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

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

and a fortiori in

By the discussion in Remark 86.5.2 we obtain a map

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

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. $\square$

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