Let $z \in M \otimes_k K$ be an element such that $\mathfrak q$ is minimal over the annihilator $J \subset R \otimes_k K$ of $z$. Choose a finitely generated subextension $K/L/k$ such that $z \in M \otimes_k L$. Since $R \otimes_k L \to R \otimes_k K$ is flat we see that $J = I(R \otimes_k K)$ where $I \subset R \otimes_k L$ is the annihilator of $z$ in the smaller ring (Lemma \ref{lemma-annihilator-flat-base-change}). Thus $\mathfrak q \cap (R \otimes_k L)$ is minimal over $I$ by going down (Lemma \ref{lemma-flat-going-down}). In this way we reduce to the case described in the next paragraph. \medskip\noindent Assume $K/k$ is a finitely generated field extension. Let $x_1, \ldots, x_r \in K$ be a transcendence basis of $K$ over $k$, see Fields, Section \ref{fields-section-transcendence}. Set $L = k(x_1, \ldots, x_r)$. Say $[K : L] = n$. Then $R \otimes_k L \to R \otimes_k K$ is a finite ring map. Hence $\mathfrak q \cap (R \otimes_k L)$ is a weakly associated prime of $M \otimes_k K$ viewed as a $R \otimes_k L$-module by Lemma \ref{lemma-weakly-ass-finite-ring-map}. Since $M \otimes_k K \cong (M \otimes_k L)^{\oplus n}$ as a $R \otimes_k L$-module, we see that $\mathfrak q \cap (R \otimes_k L)$ is a weakly associated prime of $M \otimes_k L$ (for example by using Lemma \ref{lemma-weakly-ass} and induction). In this way we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $K = k(x_1, \ldots, x_r)$ is a purely transcendental field extension. We may replace $R$ by $R_\mathfrak p$, $M$ by $M_\mathfrak p$ and $\mathfrak q$ by $\mathfrak q(R_\mathfrak p \otimes_k K)$. See Lemma \ref{lemma-localize-weakly-ass}. In this way we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $K = k(x_1, \ldots, x_r)$ is a purely transcendental field extension and $R$ is local with maximal ideal $\mathfrak p$. We claim that any $f \in R \otimes_k K$, $f \not \in \mathfrak p(R \otimes_k K)$ is a nonzerodivisor on $M \otimes_k K$. Namely, let $z \in M \otimes_k K$ be an element. There is a finite $R$-submodule $M' \subset M$ such that $z \in M' \otimes_k K$ and such that $M'$ is minimal with this property: choose a basis $\{t_\alpha\}$ of $K$ as a $k$-vector space, write $z = \sum m_\alpha \otimes t_\alpha$ and let $M'$ be the $R$-submodule generated by the $m_\alpha$. If $z \in \mathfrak p(M' \otimes_k K) = \mathfrak p M' \otimes_k K$, then $\mathfrak pM' = M'$ and $M' = 0$ by Lemma \ref{lemma-NAK} a contradiction. Thus $z$ has nonzero image $\overline{z}$ in $M'/\mathfrak p M' \otimes_k K$ But $R/\mathfrak p \otimes_k K$ is a domain as a localization of $\kappa(\mathfrak p)[x_1, \ldots, x_n]$ and $M'/\mathfrak p M' \otimes_k K$ is a free module, hence $f\overline{z} \not = 0$. This proves the claim. \medskip\noindent Finally, pick $z \in M \otimes_k K$ such that $\mathfrak q$ is minimal over the annihilator $J \subset R \otimes_k K$ of $z$. For $f \in \mathfrak p$ there exists an $n \geq 1$ and a $g \in R \otimes_k K$, $g \not \in \mathfrak q$ such that $g f^n z \in J$, i.e., $g f^n z = 0$. (This holds because $\mathfrak q$ lies over $\mathfrak p$ and $\mathfrak q$ is minimal over $J$.) Above we have seen that $g$ is a nonzerodivisor hence $f^n z = 0$. This means that $\mathfrak p$ is a weakly associated prime of $M \otimes_k K$ viewed as an $R$-module. Since $M \otimes_k K$ is a direct sum of copies of $M$ we conclude that $\mathfrak p$ is a weakly associated prime of $M$ as before.