The Stacks project

Slight generalization of [Lemme 3(a), Beauville-Laszlo]

Lemma 15.90.15. Let $(R \to R', f)$ be a glueing pair. For every $R$-module $M$, we have $\text{Tor}^ R_1(R', \mathop{\mathrm{Coker}}(M \to M_ f)) = 0$.

Proof. Set $\overline{M} = M/M[f^\infty ]$. Then $\mathop{\mathrm{Coker}}(M \to M_ f) \cong \mathop{\mathrm{Coker}}(\overline{M} \to \overline{M}_ f)$ hence we may and do assume that $f$ is a nonzerodivisor on $M$. In this case $M \subset M_ f$ and $M_ f/M = \mathop{\mathrm{colim}}\nolimits M/f^ nM$ where the transition maps are given by multiplication by $f$. Since formation of Tor groups commutes with colimits (Algebra, Lemma 10.76.2) it suffices to show that $\text{Tor}^ R_1(R', M/f^ n M) = 0$.

We first treat the case $M = R/R[f^\infty ]$. By Lemma 15.90.6 we have $M \otimes _ R R' = R'/R'[f^\infty ]$. From the short exact sequence $0 \to M \to M \to M/f^ nM \to 0$ we obtain the exact sequence

\[ \xymatrix{ \text{Tor}_1^ R(R', R/R[f^\infty ]) \ar[r] & \text{Tor}_1^ R(R', M/f^ n M) \ar[r] & R'/R'[f^\infty ] \ar[dll]_{f^ n} \\ R'/R'[f^\infty ] \ar[r] & (R'/R'[f^\infty ])/(f^ n (R'/R'[f^\infty ])) \ar[r] & 0 } \]

by Algebra, Lemma 10.75.2. Here the diagonal arrow is injective. Since the first group $\text{Tor}_1^ R(R', R/R[f^\infty ])$ is zero by Lemma 15.90.14, we deduce that $\text{Tor}_1^ R(R', M/f^ nM) = 0$ as desired.

To treat the general case, choose a surjection $F \to M$ with $F$ a free $R/R[f^\infty ]$-module, and form an exact sequence

\[ 0 \to N \to F/f^ n F \to M/f^ n M \to 0. \]

By Lemma 15.88.8 this sequence remains unchanged, and hence exact, upon tensoring with $R'$. Since $\text{Tor}^ R_1(R', F/f^ n F) = 0$ by the previous paragraph, we deduce that $\text{Tor}^ R_1(R', M/f^ n M) = 0$ as desired. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 15.90: The Beauville-Laszlo theorem

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BP1. Beware of the difference between the letter 'O' and the digit '0'.