient $M_i/M_{i-1}$ is isomorphic to $R/\mathfrak p_i$ for some prime ideal $\mathfrak p_i$ of $R$, see Lemma \ref{lemma-filter-Noetherian-module}. (By Lemma \ref{lemma-ass-filter} we have $\mathfrak p_i = \mathfrak p$ for at least one $i$.) This gives a filtration $$ 0 = M_0 \otimes_R N \subset M_1 \otimes_R N \subset \ldots \subset M_n \otimes_R N = M \otimes_R N $$ with subquotients isomorphic to $N/\mathfrak p_iN$. If $\mathfrak p_i \not = \mathfrak p$ then $\mathfrak q$ cannot be associated to the module $N/\mathfrak p_iN$ by the result of the preceding paragraph (as $\text{Ass}_R(R/\mathfrak p_i) = \{\mathfrak p_i\}$). Hence we conclude that $\mathfrak q$ is associated to $N/\mathfrak pN$ as desired.
Note that $S \otimes_R K = (R \setminus \{0\})^{-1}S$ and $N \otimes_R K = (R \setminus \{0\})^{-1}N$. For any nonzero $x \in R$ multiplication by $x$ on $N$ is injective as $N$ is flat over $R$. Hence the lemma follows from Lemma \ref{lemma-localize-ass-nonzero-divisors} combined with Lemma \ref{lemma-localize-ass} part (1).
Let $(R, \mathfrak m)$ be a reduced local ring. Suppose $x \in R$ is an element whose annihilator has radical $\mathfrak m$. If $\mathfrak m \not = 0$, then $x$ cannot be a unit, so $x \in \mathfrak m$. Then in particular $x^{1 + n} = 0$ for some $n \geq 0$. Hence $x = 0$. Which contradicts the assumption that the annihilator of $x$ is contained in $\mathfrak m$. Thus we see that $\mathfrak m = 0$, i.e., $R$ is a field. By Lemma \ref{lemma-weakly-ass-local} this implies the statement of the lemma.
We will use the characterization of weakly associated primes of Lemma \ref{lemma-weakly-ass-local}. Let $\mathfrak p$ be a prime of $R$. As localization is exact we obtain the short exact sequence $0 \to M'_{\mathfrak p} \to M_{\mathfrak p} \to M''_{\mathfrak p} \to 0$. Suppose that $m \in M_{\mathfrak p}$ is an element whose annihilator has radical $\mathfrak pR_{\mathfrak p}$. Then either the image $\overline{m}$ of $m$ in $M''_{\mathfrak p}$ is zero and $m \in M'_{\mathfrak p}$, or the radical of the annihilator of $\overline{m}$ is $\mathfrak pR_{\mathfrak p}$. This proves that $\text{WeakAss}(M) \subset \text{WeakAss}(M') \cup \text{WeakAss}(M'')$. The inclusion $\text{WeakAss}(M') \subset \text{WeakAss}(M)$ is immediate from the definitions.
If $M = (0)$ then $\text{WeakAss}(M) = \emptyset$ by definition. Conversely, suppose that $M \not = 0$. Pick a nonzero element $m \in M$. Write $I = \{x \in R \mid xm = 0\}$ the annihilator of $m$. Then $R/I \subset M$. Hence $\text{WeakAss}(R/I) \subset \text{WeakAss}(M)$ by Lemma \ref{lemma-weakly-ass}. But as $I \not = R$ we have $V(I) = \Spec(R/I)$ contains a minimal prime, see Lemmas \ref{lemma-Zariski-topology} and \ref{lemma-spec-closed}, and we win.
Suppose $f \in \mathfrak q \in \text{WeakAss}(M)$. Then there exists an element $m \in M$ such that $\mathfrak q$ is minimal over $I = \{x \in R \mid xm = 0\}$. Hence there exists a $g \in R$, $g \not \in \mathfrak q$ and $n > 0$ such that $f^ngm = 0$. Note that $gm \not = 0$ as $g \not \in I$. If we take $n$ minimal as above, then $f (f^{n - 1}gm) = 0$ and $f^{n - 1}gm \not = 0$, so $f$ is a zerodivisor on $M$. Conversely, suppose $f \in R$ is a zerodivisor on $M$. Consider the submodule $N = \{m \in M \mid fm = 0\}$. Since $N$ is not zero it has a weakly associated prime $\mathfrak q$ by Lemma \ref{lemma-weakly-ass-zero}. Clearly $f \in \mathfrak q$ and by Lemma \ref{lemma-weakly-ass} $\mathfrak q$ is a weakly associated prime of $M$.
Note that $\text{Supp}(M_{\mathfrak p}) = \{\mathfrak pR_{\mathfrak p}\}$ in $\Spec(R_{\mathfrak p})$. In particular $M_{\mathfrak p}$ is nonzero, and hence $\text{WeakAss}(M_{\mathfrak p}) \not = \emptyset$ by Lemma \ref{lemma-weakly-ass-zero}. Since $\text{WeakAss}(M_{\mathfrak p}) \subset \text{Supp}(M_{\mathfrak p})$ by Lemma \ref{lemma-weakly-ass-support} we conclude that $\text{WeakAss}(M_{\mathfrak p}) = \{\mathfrak pR_{\mathfrak p}\}$, whence $\mathfrak p \in \text{WeakAss}(M)$ by Lemma \ref{lemma-weakly-ass-local}.
Write $\mathfrak p = (g_1, \ldots, g_n)$ for some $g_i \in R$. It is enough the prove the implication ``$\Leftarrow$'' as the other implication holds in general, see Lemma \ref{lemma-weakly-ass-support}. Assume $\mathfrak p \in \text{WeakAss}(M)$. By Lemma \ref{lemma-weakly-ass-local} there exists an element $m \in M_{\mathfrak p}$ such that $I = \{x \in R_{\mathfrak p} \mid xm = 0\}$ has radical $\mathfrak pR_{\mathfrak p}$. Hence for each $i$ there exists a smallest $e_i > 0$ such that $g_i^{e_i}m = 0$ in $M_{\mathfrak p}$. If $e_i > 1$ for some $i$, then we can replace $m$ by $g_i^{e_i - 1} m \not = 0$ and decrease $\sum e_i$. Hence we may assume that the annihilator of $m \in M_{\mathfrak p}$ is $(g_1, \ldots, g_n)R_{\mathfrak p} = \mathfrak p R_{\mathfrak p}$. By Lemma \ref{lemma-associated-primes-localize} we see that $\mathfrak p \in \text{Ass}(M)$.
Let $\mathfrak p$ be an element of $\text{WeakAss}_R(M)$. Then there exists an $m \in M_{\mathfrak p}$ whose annihilator $I = \{x \in R_{\mathfrak p} \mid xm = 0\}$ has radical $\mathfrak pR_{\mathfrak p}$. Consider the annihilator $J = \{x \in S_{\mathfrak p} \mid xm = 0 \}$ of $m$ in $S_{\mathfrak p}$. As $IS_{\mathfrak p} \subset J$ we see that any minimal prime $\mathfrak q \subset S_{\mathfrak p}$ over $J$ lies over $\mathfrak p$. Moreover such a $\mathfrak q$ corresponds to a weakly associated prime of $M$ for example by Lemma \ref{lemma-weakly-ass-local}.
One of the inclusions has already been proved, see Remark \ref{remark-ass-functorial}. To prove the other assume $\mathfrak q \in \text{WeakAss}_S(M)$ and let $\mathfrak p$ be the corresponding prime of $R$. Let $m \in M$ be an element such that $\mathfrak q$ is a minimal prime over $J = \{g \in S \mid gm = 0\}$. Thus the radical of $JS_{\mathfrak q}$ is $\mathfrak qS_{\mathfrak q}$. As $R \to S$ is finite there are finitely many primes $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots, \mathfrak q_l$ over $\mathfrak p$, see Lemma \ref{lemma-finite-finite-fibres}. Pick $x \in \mathfrak q$ with $x \not \in \mathfrak q_i$ for $i > 1$, see Lemma \ref{lemma-silly}. By the above there exists an element $y \in S$, $y \not \in \mathfrak q$ and an integer $t > 0$ such that $y x^t m = 0$. Thus the element $ym \in M$ is annihilated by $x^t$, hence $ym$ maps to zero in $M_{\mathfrak q_i}$, $i = 2, \ldots, l$. To be sure, $ym$ does not map to zero in $S_{\mathfrak q}$. \medskip\noindent The ring $S_{\mathfrak p}$ is semi-local with maximal ideals $\mathfrak q_i S_{\mathfrak p}$ by going up for finite ring maps, see Lemma \ref{lemma-integral-going-up}. If $f \in \mathfrak pR_{\mathfrak p}$ then some power of $f$ ends up in $JS_{\mathfrak q}$ hence for some $n > 0$ we see that $f^t ym$ maps to zero in $M_{\mathfrak q}$. As $ym$ vanishes at the other maximal ideals of $S_{\mathfrak p}$ we conclude that $f^t ym$ is zero in $M_{\mathfrak p}$, see Lemma \ref{lemma-characterize-zero-local}. In this way we see that $\mathfrak p$ is a minimal prime over the annihilator of $ym$ in $R$ and we win.
Suppose that $m \in S^{-1}M$. Let $I = \{x \in R \mid xm = 0\}$ and $I' = \{x' \in S^{-1}R \mid x'm = 0\}$. Then $I' = S^{-1}I$ and $I \cap S = \emptyset$ unless $I = R$ (verifications omitted). Thus primes in $S^{-1}R$ minimal over $I'$ correspond bijectively to primes in $R$ minimal over $I$ and avoiding $S$. This proves the equality $\text{WeakAss}_R(S^{-1}M) = \text{WeakAss}_{S^{-1}R}(S^{-1}M)$. The second equality follows from Lemma \ref{lemma-weakly-ass-local} since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset$ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$.
As $M \subset S^{-1}M$ by assumption we obtain $\text{WeakAss}(M) \subset \text{WeakAss}(S^{-1}M)$ from Lemma \ref{lemma-weakly-ass}. Conversely, suppose that $n/s \in S^{-1}M$ is an element with annihilator $I$ and $\mathfrak p$ a prime which is minimal over $I$. Then the annihilator of $n \in M$ is $I$ and $\mathfrak p$ is a prime minimal over $I$.
Let $x \in M$ be an element of the kernel of the map. Set $N = Rx \subset M$. If $\mathfrak p$ is a weakly associated prime of $N$ we see on the one hand that $\mathfrak p \in \text{WeakAss}(M)$ (Lemma \ref{lemma-weakly-ass}) and on the other hand that $N_{\mathfrak p} \subset M_{\mathfrak p}$ is not zero. This contradiction shows that $\text{WeakAss}(N) = \emptyset$. Hence $N = 0$, i.e., $x = 0$ by Lemma \ref{lemma-weakly-ass-zero}.
Note that $S \otimes_R K = (R \setminus \{0\})^{-1}S$ and $N \otimes_R K = (R \setminus \{0\})^{-1}N$. For any nonzero $x \in R$ multiplication by $x$ on $N$ is injective as $N$ is flat over $R$. Hence the lemma follows from Lemma \ref{lemma-localize-weakly-ass-nonzero-divisors}.
Let $\mathfrak q_1, \ldots, \mathfrak q_t$ denote the minimal primes in the support of $M$. Let $\mathfrak p_1, \ldots, \mathfrak p_s$ denote the embedded associated primes of $M$. Then $\text{Ass}(M) = \{\mathfrak q_j, \mathfrak p_i\}$. There are finitely many of these, see Lemma \ref{lemma-finite-ass}. Set $I = \prod_{i = 1, \ldots, s} \mathfrak p_i$. Then $I \not \subset \mathfrak q_j$ for any $j$. Hence by Lemma \ref{lemma-silly} we can find an $f \in I$ such that $f \not \in \mathfrak q_j$ for all $j = 1, \ldots, t$. Set $M' = \Im(M \to M_f)$. This implies that $M_f \cong M'_f$. Since $M' \subset M_f$ we see that $\text{Ass}(M') \subset \text{Ass}(M_f) = \{\mathfrak q_j\}$. Thus $M'$ has no embedded associated primes. \medskip\noindent Moreover, the support of $K = \Ker(M \to M')$ is contained in $V(\mathfrak p_1) \cup \ldots \cup V(\mathfrak p_s)$, because $\text{Ass}(K) \subset \text{Ass}(M)$ (see Lemma \ref{lemma-ass}) and $\text{Ass}(K)$ contains none of the $\mathfrak q_i$ by construction. Clearly, $K$ is in fact the largest submodule of $M$ whose support is contained in $V(\mathfrak p_1) \cup \ldots \cup V(\mathfrak p_t)$. This implies that $K$ is the maximal element of the set displayed in the lemma.
We may replace $R$ by $R/I$. Hence we may assume every nonzero element of $R$ acts nontrivially on $M$. By Lemma \ref{lemma-support-closed} this implies that $\Spec(R)$ equals the support of $M$. Suppose that $\mathfrak p$ is an embedded prime of $R$. Let $x \in R$ be an element whose annihilator is $\mathfrak p$. Consider the nonzero module $N = xM \subset M$. It is annihilated by $\mathfrak p$. Hence any associated prime $\mathfrak q$ of $N$ contains $\mathfrak p$ and is also an associated prime of $M$. Then $\mathfrak q$ would be an embedded associated prime of $M$ which contradicts the assumption of the lemma.
First we do the case $c = 2$. Consider $K \subset M$ the kernel of $x_2 : M \to M$. For any $z \in K$ we know that $z = x_1 z'$ for some $z' \in M$ because $x_2$ is a nonzerodivisor on $M/x_1M$. Because $x_1$ is a nonzerodivisor on $M$ we see that $x_2 z' = 0$ as well. Hence $x_1 : K \to K$ is surjective. Thus $K = 0$ by Nakayama's Lemma \ref{lemma-NAK}. Next, consider multiplication by $x_1$ on $M/x_2M$. If $z \in M$ maps to an element $\overline{z} \in M/x_2M$ in the kernel of this map, then $x_1 z = x_2 y$ for some $y \in M$. But then since $x_1, x_2$ is a regular sequence we see that $y = x_1 y'$ for some $y' \in M$. Hence $x_1 ( z - x_2 y' ) =0$ and hence $z = x_2 y'$ and hence $\overline{z} = 0$ as desired. \medskip\noindent For the general case, observe that any permutation is a composition of transpositions of adjacent indices. Hence it suffices to prove that $$ x_1, \ldots, x_{i-2}, x_i, x_{i-1}, x_{i + 1}, \ldots, x_c $$ is an $M$-regular sequence. This follows from the case we just did applied to the module $M/(x_1, \ldots, x_{i-2})$ and the length $2$ regular sequence $x_{i-1}, x_i$.
By Lemma \ref{lemma-snake}, if $f_1 : M_1 \to M_1$ and $f_1 : M_3 \to M_3$ are injective, then so is $f_1 : M_2 \to M_2$ and we obtain a short exact sequence $$ 0 \to M_1/f_1M_1 \to M_2/f_1M_2 \to M_3/f_1M_3 \to 0 $$ The lemma follows from this and induction on $r$. Some details omitted.
We will prove this by induction on $r$. If $r = 1$ this follows from the following two easy facts: (a) a power of a nonzerodivisor on $M$ is a nonzerodivisor on $M$ and (b) a divisor of a nonzerodivisor on $M$ is a nonzerodivisor on $M$. If $r > 1$, then by induction applied to $M/f_1M$ we have that $f_1, f_2, \ldots, f_r$ is an $M$-regular sequence if and only if $f_1, f_2^{e_2}, \ldots, f_r^{e_r}$ is an $M$-regular sequence. Thus it suffices to show, given $e > 0$, that $f_1^e, f_2, \ldots, f_r$ is an $M$-regular sequence if and only if $f_1, \ldots, f_r$ is an $M$-regular sequence. We will prove this by induction on $e$. The case $e = 1$ is trivial. Since $f_1$ is a nonzerodivisor under both assumptions (by the case $r = 1$) we have a short exact sequence $$ 0 \to M/f_1M \xrightarrow{f_1^{e - 1}} M/f_1^eM \to M/f_1^{e - 1}M \to 0 $$ Suppose that $f_1, f_2, \ldots, f_r$ is an $M$-regular sequence. Then by induction the elements $f_2, \ldots, f_r$ are $M/f_1M$ and $M/f_1^{e - 1}M$-regular sequences. By Lemma \ref{lemma-regular-sequence-short-exact-sequence} $f_2, \ldots, f_r$ is $M/f_1^eM$-regular. Hence $f_1^e, f_2, \ldots, f_r$ is $M$-regular. Conversely, suppose that $f_1^e, f_2, \ldots, f_r$ is an $M$-regular sequence. Then $f_2 : M/f_1^eM \to M/f_1^eM$ is injective, hence $f_2 : M/f_1M \to M/f_1M$ is injective, hence by induction(!) $f_2 : M/f_1^{e - 1}M \to M/f_1^{e - 1}M$ is injective, hence $$ 0 \to M/(f_1, f_2)M \xrightarrow{f_1^{e - 1}} M/(f_1^e, f_2)M \to M/(f_1^{e - 1}, f_2)M \to 0 $$ is a short exact sequence by Lemma \ref{lemma-snake}. This proves the converse for $r = 2$. If $r > 2$, then we have $f_3 : M/(f_1^e, f_2)M \to M/(f_1^e, f_2)M$ is injective, hence $f_3 : M/(f_1, f_2)M \to M/(f_1, f_2)M$ is injective, and so on. Some details omitted.
Set $J = (f_1, \ldots, f_c)$. Let us show that $f_1$ is a nonzerodivisor on $M$. Suppose $x \in M$ is not zero. By Krull's intersection theorem there exists an integer $r$ such that $x \in J^rM$ but $x \not \in J^{r + 1}M$, see Lemma \ref{lemma-intersect-powers-ideal-module-zero}. Then $f_1 x \in J^{r + 1}M$ is an element whose class in $J^{r + 1}M/J^{r + 2}M$ is nonzero by the assumed structure of $\bigoplus J^nM/J^{n + 1}M$. Whence $f_1x \not = 0$. \medskip\noindent Now we can finish the proof by induction on $c$ using Lemma \ref{lemma-truncate-quasi-regular}.
Consider the ring map $P = \mathbf{Z}[t_1, \ldots, t_n] \to R$ sending $t_i$ to $a_i$. Set $J = (t_1, \ldots, t_n)$. By Example \ref{example-affine-blowup-algebra-polynomial} we have $P[\frac{J}{t_1}] = P[x_2, \ldots, x_n]/(t_1 x_2 - t_2, \ldots, t_1 x_n - t_n)$. Apply Lemma \ref{lemma-blowup-base-change} to the map $P \to A$ to conclude.
The map is given by sending $x/a^n$ for $x \in I^n$ to $f^nx/(fa)^n$. It is straightforward to check this map is well defined and surjective. Since $af$ is a nonzero divisor in $R''$ (Lemma \ref{lemma-affine-blowup}) we see that the set of $f$-power torsion elements are mapped to zero. Conversely, if $x \in R'$ and $f^n x \not = 0$ for all $n > 0$, then $(af)^n x \not = 0$ for all $n$ as $a$ is a nonzero divisor in $R'$. It follows that the image of $x$ in $R''$ is not zero by the description of $R''$ following Definition \ref{definition-blow-up}.