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 $\overline{a}' \in \overline{\mathfrak q}'$ 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 $a' : N'_{\mathfrak p'} \to N'_{\mathfrak p'}$ is $R'_{\mathfrak p'}$-universally injective. In particular we see that $a' : M' \otimes_{R'} N' \to M' \otimes_{R'} N'$ is injective after localizing 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 $\mathfrak q \in \text{WeakAss}_S(M \otimes_R N)$. 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 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 $g_i \in S$ mapping to $\overline{g}_i$. Then $\mathfrak q = \mathfrak pS + g_1S + \ldots + g_nS$. Consider the map $$ \Psi : N/IN \longrightarrow (N/IN)^{\oplus n}, \quad z \longmapsto (g_1z, \ldots, g_nz). $$ 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.