diff --git a/flat.tex b/flat.tex
index 9086eff27..48e85a1b8 100644
--- a/flat.tex
+++ b/flat.tex
@@ -3628,93 +3628,285 @@ \section{Flat finite type modules, Part II}
 \label{section-finite-type-flat-II}
 
 \noindent
-The following lemma will be superseded by the stronger
+We will need the following lemma.
+
+\begin{lemma}
+\label{lemma-weak-bourbaki-pre-pre}
+Let $R \to S$ be a ring map of finite presentation. Let $N$ be a
+finitely presented $S$-module. Let $\mathfrak q \subset S$ be a prime ideal
+lying over $\mathfrak p \subset R$. Set
+$\overline{S} = S \otimes_R \kappa(\mathfrak p)$,
+$\overline{\mathfrak q} = \mathfrak q \overline{S}$, and
+$\overline{N} = N \otimes_R \kappa(\mathfrak p)$. Then
+we can find a $g \in S$ with
+$g \not \in \mathfrak q$ such that
+$\overline{g} \in \mathfrak r$ for all
+$\mathfrak r \in \text{Ass}_{\overline{S}}(\overline{N})$
+such that $\mathfrak r \not \subset \overline{\mathfrak q}$.
+\end{lemma}
+
+\begin{proof}
+Namely, if $\text{Ass}_{\overline{S}}(\overline{N}) =
+\{\mathfrak r_1, \ldots, \mathfrak r_n\}$
+(finiteness by Algebra, Lemma \ref{algebra-lemma-finite-ass}),
+then after renumbering we may assume that
+$$
+\mathfrak r_1 \subset \overline{\mathfrak q},
+\ldots,
+\mathfrak r_r \subset \overline{\mathfrak q}, \quad
+\mathfrak r_{r + 1} \not \subset \overline{\mathfrak q},
+\ldots,
+\mathfrak r_n \not \subset \overline{\mathfrak q}
+$$
+Since $\overline{\mathfrak q}$ is a prime ideal we see that the product
+$\mathfrak r_{r + 1} \ldots \mathfrak r_n$ is not contained in
+$\overline{\mathfrak q}$ and hence we can pick an element
+$a$ of $\overline{S}$ contained in
+$\mathfrak r_{r + 1}, \ldots, \mathfrak r_n$ but not in
+$\overline{\mathfrak q}$.
+If there exists $g \in S$ mapping to $a$, then $g$
+works. In general we can find a nonzero element
+$\lambda \in \kappa(\mathfrak p)$
+such that $\lambda a$ is the image of a $g \in S$.
+\end{proof}
+
+\noindent
+The following lemma has a sligthly stronger variant
 Lemma \ref{lemma-weak-bourbaki}
 below.
 
 \begin{lemma}
 \label{lemma-weak-bourbaki-pre}
-Let $(R, \mathfrak m)$ be a local ring.
-Let $R \to S$ be of finite presentation.
-Let $N$ be a finitely presented $S$-module which is free as an $R$-module.
-Let $M$ be an $R$-module.
-Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak m$.
+Let $R \to S$ be a ring map of finite presentation.
+Let $N$ be a finitely presented $S$-module
+which is flat as an $R$-module. Let $M$ be an $R$-module.
+Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p \subset R$.
 Then
-\begin{enumerate}
-\item if $\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)$
-then $\mathfrak m \in \text{WeakAss}_R(M)$ and
-$\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$,
-\item if $\mathfrak m \in \text{WeakAss}_R(M)$ and
-$\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$
-is a maximal element then $\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)$.
-\end{enumerate}
-Here $\overline{S} = S/\mathfrak m S$,
+$$
+\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)
+\Leftrightarrow
+\Big(
+\mathfrak p \in \text{WeakAss}_R(M)
+\text{ and }
+\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})
+\Big)
+$$
+Here $\overline{S} = S \otimes_R \kappa(\mathfrak p)$,
 $\overline{\mathfrak q} = \mathfrak q \overline{S}$, and
-$\overline{N} = N/\mathfrak m N$.
+$\overline{N} = N \otimes_R \kappa(\mathfrak p)$.
 \end{lemma}
 
 \begin{proof}
+Pick $g \in S$ as in Lemma \ref{lemma-weak-bourbaki-pre-pre}.
+Apply Proposition \ref{proposition-finite-presentation-flat-at-point}
+to the morphism of schemes $\Spec(S_g) \to \Spec(R)$, the quasi-coherent
+module associated to $N_g$, and the points
+corresponding to the primes $\mathfrak qS_g$ and $\mathfrak p$. Translating
+into algebra we obtain a commutative diagram of rings
+$$
+\xymatrix{
+S \ar[r] & S_g \ar[r] & S' \\
+& R \ar[lu] \ar[u] \ar[r] & R' \ar[u]
+}
+\quad\quad
+\xymatrix{
+\mathfrak q \ar@{-}[r] \ar@{-}[rd] &
+\mathfrak qS_g \ar@{-}[d] \ar@{-}[r] & \mathfrak q' \ar@{-}[d] \\
+& \mathfrak p \ar@{-}[r] & \mathfrak p'
+}
+$$
+endowed with primes as shown, the horizontal arrows are \'etale,
+and $N \otimes_S S'$ is projective as an $R'$-module. Set
+$N' = N \otimes_S S'$, $M' = M \otimes_R R'$,
+$\overline{S}' = S' \otimes_{R'} \kappa(\mathfrak q')$,
+$\overline{\mathfrak q}' = \mathfrak q' \overline{S}'$,
+and
+$$
+\overline{N}' = N' \otimes_{R'} \kappa(\mathfrak p') =
+\overline{N} \otimes_{\overline{S}} \overline{S}'
+$$
+By Lemma \ref{lemma-etale-weak-assassin-up-down} we have
+\begin{align*}
+\text{WeakAss}_{S'}(M' \otimes_{R'} N') & =
+(\Spec(S') \to \Spec(S))^{-1}\text{WeakAss}_S(M \otimes_R N) \\
+\text{WeakAss}_{R'}(M') & =
+(\Spec(R') \to \Spec(R))^{-1}\text{WeakAss}_R(M) \\
+\text{Ass}_{\overline{S}'}(\overline{N}') & =
+(\Spec(\overline{S}') \to \Spec(\overline{S}))^{-1}
+\text{Ass}_{\overline{S}}(\overline{N})
+\end{align*}
+Use Algebra, Lemma \ref{algebra-lemma-ass-weakly-ass}
+for $\overline{N}$ and $\overline{N}'$. In particular we have
+\begin{align*}
+\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)
+& \Leftrightarrow
+\mathfrak q' \in \text{WeakAss}_{S'}(M' \otimes_{R'} N') \\
+\mathfrak p \in \text{WeakAss}_R(M)
+& \Leftrightarrow
+\mathfrak p' \in \text{WeakAss}_{R'}(M') \\
+\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})
+& \Leftrightarrow
+\overline{\mathfrak q}' \in \text{WeakAss}_{\overline{S}'}(\overline{N}')
+\end{align*}
+Our careful choice of $g$ and the formula for
+$\text{Ass}_{\overline{S}'}(\overline{N}')$ above shows that
+\begin{equation}
+\label{equation-key-observation}
+\text{if }\mathfrak r' \in \text{Ass}_{\overline{S}'}(\overline{N}')
+\text{ lies over }\mathfrak r \subset \overline{S}\text{ then }
+\mathfrak r \subset \overline{\mathfrak q}
+\end{equation}
+This will be a key observation later in the proof. We will use
+the characterization of weakly associated primes given in
+Algebra, Lemma \ref{algebra-lemma-weakly-ass-local} without further mention.
+
+\medskip\noindent
 Suppose that
 $\overline{\mathfrak q} \not \in \text{Ass}_{\overline{S}}(\overline{N})$.
+Then
+$\overline{\mathfrak q}' \not \in \text{Ass}_{\overline{S}'}(\overline{N}')$.
 By
 Algebra, Lemmas \ref{algebra-lemma-ass-zero-divisors},
 \ref{algebra-lemma-finite-ass}, and
 \ref{algebra-lemma-silly}
-there exists an element $\overline{g} \in \overline{\mathfrak q}$
-which is not a zerodivisor on $\overline{N}$. Let $g \in \mathfrak q$
-be an element which maps to $\overline{g}$ in $\overline{\mathfrak q}$. By
+there exists an element $\overline{a}' \in \overline{\mathfrak q}'$
+which is not a zerodivisor on $\overline{N}'$.
+After replacing $\overline{a}'$ by $\lambda \overline{a}'$ for some nonzero
+$\lambda \in \kappa(\mathfrak p)$ we can find
+$a' \in \mathfrak q'$ mapping to $\overline{a}'$. By
 Lemma \ref{lemma-invert-universally-injective}
-the map $g : N \to N$ is $R$-universally injective. In particular
-we see that $g : M \otimes_R N \to M \otimes_R N$ is injective.
-Clearly this implies that
-$\mathfrak q \not \in \text{WeakAss}_S(M \otimes_R N)$.
+the map $a' : N'_{\mathfrak p'} \to N'_{\mathfrak p'}$ is
+$R'_{\mathfrak p'}$-universally injective. In particular
+we see that $a' : M' \otimes_{R'} N' \to M' \otimes_{R'} N'$ is
+injective after localizating at $\mathfrak p'$ and hence after
+localizing at $\mathfrak q'$. Clearly this implies that
+$\mathfrak q' \not \in \text{WeakAss}_{S'}(M' \otimes_{R'} N')$.
 We conclude that $\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)$ implies
 $\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$.
 
 \medskip\noindent
-Assume $\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)$.
-Let $z \in M \otimes_R N$ be an element whose annihilator in $S$
-has radical $\mathfrak q$. As $N$ is a free $R$-module, we can find
-a finite free direct summand $F \subset N$ such that
-$z \in M \otimes_R F$. The radical of the annihilator of
-$z \in M \otimes_R F$ in $R$ is $\mathfrak m$ (by our assumption on $z$
-and because $\mathfrak q$ lies over $\mathfrak m$). Hence we see that
-$\mathfrak m \in \text{WeakAss}(M \otimes_R F)$ which implies
-that $\mathfrak m \in \text{WeakAss}(M)$ by
-Algebra, Lemma \ref{algebra-lemma-weakly-ass}.
-This finishes the proof of (1).
+Assume $\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)$. We want
+to show $\mathfrak p \in \text{WeakAss}_S(M)$.
+Let $z \in M \otimes_R N$ be an element such that $\mathfrak q$
+is minimal over $J = \text{Ann}_S(z)$.
+Let $f_i \in \mathfrak p$, $i \in I$ be a set of generators of the
+ideal $\mathfrak p$. Since $\mathfrak q$ lies over $\mathfrak p$, for every $i$
+we can choose an $n_i \geq 1$ and $g_i \in S$, $g_i \not \in \mathfrak q$
+with $g_i f_i^{n_i} \in J$, i.e., $g_i f_i^{n_i} z = 0$.
+Let $z' \in (M' \otimes_{R'} N')_{\mathfrak p'}$ be the image of $z$.
+Observe that $z'$ is nonzero because $z$ has nonzero image in
+$(M \otimes_R N)_\mathfrak q$ and because $S_\mathfrak q \to S'_{\mathfrak q'}$
+is faithfully flat. We claim that $f_i^{n_i} z' = 0$.
+
+\medskip\noindent
+Proof of the claim: Let $g'_i \in S'$ be the image of $g_i$.
+By the key observation (\ref{equation-key-observation})
+we find that the image $\overline{g}'_i \in \overline{S}'$
+is not contained in $\mathfrak r'$ for any
+$\mathfrak r' \in \text{Ass}_{\overline{S}'}(\overline{N})$.
+Hence by Lemma \ref{lemma-invert-universally-injective}
+we see that $g'_i : N'_{\mathfrak p'} \to N'_{\mathfrak p'}$ is
+$R'_{\mathfrak p'}$-universally injective. In particular
+we see that $g'_i : M' \otimes_{R'} N' \to M' \otimes_{R'} N'$ is
+injective after localizating at $\mathfrak p'$. The claim
+follows because $g_i f_i^{n_i} z' = 0$.
+
+\medskip\noindent
+Our claim shows that the annihilator of $z'$ in $R'_{\mathfrak p'}$
+contains the elements $f_i^{n_i}$. As $R \to R'$ is \'etale we have
+$\mathfrak p'R'_{\mathfrak p'} = \mathfrak pR'_{\mathfrak p'}$
+by Algebra, Lemma \ref{algebra-lemma-etale-at-prime}.
+Hence the annihilator of $z'$ in $R'_{\mathfrak p'}$ has radical equal to
+$\mathfrak p' R_{\mathfrak p'}$ (here we use $z'$ is not zero).
+On the other hand
+$$
+z' \in (M' \otimes_{R'} N')_{\mathfrak p'} =
+M'_{\mathfrak p'} \otimes_{R'_{\mathfrak p'}} N'_{\mathfrak p'}
+$$
+The module $N'_{\mathfrak p'}$ is projective over the local ring
+$R'_{\mathfrak p'}$ and hence free
+(Algebra, Theorem \ref{algebra-theorem-projective-free-over-local-ring}).
+Thus we can find a finite free direct summand $F' \subset N'_{\mathfrak p'}$
+such that $z' \in M'_{\mathfrak p'} \otimes_{R'_{\mathfrak p'}} F'$.
+If $F'$ has rank $n$, then we deduce that
+$\mathfrak p' R'_{\mathfrak p'} \in
+\text{WeakAss}_{R'_{\mathfrak p'}}({M'_{\mathfrak p'}}^{\oplus n})$.
+This implies
+$\mathfrak p'R'_{\mathfrak p'} \in \text{WeakAss}(M'_{\mathfrak p'})$
+for example by Algebra, Lemma \ref{algebra-lemma-weakly-ass}.
+Then $\mathfrak p' \in \text{WeakAss}_{R'}(M')$
+which in turn gives $\mathfrak p \in \text{WeakAss}_R(M)$.
+This finishes the proof of the implication
+``$\Rightarrow$'' of the equivalence of the lemma.
+
+\medskip\noindent
+Assume that $\mathfrak p \in \text{WeakAss}_R(M)$ and
+$\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$.
+We want to show that $\mathfrak q$ is weakly associated to $M \otimes_R N$.
+Note that $\overline{\mathfrak q}'$ is a maximal element of
+$\text{Ass}_{\overline{S}'}(\overline{N}')$.
+This is a consequence of (\ref{equation-key-observation})
+and the fact that there are no inclusions among the primes
+of $\overline{S}'$ lying over $\overline{\mathfrak q}$
+(as fibres of \'etale morphisms are discrete
+Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}).
+Thus, after replacing $R, S, \mathfrak p, \mathfrak q, M, N$ by
+$R', S', \mathfrak p', \mathfrak q', M', N'$
+we may assume, in addition to the assumptions of the lemma, that
+\begin{enumerate}
+\item $\mathfrak p \in \text{WeakAss}_R(M)$,
+\item $\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$,
+\item $N$ is projective as an $R$-module, and
+\item $\overline{\mathfrak q}$ is maximal in
+$\text{Ass}_{\overline{S}}(\overline{N})$.
+\end{enumerate}
+There is one more reduction, namely, we may replace
+$R, S, M, N$ by their localizations at $\mathfrak p$.
+This leads to one more condition, namely,
+\begin{enumerate}
+\item[(5)] $R$ is a local ring with maximal ideal $\mathfrak p$.
+\end{enumerate}
+We will finish by showing that (1) -- (5) imply
+$\mathfrak q \in \text{WeakAss}(M \otimes_R N)$.
 
 \medskip\noindent
-Assume that $\mathfrak m \in \text{WeakAss}_R(M)$ and
-$\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$
-is a maximal element.
-Let $y \in M$ be an element whose annihilator $I = \text{Ann}_R(y)$
-has radical $\mathfrak m$. Then $R/I \subset M$ and by flatness of $N$
-over $R$ we get $N/IN = R/I \otimes_R N \subset M \otimes_R N$. Hence
-it is enough to show that $\mathfrak q \in \text{WeakAss}(N/IN)$.
+Since $R$ is local and $\mathfrak p \in \text{WeakAss}_R(M)$
+we can pick a $y \in M$ whose annihilator $I$ has radical
+equal to $\mathfrak p$.
 Write $\overline{\mathfrak q} = (\overline{g}_1, \ldots, \overline{g}_n)$
-for some $\overline{g}_i \in \overline{S}$. Choose lifts
-$g_i \in \mathfrak q$. Consider the map
+for some $\overline{g}_i \in \overline{S}$. Choose $g_i \in S$
+mapping to $\overline{g}_i$.
+Then $\mathfrak q = \mathfrak pS + g_1S + \ldots + g_nS$.
+Consider the map
 $$
-\Psi : N/IN \longrightarrow N/IN^{\oplus n}, \quad
+\Psi : N/IN \longrightarrow (N/IN)^{\oplus n}, \quad
 z \longmapsto (g_1z, \ldots, g_nz).
 $$
-We may think of this as a map of free $R/I$-modules. As the ring
-$R/I$ is auto-associated (since $\mathfrak m/I$ is locally nilpotent)
-and since $\Psi \otimes R/\mathfrak m$ isn't injective (since
-$\overline{\mathfrak q} \in \text{Ass}(\overline{N})$) we see by
-More on Algebra, Lemma \ref{more-algebra-lemma-P-fPD-zero}
-that $\Psi$ isn't injective. Pick $z \in N/IN$ nonzero in the kernel
-of $\Psi$. The annihilator of $z$ contains $I$ and $g_i$, whence
-its radical $J = \sqrt{\text{Ann}_S(z)}$ contains $\mathfrak q$.
-Let $\mathfrak q' \supset J$ be a minimal prime over $J$.
-Then $\mathfrak q' \in \text{WeakAss}(M \otimes_R N)$ (by definition)
-and by (1) we see that
-$\overline{\mathfrak q}' \in \text{Ass}(\overline{N})$.
-Then since $\mathfrak q \subset \mathfrak q'$ by construction the
-maximality of $\overline{\mathfrak q}$ implies $\mathfrak q = \mathfrak q'$
-whence $\mathfrak q \in \text{WeakAss}(M \otimes_R N)$.
-This proves part (2) of the lemma.
+This is a homomorphism of projective $R/I$-modules.
+The local ring $R/I$ is auto-associated
+(More on Algebra, Definition \ref{more-algebra-definition-auto-ass})
+as $\mathfrak p/I$ is locally nilpotent.
+The map $\Psi \otimes \kappa(\mathfrak p)$ is not injective, because
+$\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$.
+Hence More on Algebra, Lemma \ref{more-algebra-lemma-P-fPD-zero}
+implies $\Psi$ is not injective. Pick $z \in N/IN$
+nonzero in the kernel of $\Psi$. The annihilator $J = \text{Ann}_S(z)$
+contains $IS$ and $g_i$ by construction. Thus
+$\sqrt{J} \subset S$ contains $\mathfrak q$.
+Let $\mathfrak s \subset S$ be a prime minimal over $J$.
+Then $\mathfrak q \subset \mathfrak s$,
+$\mathfrak s$ lies over $\mathfrak p$, and
+$\mathfrak s \in \text{WeakAss}_S(N/IN)$.
+The last fact by definition of weakly associated primes.
+Apply the ``$\Rightarrow$'' part of the lemma (which we've already proven)
+to the ring map $R \to S$ and the modules $R/I$ and $N$
+to conclude that
+$\overline{\mathfrak s} \in \text{Ass}_{\overline{S}}(\overline{N})$.
+Since $\overline{\mathfrak q} \subset \overline{\mathfrak s}$
+the maximality of $\overline{\mathfrak q}$, see condition (4) above,
+implies that $\overline{\mathfrak q} = \overline{\mathfrak s}$.
+This shows that $\mathfrak q = \mathfrak s$ and we conlude
+what we want.
 \end{proof}
 
 \begin{lemma}
@@ -3735,6 +3927,7 @@ \section{Flat finite type modules, Part II}
 \end{lemma}
 
 \begin{proof}
+In this paragraph we reduce to $f$ being of finite presentation.
 The question is local on $X$ and $S$, hence we may assume $X$ and $S$
 are affine. Write $X = \Spec(B)$, $S = \Spec(A)$ and write
 $B = A[x_1, \ldots, x_n]/I$. In other words we obtain a closed immersion
@@ -3755,71 +3948,40 @@ \section{Flat finite type modules, Part II}
 if $t \in \text{Ass}_{\mathbf{A}^n_s}((i_*\mathcal{F})_s)$ (see
 Algebra, Lemma \ref{algebra-lemma-ass-quotient-ring}).
 Hence it suffices to prove the lemma in case $X = \mathbf{A}^n_S$.
-In particular we may assume that $X \to S$ is of finite presentation.
+Thus we may assume that $X \to S$ is of finite presentation.
 
 \medskip\noindent
-Recall that $\text{Ass}_{X_s}(\mathcal{F}_s)$ is a locally finite subset
-of the locally Noetherian scheme $X_s$, see
-Divisors, Lemma \ref{divisors-lemma-finite-ass}.
-After replacing $X$ by a suitable affine neighbourhood of $x$ we may
-assume that
-\begin{itemize}
-\item[$(*)$] if $x' \in \text{Ass}_{X_s}(\mathcal{F}_s)$ and $x \leadsto x'$
-then $x = x'$.
-\end{itemize}
-(Proof omitted. Hint: using
-Algebra, Lemma \ref{algebra-lemma-silly}
-invert a function which does not vanish at $x$ but does vanish
-in all the finitely many points of $\text{Ass}_{X_s}(\mathcal{F}_s)$
-which are specializations of $x$ but not equal to $x$.)
-In words, no point of $\text{Ass}_{X_s}(\mathcal{F}_s)$
-is a proper specialization of $x$.
+In this paragraph we reduce to $\mathcal{F}$ being of finite presentation
+and flat over $S$.
+Choose an elementary \'etale neighbourhood $e : (S', s') \to (S, s)$
+and an open $V \subset X \times_S \Spec(\mathcal{O}_{S', s'})$
+as in Proposition \ref{proposition-finite-type-flat-at-point}.
+Let $x' \in X' = X \times_S S'$ be the unique point mapping to $x$
+and $s'$. Then it suffices to prove the statement for
+$X' \to S'$, $x'$, $s'$, $(X' \to X)^*\mathcal{F}$, and $e^*\mathcal{G}$, see
+Lemma \ref{lemma-etale-weak-assassin-up-down}.
+Let $v \in V$ the unique point mapping to $x'$
+and let $s' \in \Spec(\mathcal{O}_{S', s'})$ be the closed point.
+Then $\mathcal{O}_{V, v} = \mathcal{O}_{X', x'}$
+and $\mathcal{O}_{\Spec(\mathcal{O}_{S', s'}), s'} =
+\mathcal{O}_{S', s'}$ and similarly for the stalks of pullbacks of
+$\mathcal{F}$ and $\mathcal{G}$.
+Also $V_{s'} \subset X'_{s'}$ is an open subscheme.
+Since the condition of being a weakly associated point
+depend only on the stalk of the sheaf, we may
+replace
+$X' \to S'$, $x'$, $s'$, $(X' \to X)^*\mathcal{F}$, and $e^*\mathcal{G}$
+by
+$V \to \Spec(\mathcal{O}_{S', s'})$, $v$, $s'$, $(V \to X)^*\mathcal{F}$,
+and $(\Spec(\mathcal{O}_{S', s'}) \to S)^*\mathcal{G}$.
+Thus we may assume that $f$ is of finite presentation and
+$\mathcal{F}$ of finite presentation and flat over $S$.
 
 \medskip\noindent
-Suppose given a commutative diagram
-$$
-\xymatrix{
-(X, x) \ar[d] & (X', x') \ar[l]^g \ar[d] \\
-(S, s) & (S', s') \ar[l]_e
-}
-$$
-of pointed schemes whose horizontal arrows are elementary \'etale
-neighbourhoods. Then it suffices to prove the statement for
-$x'$, $s'$, $g^*\mathcal{F}$ and $e^*\mathcal{G}$, see
-Lemma \ref{lemma-etale-weak-assassin-up-down}.
-Note that property $(*)$ is preserved by such an \'etale localization
-by the same lemma (if there is a proper specialization
-$x' \leadsto x''$ on $X'_{s'}$ then this maps to a proper
-specialization on $X_s$ because the fibres of an \'etale morphism
-are discrete). We may also replace $S$ by the spectrum of its local ring
-as the condition of being an associated point of a quasi-coherent sheaf
-depends only on the stalk of the sheaf. Again property $(*)$ is
-preserved by this as well. Thus we may first apply
-Proposition \ref{proposition-finite-type-flat-at-point}
-to reduce to the case where $\mathcal{F}$ is of finite presentation
-and flat over $S$, whereupon we may use
-Proposition \ref{proposition-finite-presentation-flat-at-point}
-to reduce to the case that $X \to S$ is a morphism of affines
-and $\Gamma(X, \mathcal{F})$ is a finitely presented
-$\Gamma(X, \mathcal{O}_X)$-module which is projective as a
-$\Gamma(S, \mathcal{O}_S)$-module. Localizing $S$ once more we
-may assume that $\Gamma(S, \mathcal{O}_S)$ is a local ring such that
-$s$ corresponds to the maximal ideal. In this case
-Algebra, Theorem \ref{algebra-theorem-projective-free-over-local-ring}
-guarantees that $\Gamma(X, \mathcal{F})$ is free as an
-$\Gamma(S, \mathcal{O}_S)$-module. The implication
-$x \in \text{WeakAss}_X(\mathcal{F} \otimes_{\mathcal{O}_X} f^*\mathcal{G})
-\Rightarrow
-s \in \text{WeakAss}_S(\mathcal{G})
-\text{ and }
-x \in \text{Ass}_{X_s}(\mathcal{F}_s)$ follows from part (1) of
-Lemma \ref{lemma-weak-bourbaki-pre}.
-The converse implication follows from
-part (2) of Lemma \ref{lemma-weak-bourbaki-pre}
-as property $(*)$ insures that the prime corresponding
-to $x$ gives rise to a maximal element of
-$\text{Ass}_{\overline{S}}(\overline{N})$ exactly as in the statement
-of part (2) of
+Assume $f$ is of finite presentation and
+$\mathcal{F}$ of finite presentation and flat over $S$.
+After shrinking $X$ and $S$ to affine neighbourhoods
+of $x$ and $s$, this case is handled by
 Lemma \ref{lemma-weak-bourbaki-pre}.
 \end{proof}