From 95fb6c6ec6ae0991386d8ea5cf8669bc507b95e5 Mon Sep 17 00:00:00 2001 From: Aise Johan de Jong Date: Fri, 21 Oct 2016 20:47:52 -0400 Subject: [PATCH] Fix snafu in flat.tex The proof of a small part of lemma-weak-bourbaki-pre was erroneous... The problem was in the sentence "Let $z \in M \otimes_R N$ be an element whose annihilator in $S$ has radical $\mathfrak q$." In general weakly associated primes don't automatically have this property, only after you localize does it become true. To work around it we invented a trick where you choose a z such that \mathfrak q is minimal over its annihilator and then you look at what happens after you map z up into a suitably chosen overring (using the machinery of RG). If you look at the proof of the corresponding fact in RG then you don't find any explanation of how to deal with this whatsoever, so this could mean there is a much simpler way to deal with it. Anyway, I am quite convinced the current argument is correct, so I will leave it like this for now. --- flat.tex | 404 ++++++++++++++++++++++++++++++++++++++----------------- 1 file changed, 283 insertions(+), 121 deletions(-) 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\section{Flat finite type modules, Part II} \label{section-finite-type-flat-II} \noindent -The following lemma will be superseded by the stronger +We will need the following lemma.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}