The Stacks project

Lemma 15.88.9. Assume $\varphi : R \to S$ is a flat ring map and $I = (f_1, \ldots , f_ t) \subset R$ is an ideal such that $R/I \to S/IS$ is an isomorphism. Let $M$ be an $R$-module. Then the complex (15.88.8.1) is exact.

Proof. First proof. Denote $\check{\mathcal{C}}_ R \to \check{\mathcal{C}}_ S$ the quasi-isomorphism of extended alternating Čech complexes of Lemma 15.88.4. Since these complexes are bounded with flat terms, we see that $M \otimes _ R \check{\mathcal{C}}_ R \to M \otimes _ R \check{\mathcal{C}}_ S$ is a quasi-isomorphism too (Lemmas 15.58.9 and 15.58.14). Now the complex (15.88.8.1) is a truncation of the cone of the map $M \otimes _ R \check{\mathcal{C}}_ R \to M \otimes _ R \check{\mathcal{C}}_ S$ and we win.

Second computational proof. Let $m \in M$. If $\alpha (m) = 0$, then $m \in M[I^\infty ]$, see Lemma 15.87.3. Pick $n$ such that $I^ n m = 0$ and consider the map $\varphi : R/I^ n \to M$. If $m \otimes 1 = 0$, then $\varphi \otimes 1_ S = 0$, hence $\varphi = 0$ (see Lemma 15.88.3) hence $m = 0$. In this way we see that $\alpha $ is injective.

Let $(m', m'_1, \ldots , m'_ t) \in \mathop{\mathrm{Ker}}(\beta )$. Write $m'_ i = m_ i/f_ i^ n$ for some $n > 0$ and $m_ i \in M$. We may, after possibly enlarging $n$ assume that $f_ i^ n m' = m_ i \otimes 1$ in $M \otimes _ R S$ and $f_ j^ nm_ i - f_ i^ nm_ j = 0$ in $M$. In particular we see that $(m_1, \ldots , m_ t)$ defines an element $\xi $ of $H_1(M, (f_1^ n, \ldots , f_ t^ n))$. Since $H_1(M, (f_1^ n, \ldots , f_ t^ n))$ is annihilated by $I^{tn + 1}$ (see Lemma 15.88.7) and since $R \to S$ is flat we see that

\[ H_1(M, (f_1^ n, \ldots , f_ t^ n)) = H_1(M, (f_1^ n, \ldots , f_ t^ n)) \otimes _ R S = H_1(M \otimes _ R S, (f_1^ n, \ldots , f_ t^ n)) \]

by Lemma 15.88.2 The existence of $m'$ implies that $\xi $ maps to zero in the last group, i.e., the element $\xi $ is zero. Thus there exists an $m \in M$ such that $m_ i = f_ i^ n m$. Then $(m', m'_1, \ldots , m'_ t) - \alpha (m) = (m'', 0, \ldots , 0)$ for some $m'' \in (M \otimes _ R S)[(IS)^\infty ]$. By Lemma 15.88.3 we conclude that $m'' \in M[I^\infty ]$ and we win. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.88: Formal glueing of module categories

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Your email address will not be published.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 05EK. Beware of the difference between the letter 'O' and the digit '0'.