Processing math: 100%

The Stacks project

Lemma 88.5.4. Let I = (a) be a principal ideal of a Noetherian ring A. Let B be an object of (88.2.0.2). Assume given an integer c \geq 0 such that \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N) is annihilated by a^ c for all B-modules N. Let C be an I-adically complete A-algebra such that a is a nonzerodivisor on C. Let n > 2c. For any A-algebra map \psi _ n : B \to C/a^ nC there exists an A-algebra map \varphi : B \to C such that \psi _ n \bmod a^{n - c}C = \varphi \bmod a^{n - c}C.

Proof. Consider the obstruction class

o(\psi _ n) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , a^ nC/a^{2n}C)

of Remark 88.5.2. Since a is a nonzerodivisor on C the map a^ c : a^ nC/a^{2n}C \to a^ nC/a^{2n}C is isomorphic to the map a^ nC/a^{2n}C \to a^{n - c}C/a^{2n - c}C in the category of C-modules. Hence by our assumption on \mathop{N\! L}\nolimits _{B/A}^\wedge we conclude that the class o(\psi _ n) maps to zero in

\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , a^{n - c}C/a^{2n - c}C)

and a fortiori in

\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , a^{n - c}C/a^{2n - 2c}C)

By the discussion in Remark 88.5.2 we obtain a map

\psi _{2n - 2c} : B \to C/a^{2n - 2c}C

which agrees with \psi _ n modulo a^{n - c}C. Observe that 2n - 2c > n because n > 2c.

We may repeat this procedure. Starting with n_0 = n and \psi ^0 = \psi _ n we end up getting a strictly increasing sequence of integers

n_0 < n_1 < n_2 < \ldots

and A-algebra homorphisms \psi ^ i : B \to C/a^{n_ i}C such that \psi ^{i + 1} and \psi ^ i agree modulo a^{n_ i - c}C. Since C is I-adically complete we can take \varphi to be the limit of the maps \psi ^ i \bmod a^{n_ i - c}C : B \to C/a^{n_ i - c}C and the lemma follows. \square


Comments (0)


Your email address will not be published.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.