Lemma 86.5.4. Let $I = (a)$ be a principal ideal of a Noetherian ring $A$. Let $B$ be an object of (86.2.0.2). 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 86.5.2. 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 86.5.2 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. $\square$

## 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Your email address will not be published.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 0AK6. Beware of the difference between the letter 'O' and the digit '0'.