Lemma 15.18.1. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Let $R \to R'$ be a ring map and $IR' \subset I' \subset R'$ an ideal. If (15.18.0.1) holds for $(R \to S, I, M)$, then (15.18.0.1) holds for $(R' \to S \otimes _ R R', I', M \otimes _ R R')$.
Proof. Assume (15.18.0.1) holds for $(R \to S, I \subset R, M)$. Let $I'(S \otimes _ R R') \subset \mathfrak q'$ be a prime of $S \otimes _ R R'$. Let $\mathfrak q \subset S$ be the corresponding prime of $S$. Then $IS \subset \mathfrak q$. Note that $(M \otimes _ R R')_{\mathfrak q'}$ is a localization of the base change $M_{\mathfrak q} \otimes _ R R'$. Hence $(M \otimes _ R R')_{\mathfrak q'}$ is flat over $R'$ as a localization of a flat module, see Algebra, Lemmas 10.39.7 and 10.39.18. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)