Lemma 86.7.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $t$ be the minimal number of generators for $I$. Let $C$ be a Noetherian $I$-adically complete $A$-algebra. There exists an integer $d \geq 0$ depending only on $I \subset A \to C$ with the following property: given

1. $c \geq 0$ and $B$ in (86.2.0.2) such that for $a \in I^ c$ multiplication by $a$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$,

2. an integer $n > 2t\max (c, d)$,

3. an $A/I^ n$-algebra map $\psi _ n : B/I^ nB \to C/I^ nC$,

there exists a map $\varphi : B \to C$ of $A$-algebras such that $\psi _ n \bmod I^{m - c} = \varphi \bmod I^{m - c}$ with $m = \lfloor \frac{n}{t} \rfloor$.

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. $\square$

Comment #5883 by on

The proof of this lemma is wrong unfortunately. Please refrain from using this lemma for now.

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).