The Stacks project

Proof. 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 10.40.4). Thus $\mathfrak q \cap (R \otimes _ k L)$ is minimal over $I$ by going down (Lemma 10.39.19). In this way we reduce to the case described in the next paragraph.

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 9.26. 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 10.66.13. 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 10.66.4 and induction). In this way we reduce to the case discussed in the next paragraph.

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 10.66.15. In this way we reduce to the case discussed in the next paragraph.

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 10.20.1 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.

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. $\square$