Lemma 15.90.18. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module which is not necessarily glueable for $(R \to R', f)$. Then $M$ is flat over $R$ if and only if $M \otimes _ R R'$ is flat over $R'$ and $M_ f$ is flat over $R_ f$.
Proof. One direction of the lemma follows from Algebra, Lemma 10.39.7. For the other direction, assume $M \otimes _ R R'$ is flat over $R'$ and $M_ f$ is flat over $R_ f$. Let $\tilde M$ be as in Remark 15.90.17. If $\tilde M$ is flat over $R$, then applying Algebra, Lemma 10.39.12 to the short exact sequence $0 \to \mathop{\mathrm{Ker}}(M \to \tilde M) \to M \to \tilde M \to 0$ we find that $\mathop{\mathrm{Ker}}(M \to \tilde M) \otimes _ R (R' \oplus R_ f)$ is zero. Hence $M = \tilde M$ by Lemma 15.90.2 and we conclude. In other words, we may replace $M$ by $\tilde M$ and assume $M$ is glueable for $(R \to R', f)$. Let $N$ be a second $R$-module. It suffices to prove that $\text{Tor}_1^ R(M, N) = 0$, see Algebra, Lemma 10.75.8.
The long the exact sequence of Tors associated to the short exact sequence $0 \to R \to R' \oplus R_ f \to (R')_ f \to 0$ and $N$ gives an exact sequence
and isomorphisms $\text{Tor}_ i^ R(R', N) = \text{Tor}_ i^ R((R')_ f, N)$ for $i \geq 2$. Since $\text{Tor}_ i^ R((R')_ f, N) = \text{Tor}_ i^ R(R', N)_ f$ we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^ R(R', N)$ and invertible on $\text{Tor}_ i^ R(R', N)$ for $i \geq 2$. Since $M \otimes _ R R'$ is flat over $R'$ we have
by the spectral sequence of Example 15.62.2. Writing $M \otimes _ R R'$ as a filtered colimit of finite free $R'$-modules (Algebra, Theorem 10.81.4) we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^ R(M \otimes _ R R', N)$ and invertible on $\text{Tor}_ i^ R(M \otimes _ R R', N)$. Next, we consider the exact sequence $0 \to M \to M \otimes _ R R' \oplus M_ f \to M \otimes _ R (R')_ f \to 0$ coming from the fact that $M$ is glueable and the associated long exact sequence of $\text{Tor}$. The relevant part is
We conclude that $\text{Tor}_1^ R(M, N) = 0$ by our remarks above on the action on $f$ on $\text{Tor}_ i^ R(M \otimes _ R R', N)$. $\square$
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Your email address will not be published.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)
There are also: