hcju/stacksqa huggingface dataset:
We may and do assume $\mathfrak p \not = \mathfrak m$ and $M$ not zero. Choose a maximal chain of primes $\mathfrak p = \mathfrak p_c \subset \mathfrak p_{c - 1} \subset \ldots \subset \mathfrak p_1 \subset \mathfrak m$. If we prove the result for $M_{\mathfrak p_1}$ over $R_{\mathfrak p_1}$, then the lemma will follow by induction on $c$. Thus we may assume that there is no prime strictly between $\mathfrak p$ and $\mathfrak m$. Note that $\dim(\text{Supp}(M_\mathfrak p)) \leq \dim(\text{Supp}(M)) - 1$ because any chain of primes in the support of $M_\mathfrak p$ can be extended by one more prime (namely $\mathfrak m$) in the support of $M$. On the other hand, we have $\text{depth}(M_\mathfrak p) \geq \text{depth}(M) - \dim(R/\mathfrak p) = \text{depth}(M) - 1$ by Lemma \ref{lemma-depth-localization} and our choice of $\mathfrak p$. Thus $\text{depth}(M_\mathfrak p) \geq \dim(\text{Supp}(M_\mathfrak p)$ as desired (the other inequality is Lemma \ref{lemma-bound-depth}).
By induction on the number of variables it suffices to prove this for $M[x] = M \otimes_R R[x]$ over $R[x]$. Let $\mathfrak m \subset R[x]$ be a maximal ideal, and let $\mathfrak p = R \cap \mathfrak m$. Let $f_1, \ldots, f_d$ be a $M_\mathfrak p$-regular sequence in the maximal ideal of $R_{\mathfrak p}$ of length $d = \dim(\text{Supp}(M_{\mathfrak p}))$. Note that since $R[x]$ is flat over $R$ the localization $R[x]_{\mathfrak m}$ is flat over $R_{\mathfrak p}$. Hence, by Lemma \ref{lemma-flat-increases-depth}, the sequence $f_1, \ldots, f_d$ is a $M[x]_{\mathfrak m}$-regular sequence of length $d$ in $R[x]_{\mathfrak m}$. The quotient $$ Q = M[x]_{\mathfrak m}/(f_1, \ldots, f_d)M[x]_{\mathfrak m} = M_{\mathfrak p}/(f_1, \ldots, f_d)M_{\mathfrak p} \otimes_{R_\mathfrak p} R[x]_{\mathfrak m} $$ has support equal to the primes lying over $\mathfrak p$ because $R_\mathfrak p \to R[x]_\mathfrak m$ is flat and the support of $M_{\mathfrak p}/(f_1, \ldots, f_d)M_{\mathfrak p}$ is equal to $\{\mathfrak p\}$ (details omitted; hint: follows from Lemmas \ref{lemma-annihilator-flat-base-change} and \ref{lemma-support-closed}). Hence the dimension is $1$. To finish the proof it suffices to find an $f \in \mathfrak m$ which is a nonzerodivisor on $Q$. Since $\mathfrak m$ is a maximal ideal, the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak m)$ is finite (Theorem \ref{theorem-nullstellensatz}). Hence we can find $f \in \mathfrak m$ which viewed as a polynomial in $x$ has leading coefficient not in $\mathfrak p$. Such an $f$ acts as a nonzerodivisor on $$ M_{\mathfrak p}/(f_1, \ldots, f_d)M_{\mathfrak p} \otimes_R R[x] = \bigoplus\nolimits_{n \geq 0} M_{\mathfrak p}/(f_1, \ldots, f_d)M_{\mathfrak p} \cdot x^n $$ and hence acts as a nonzerodivisor on $Q$.
By induction on $\dim(B)$ it suffices to prove: If $\dim(B) \geq 1$, then we can find an element $f$ of $A$ which maps to a nonzerodivisor in $B$. By Lemma \ref{lemma-reformulate-CM} it suffices to find $f \in A$ whose image in $B$ is not contained in any of the finitely many minimal primes $\mathfrak q_1, \ldots, \mathfrak q_r$ of $B$. By the assumption that $\mathfrak m_B = \sqrt{\varphi(\mathfrak m_A) B}$ we see that $\mathfrak m_A \not \subset \varphi^{-1}(\mathfrak q_i)$. Hence we can find $f$ by Lemma \ref{lemma-silly}.
Let $A$ be a ring and let $S \subset A$ be a multiplicative subset. The description of $\Spec(S^{-1}A)$ in Lemma \ref{lemma-spec-localization} shows that if $A$ is catenary, then so is $S^{-1}A$. If $S^{-1}A \to C$ is of finite type, then $C = S^{-1}B$ for some finite type ring map $A \to B$. Hence if $A$ is Noetherian and universally catenary, then $B$ is catenary and we see that $C$ is catenary too. This proves the lemma.
If $B$ is a finite type $A$-algebra, then $B$ is Noetherian by Lemma \ref{lemma-Noetherian-permanence}. Any finite type $B$-algebra is a finite type $A$-algebra and hence catenary by our assumption that $A$ is universally catenary. Thus $B$ is universally catenary. Any localization of $B$ is universally catenary by Lemma \ref{lemma-localization-catenary} and this finishes the proof.
The implication (1) $\Rightarrow$ (2) follows from Lemma \ref{lemma-localization-catenary} in both cases. The implication (2) $\Rightarrow$ (3) is immediate in both cases. Assume $R_\mathfrak m$ is catenary for all maximal ideals $\mathfrak m$ of $R$. If $\mathfrak p \subset \mathfrak q$ are primes in $R$, then choose a maximal ideal $\mathfrak q \subset \mathfrak m$. Chains of primes ideals between $\mathfrak p$ and $\mathfrak q$ are in 1-to-1 correspondence with chains of prime ideals between $\mathfrak pR_\mathfrak m$ and $\mathfrak qR_\mathfrak m$ hence we see $R$ is catenary. Assume $R$ is Noetherian and $R_\mathfrak m$ is universally catenary for all maximal ideals $\mathfrak m$ of $R$. Let $R \to S$ be a finite type ring map. Let $\mathfrak q$ be a prime ideal of $S$ lying over the prime $\mathfrak p \subset R$. Choose a maximal ideal $\mathfrak p \subset \mathfrak m$ in $R$. Then $R_\mathfrak p$ is a localization of $R_\mathfrak m$ hence universally catenary by Lemma \ref{lemma-localization-catenary}. Then $S_\mathfrak p$ is catenary as a finite type ring over $R_\mathfrak p$. Hence $S_\mathfrak q$ is catenary as a localization. Thus $S$ is catenary by the first case treated above.
Since a polynomial algebra over $R$ is Cohen-Macaulay, by Lemma \ref{lemma-CM-polynomial-algebra}, it suffices to show that a Cohen-Macaulay ring is catenary. Let $R$ be Cohen-Macaulay and $\mathfrak p \subset \mathfrak q$ primes of $R$. By definition $R_{\mathfrak q}$ and $R_{\mathfrak p}$ are Cohen-Macaulay. Take a maximal chain of primes $\mathfrak p = \mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_n = \mathfrak q$. Next choose a maximal chain of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_m = \mathfrak p$. By Lemma \ref{lemma-maximal-chain-CM} we have $n + m = \dim(R_{\mathfrak q})$. And we have $m = \dim(R_{\mathfrak p})$ by the same lemma. Hence $n = \dim(R_{\mathfrak q}) - \dim(R_{\mathfrak p})$ is independent of choices. \medskip\noindent To prove the more general statement, argue exactly as above but using Lemmas \ref{lemma-maximal-CM-polynomial-algebra} and \ref{lemma-maximal-chain-maximal-CM}.
If $A$ is catenary, then $\Spec(A)$ has a dimension function $\delta$ by Topology, Lemma \ref{topology-lemma-locally-dimension-function} (and Lemma \ref{lemma-catenary}). We may assume $\delta(\mathfrak m) = 0$. Then we see that $$ \delta(\mathfrak p) = \text{codim}(V(\mathfrak m), V(\mathfrak p)) = \dim(A/\mathfrak p) $$ by Topology, Lemma \ref{topology-lemma-dimension-function-catenary}. In this way we see that (1) implies (2). The reverse implication follows from Topology, Lemma \ref{topology-lemma-dimension-function-catenary} as well.
Let $x_1, \ldots, x_d$ be a minimal set of generators for the maximal ideal $\mathfrak m$, see Definition \ref{definition-regular-local}. There is a surjection $\kappa[X_1, \ldots, X_d] \to \bigoplus \mathfrak m^n/\mathfrak m^{n + 1}$, which maps $X_i$ to the class of $x_i$ in $\mathfrak m/\mathfrak m^2$. Since $d(R) = d$ by Proposition \ref{proposition-dimension} we know that the numerical polynomial $n \mapsto \dim_\kappa \mathfrak m^n/\mathfrak m^{n + 1}$ has degree $d - 1$. By Lemma \ref{lemma-quotient-smaller-d} we conclude that the surjection $\kappa[X_1, \ldots, X_d] \to \bigoplus \mathfrak m^n/\mathfrak m^{n + 1}$ is an isomorphism.
We will use that $\bigcap \mathfrak m^n = 0$ by Lemma \ref{lemma-intersect-powers-ideal-module-zero}. Let $f, g \in R$ such that $fg = 0$. Suppose that $f \in \mathfrak m^a$ and $g \in \mathfrak m^b$, with $a, b$ maximal. Since $fg = 0 \in \mathfrak m^{a + b + 1}$ we see from the result of Lemma \ref{lemma-regular-graded} that either $f \in \mathfrak m^{a + 1}$ or $g \in \mathfrak m^{b + 1}$. Contradiction.
Note that $R/x_1R$ is a Noetherian local ring of dimension $\geq d - 1$ by Lemma \ref{lemma-one-equation} with $x_2, \ldots, x_d$ generating the maximal ideal. Hence it is a regular local ring by definition. Since $R$ is a domain by Lemma \ref{lemma-regular-domain} $x_1$ is a nonzerodivisor.
Say $\dim(R) = d$ and $\dim(R/I) = d - c$. Denote $\overline{\mathfrak m} = \mathfrak m/I$ the maximal ideal of $R/I$. Let $\kappa = R/\mathfrak m$. We have $$ \dim_\kappa((I + \mathfrak m^2)/\mathfrak m^2) = \dim_\kappa(\mathfrak m/\mathfrak m^2) - \dim(\overline{\mathfrak m}/\overline{\mathfrak m}^2) = d - (d - c) = c $$ by the definition of a regular local ring. Hence we can choose $x_1, \ldots, x_c \in I$ whose images in $\mathfrak m/\mathfrak m^2$ are linearly independent and supplement with $x_{c + 1}, \ldots, x_d$ to get a minimal system of generators of $\mathfrak m$. The induced map $R/(x_1, \ldots, x_c) \to R/I$ is a surjection between regular local rings of the same dimension (Lemma \ref{lemma-regular-ring-CM}). It follows that the kernel is zero, i.e., $I = (x_1, \ldots, x_c)$. Namely, if not then we would have $\dim(R/I) < \dim(R/(x_1, \ldots, x_c))$ by Lemmas \ref{lemma-regular-domain} and \ref{lemma-one-equation}.
Let $m_1, \ldots, m_r$ be elements of $M$ which map to a $R/xR$-basis of $M/xM$. By Nakayama's Lemma \ref{lemma-NAK} $m_1, \ldots, m_r$ generate $M$. If $\sum a_i m_i = 0$ is a relation, then $a_i \in xR$ for all $i$. Hence $a_i = b_i x$ for some $b_i \in R$. Hence the kernel $K$ of $R^r \to M$ satisfies $xK = K$ and hence is zero by Nakayama's lemma.
Let $M$ be a maximal Cohen-Macaulay module over $R$. Let $x \in \mathfrak m$ be part of a regular sequence generating $\mathfrak m$. Then $x$ is a nonzerodivisor on $M$ by Proposition \ref{proposition-CM-module}, and $M/xM$ is a maximal Cohen-Macaulay module over $R/xR$. By induction on $\dim(R)$ we see that $M/xM$ is free. We win by Lemma \ref{lemma-free-mod-x}.
Let $\mathfrak m \subset R$ be the maximal ideal; it is the colimit of the maximal ideal $\mathfrak m_i \subset R_i$. We prove the lemma by induction on $d = \dim \mathfrak m/\mathfrak m^2$. If $d = 0$, then $R = R/\mathfrak m$ is a field and $R$ is a regular local ring. If $d > 0$ pick an $x \in \mathfrak m$, $x \not \in \mathfrak m^2$. For some $i$ we can find an $x_i \in \mathfrak m_i$ mapping to $x$. Note that $R/xR = \colim_{i' \geq i} R_{i'}/x_iR_{i'}$ is a Noetherian local ring. By Lemma \ref{lemma-regular-ring-CM} we see that $R_{i'}/x_iR_{i'}$ is a regular local ring. Hence by induction we see that $R/xR$ is a regular local ring. Since each $R_i$ is a domain (Lemma \ref{lemma-regular-graded}) we see that $R$ is a domain. Hence $x$ is a nonzerodivisor and we conclude that $R$ is a regular local ring by Lemma \ref{lemma-regular-mod-x}.
Since $S_{\mathfrak p} = R_{\mathfrak p} \otimes_R S$ (see Lemma \ref{lemma-tensor-localization}) we see that (1) implies (2) by Lemma \ref{lemma-base-change-epimorphism}. Conversely, assume that (2) holds. Let $a, b : S \to A$ be two ring maps from $S$ to a ring $A$ equalizing the map $R \to S$. By assumption we see that for every prime $\mathfrak p$ of $R$ the induced maps $a_{\mathfrak p}, b_{\mathfrak p} : S_{\mathfrak p} \to A_{\mathfrak p}$ are the same. Hence $a = b$ as $A \subset \prod_{\mathfrak p} A_{\mathfrak p}$, see Lemma \ref{lemma-characterize-zero-local}.
(This lemma seems to have been reproved many times in the literature, and has many different proofs.) It is clear that a surjective ring map is an epimorphism. Suppose that $R \to S$ is a finite ring map such that $S \otimes_R S \to S$ is an isomorphism. Our goal is to show that $R \to S$ is surjective. Assume $S/R$ is not zero. The exact sequence $R \to S \to S/R \to 0$ leads to an exact sequence $$ R \otimes_R S \to S \otimes_R S \to S/R \otimes_R S \to 0. $$ Our assumption implies that the first arrow is an isomorphism, hence we conclude that $S/R \otimes_R S = 0$. Hence also $S/R \otimes_R S/R = 0$. By Lemma \ref{lemma-trivial-filter-finite-module} there exists a surjection of $R$-modules $S/R \to R/I$ for some proper ideal $I \subset R$. Hence there exists a surjection $S/R \otimes_R S/R \to R/I \otimes_R R/I = R/I \not = 0$, contradiction.
Let $\mathfrak p$ be a prime of $R$. The fibre of the map is the spectrum of the fibre ring $S \otimes_R \kappa(\mathfrak p)$. By Lemma \ref{lemma-base-change-epimorphism} the map $\kappa(\mathfrak p) \to S \otimes_R \kappa(\mathfrak p)$ is an epimorphism, and hence by Lemma \ref{lemma-epimorphism-over-field} we have either $S \otimes_R \kappa(\mathfrak p) = 0$ or $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak p)$ which proves (1) and (2).
It is clear that (2) implies (1). Conversely, suppose that $g \otimes 1 = 1 \otimes g$. Choose generators $\{s_i\}_{i \in I}$ of $S$ as an $R$-module with $0, 1 \in I$ and $s_0 = 1$ and $s_1 = g$. Apply Lemma \ref{lemma-relations} to the relation $g \otimes s_0 + (-1) \otimes s_1 = 0$. We see that there exist $a_{i, j} \in R$ such that $g = \sum_i a_{i, 0} s_i$, $-1 = \sum_i a_{i, 1} s_i$, and for $j \not = 0, 1$ we have $0 = \sum_i a_{i, j} s_i$, and moreover for all $i$ we have $\sum_j a_{i, j}s_j = 0$. Then we have $$ \sum\nolimits_{i, j \not = 0} a_{i, j} s_i s_j = -g + a_{0, 0} $$ and for each $j \not = 0$ we have $\sum_{i \not = 0} a_{i, j}s_i \in R$. This proves that $-g + a_{0, 0}$ can be written as in (2). It follows that $g$ can be written as in (2). Details omitted. Hint: Show that the set of elements of $S$ which have an expression as in (2) form an $R$-subalgebra of $S$.
The condition that $R \to S$ is an epimorphism means that each $g \in S$ satisfies $g \otimes 1 = 1 \otimes g$, see Lemma \ref{lemma-epimorphism}. We are going to use the notation introduced in Remark \ref{remark-matrices-associated-to-elements-epicenter}. Suppose that $g, g' \in S$ and suppose that $(P, U, V)$ is an $n$-triple which is associated to both $g$ and $g'$. Then we claim that $g = g'$. Namely, write $(P, U, V) = (X, YX, XZ)$ for a matrix factorization $(g) = YXZ$ of $g$ and write $(P, U, V) = (X', Y'X', X'Z')$ for a matrix factorization $(g') = Y'X'Z'$ of $g'$. Then we see that $$ (g) = YXZ = UZ = Y'X'Z = Y'PZ = Y'XZ = Y'V = Y'X'Z' = (g') $$ and hence $g = g'$. This implies that the cardinality of $S$ is bounded by the number of possible triples, which has cardinality at most $\sup_{n \in \mathbf{N}} |R|^n$. If $R$ is infinite then this is at most $|R|$, see \cite[Ch. I, 10.13]{Kunen}. \medskip\noindent If $R$ is a finite ring then the argument above only proves that $S$ is at worst countable. In fact in this case $R$ is Artinian and the map $R \to S$ is surjective. We omit the proof of this case.
Let $\varphi : N_1 \to N_2$ be an $R$-linear map. For any $x \in N_1$ consider the map $S \otimes_R S \to N_2$ defined by the rule $g \otimes g' \mapsto g\varphi(g'x)$. Since both maps $S \to S \otimes_R S$ are isomorphisms (Lemma \ref{lemma-epimorphism}), we conclude that $g \varphi(g'x) = gg'\varphi(x) = \varphi(gg' x)$. Thus $\varphi$ is $S$-linear.
For any ideal $J$ of $R$ we have the short exact sequence $0 \to J \to R \to R/J \to 0$. Tensoring with $R/I$ we get an exact sequence $J \otimes_R R/I \to R/I \to R/I + J \to 0$ and $J \otimes_R R/I = J/JI$. Thus the equivalence of (1), (2), and (3) follows from Lemma \ref{lemma-flat}. Moreover, these imply (4). \medskip\noindent The implication (4) $\Rightarrow$ (5) is trivial. Assume (5) and let $x_1, \ldots, x_n \in I$. Choose $y_i \in I$ such that $x_i = y_ix_i$. Let $y \in I$ be the element such that $1 - y = \prod_{i = 1, \ldots, n} (1 - y_i)$. Then $x_i = yx_i$ for all $i = 1, \ldots, n$. Hence (6) holds, and it follows that (5) $\Leftrightarrow$ (6). \medskip\noindent Assume (5). Let $x \in I$. Then $x = yx$ for some $y \in I$. Hence $x(1 - y) = 0$, which shows that $x$ maps to zero in $(1 + I)^{-1}R$. Of course the kernel of the map $R \to (1 + I)^{-1}R$ is always contained in $I$. Hence we see that (5) implies (9). Assume (9). Then for any $x \in I$ we see that $x(1 - y) = 0$ for some $y \in I$. In other words, $x = yx$. We conclude that (5) is equivalent to (9). \medskip\noindent Assume (5). Let $\mathfrak p$ be a prime of $R$. If $\mathfrak p \not \in V(I)$, then $IR_{\mathfrak p} = R_{\mathfrak p}$. If $\mathfrak p \in V(I)$, in other words, if $I \subset \mathfrak p$, then $x \in I$ implies $x(1 - y) = 0$ for some $y \in I$, implies $x$ maps to zero in $R_{\mathfrak p}$, i.e., $IR_{\mathfrak p} = 0$. Thus we see that (7) holds. \medskip\noindent Assume (7). Then $(R/I)_{\mathfrak p}$ is either $0$ or $R_{\mathfrak p}$ for any prime $\mathfrak p$ of $R$. Hence by Lemma \ref{lemma-flat-localization} we see that (1) holds. At this point we see that all of (1) -- (7) and (9) are equivalent. \medskip\noindent As $IR_{\mathfrak p} = I_{\mathfrak p}$ we see that (7) implies (8). Finally, if (8) holds, then this means exactly that $I_{\mathfrak p}$ is the zero module if and only if $\mathfrak p \in V(I)$, which is clearly saying that (7) holds. Now (1) -- (9) are equivalent. \medskip\noindent Assume (1) -- (9) hold. Then $R/I \subset (1 + I)^{-1}R$ by (9) and the map $R/I \to (1 + I)^{-1}R$ is also surjective by the description of localizations at primes afforded by (7). Hence (11) holds. \medskip\noindent The implication (11) $\Rightarrow$ (10) is trivial. And (10) implies that (1) holds because a localization of $R$ is flat over $R$, see Lemma \ref{lemma-flat-localization}.
Let $I$ be a pure ideal. Then since $R \to R/I$ is flat, by going up generalizations lift along the map $\Spec(R/I) \to \Spec(R)$. Hence $V(I)$ is closed under generalizations. This shows that the map is well defined. By Lemma \ref{lemma-pure-ideal-determined-by-zero-set} the map is injective. Suppose that $Z \subset \Spec(R)$ is closed and closed under generalizations. Let $J \subset R$ be the radical ideal such that $Z = V(J)$. Let $I = \{x \in R : x \in xJ\}$. Note that $I$ is an ideal: if $x, y \in I$ then there exist $f, g \in J$ such that $x = xf$ and $y = yg$. Then $$ x + y = (x + y)(f + g - fg) $$ Verification left to the reader. We claim that $I$ is pure and that $V(I) = V(J)$. If the claim is true then the map of the lemma is surjective and the lemma holds. \medskip\noindent Note that $I \subset J$, so that $V(J) \subset V(I)$. Let $I \subset \mathfrak p$ be a prime. Consider the multiplicative subset $S = (R \setminus \mathfrak p)(1 + J)$. By definition of $I$ and $I \subset \mathfrak p$ we see that $0 \not \in S$. Hence we can find a prime $\mathfrak q$ of $R$ which is disjoint from $S$, see Lemmas \ref{lemma-localization-zero} and \ref{lemma-spec-localization}. Hence $\mathfrak q \subset \mathfrak p$ and $\mathfrak q \cap (1 + J) = \emptyset$. This implies that $\mathfrak q + J$ is a proper ideal of $R$. Let $\mathfrak m$ be a maximal ideal containing $\mathfrak q + J$. Then we get $\mathfrak m \in V(J)$ and hence $\mathfrak q \in V(J) = Z$ as $Z$ was assumed to be closed under generalization. This in turn implies $\mathfrak p \in V(J)$ as $\mathfrak q \subset \mathfrak p$. Thus we see that $V(I) = V(J)$. \medskip\noindent Finally, since $V(I) = V(J)$ (and $J$ radical) we see that $J = \sqrt{I}$. Pick $x \in I$, so that $x = xy$ for some $y \in J$ by definition. Then $x = xy = xy^2 = \ldots = xy^n$. Since $y^n \in I$ for some $n > 0$ we conclude that property (5) of Lemma \ref{lemma-pure} holds and we see that $I$ is indeed pure.
If (1) holds, then $I = I \cap I = I^2$ by Lemma \ref{lemma-pure}. Hence $I$ is generated by an idempotent by Lemma \ref{lemma-ideal-is-squared-union-connected}. Thus (1) $\Rightarrow$ (2). If (2) holds, then $I = (e)$ and $R = (1 - e) \oplus (e)$ as an $R$-module hence $R/I$ is flat and $I$ is pure and $V(I) = D(1 - e)$ is open. Thus (2) $\Rightarrow$ (1) $+$ (3). Finally, assume (3). Then $V(I)$ is open and closed, hence $V(I) = D(1 - e)$ for some idempotent $e$ of $R$, see Lemma \ref{lemma-disjoint-decomposition}. The ideal $J = (e)$ is a pure ideal such that $V(J) = V(I)$ hence $I = J$ by Lemma \ref{lemma-pure-ideal-determined-by-zero-set}. In this way we see that (3) $\Rightarrow$ (2). By Lemma \ref{lemma-finite-projective} we see that (4) is equivalent to the assertion that $I$ is pure and $R/I$ finitely presented. Moreover, $R/I$ is finitely presented if and only if $I$ is finitely generated, see Lemma \ref{lemma-extension}. Hence (4) is equivalent to (1).
If any finite flat $R$-module is finite locally free then the support of $R/I$ where $I$ is a pure ideal is open. Hence the implication (2) $\Rightarrow$ (1) follows from Lemma \ref{lemma-pure-ideal-determined-by-zero-set}. \medskip\noindent For the converse assume that $R$ satisfies (1). Let $M$ be a finite flat $R$-module. The support $Z = \text{Supp}(M)$ of $M$ is closed, see Lemma \ref{lemma-support-closed}. On the other hand, if $\mathfrak p \subset \mathfrak p'$, then by Lemma \ref{lemma-finite-flat-local} the module $M_{\mathfrak p'}$ is free, and $M_{\mathfrak p} = M_{\mathfrak p'} \otimes_{R_{\mathfrak p'}} R_{\mathfrak p}$ Hence $\mathfrak p' \in \text{Supp}(M) \Rightarrow \mathfrak p \in \text{Supp}(M)$, in other words, the support is closed under generalization. As $R$ satisfies (1) we see that the support of $M$ is open and closed. Suppose that $M$ is generated by $r$ elements $m_1, \ldots, m_r$. The modules $\wedge^i(M)$, $i = 1, \ldots, r$ are finite flat $R$-modules also, because $\wedge^i(M)_{\mathfrak p} = \wedge^i(M_{\mathfrak p})$ is free over $R_{\mathfrak p}$. Note that $\text{Supp}(\wedge^{i + 1}(M)) \subset \text{Supp}(\wedge^i(M))$. Thus we see that there exists a decomposition $$ \Spec(R) = U_0 \amalg U_1 \amalg \ldots \amalg U_r $$ by open and closed subsets such that the support of $\wedge^i(M)$ is $U_r \cup \ldots \cup U_i$ for all $i = 0, \ldots, r$. Let $\mathfrak p$ be a prime of $R$, and say $\mathfrak p \in U_i$. Note that $\wedge^i(M) \otimes_R \kappa(\mathfrak p) = \wedge^i(M \otimes_R \kappa(\mathfrak p))$. Hence, after possibly renumbering $m_1, \ldots, m_r$ we may assume that $m_1, \ldots, m_i$ generate $M \otimes_R \kappa(\mathfrak p)$. By Nakayama's Lemma \ref{lemma-NAK} we get a surjection $$ R_f^{\oplus i} \longrightarrow M_f, \quad (a_1, \ldots, a_i) \longmapsto \sum a_im_i $$ for some $f \in R$, $f \not \in \mathfrak p$. We may also assume that $D(f) \subset U_i$. This means that $\wedge^i(M_f) = \wedge^i(M)_f$ is a flat $R_f$ module whose support is all of $\Spec(R_f)$. By the above it is generated by a single element, namely $m_1 \wedge \ldots \wedge m_i$. Hence $\wedge^i(M)_f \cong R_f/J$ for some pure ideal $J \subset R_f$ with $V(J) = \Spec(R_f)$. Clearly this means that $J = (0)$, see Lemma \ref{lemma-pure-ideal-determined-by-zero-set}. Thus $m_1 \wedge \ldots \wedge m_i$ is a basis for $\wedge^i(M_f)$ and it follows that the displayed map is injective as well as surjective. This proves that $M$ is finite locally free as desired.
We prove this by induction on $d$. If $d = 0$, then $M$ is projective. In this case there is a splitting $F_0 = \Ker(F_0 \to M) \oplus M$, and hence $\Ker(F_0 \to M)$ is projective. This finishes the proof if $e = 0$, and if $e > 0$, then replacing $M$ by $\Ker(F_0 \to M)$ we decrease $e$. \medskip\noindent Next assume $d > 0$. Let $0 \to P_d \to P_{d-1} \to \ldots \to P_0 \to M \to 0$ be a minimal length finite resolution with $P_i$ projective. According to Schanuel's Lemma \ref{lemma-Schanuel} we have $P_0 \oplus \Ker(F_0 \to M) \cong F_0 \oplus \Ker(P_0 \to M)$. This proves the case $d = 1$, $e = 0$, because then the right hand side is $F_0 \oplus P_1$ which is projective. Hence now we may assume $e > 0$. The module $F_0 \oplus \Ker(P_0 \to M)$ has the finite projective resolution $$ 0 \to P_d \to P_{d-1} \to \ldots \to P_2 \to P_1 \oplus F_0 \to \Ker(P_0 \to M) \oplus F_0 \to 0 $$ of length $d - 1$. By induction applied to the exact sequence $$ F_e \to F_{e-1} \to \ldots \to F_2 \to P_0 \oplus F_1 \to P_0 \oplus \Ker(F_0 \to M) \to 0 $$ of length $e - 1$ we conclude $\Ker(F_e \to F_{e - 1})$ is projective (if $e \geq 2$) or that $\Ker(F_1 \oplus P_0 \to F_0 \oplus P_0)$ is projective. This implies the lemma.