**Proof.**
We prove this lemma by induction on the number of generators of $I$. Say $I = (a_1, \ldots , a_ t)$. If $t = 0$, then $I = 0$ and there is nothing to prove. If $t = 1$, then the lemma follows from Lemma 86.6.3 because $2\max (c, d) \geq \max (2c, c + d)$. Assume $t > 1$.

Set $m = \lfloor \frac{n}{t} \rfloor $ as in the lemma. Set $\bar A = A/(a_ t^ m)$. Consider the ideal $\bar I = (\bar a_1, \ldots , \bar a_{t - 1})$ in $\bar A$. Set $\bar C = C/(a_ t^ m)$. Note that $\bar C$ is a $\bar I$-adically complete Noetherian $\bar A$-algebra (use Algebra, Lemmas 10.96.1 and 10.95.9). Let $\bar d$ be the integer for $\bar I \subset \bar A \to \bar C$ which exists by induction hypothesis.

Let $d_1 \geq 0$ be an integer such that $C[a_ t^\infty ] \cap a_ t^{d_1}C = 0$ as in Lemma 86.6.3 (see discussion following the lemma and before the proof).

We claim the lemma holds with $d = \max (\bar d, d_1)$. To see this, let $c, B, n, \psi _ n$ be as in the lemma.

Note that $\bar I \subset I\bar A$. Hence by Lemma 86.4.3 multiplication by an element of $\bar I^ c$ on the cotangent complex of $\bar B = B/(a_ t^ m)$ is zero in $D(\bar B)$. Also, we have

\[ \bar I^{n - m + 1} \supset I^ n \bar A \]

Thus $\psi _ n$ gives rise to a map

\[ \bar\psi _{n - m + 1} : \bar B/\bar I^{n - m + 1}\bar B \longrightarrow \bar C/\bar I^{n - m + 1}\bar C \]

Since $n > 2t\max (c, d)$ and $d \geq \bar d$ we see that

\[ n - m + 1 \geq (t - 1)n/t > 2(t - 1)\max (c, d) \geq 2(t - 1)\max (c, \bar d) \]

Hence we can find a morphism $\varphi _ m : \bar B \to \bar C$ agreeing with $\bar\psi _{n - m + 1}$ modulo the ideal $\bar I^{m' - c}$ where $m' = \lfloor \frac{n - m + 1}{t - 1} \rfloor $.

Since $m \geq n/t > 2\max (c, d) \geq 2\max (c, d_1) \geq \max (2c, c+ d_1)$, we can apply Lemma 86.6.3 for the ring map $A \to B$ and the ideal $(a_ t)$ to find a morphism $\varphi : B \to C$ agreeing modulo $a_ t^{m - c}$ with $\varphi _ m$.

All in all we find $\varphi : B \to C$ which agrees with $\psi _ n$ modulo

\[ (a_ t^{m - c}) + (a_1, \ldots , a_{t - 1})^{m' - c} \subset I^{\min (m - c, m' - c)} \]

We leave it to the reader to see that $\min (m - c, m' - c) = m - c$. This concludes the proof.
$\square$

## Comments (1)

Comment #5883 by Johan on