Lemma 88.6.1. Let $I$ be an ideal of a Noetherian ring $A$. Let $r \geq 0$ and write $P = A[x_1, \ldots , x_ r]$ the $I$-adic completion. Consider a resolution
\[ P^{\oplus t} \xrightarrow {K} P^{\oplus m} \xrightarrow {g_1, \ldots , g_ m} P \to B \to 0 \]
of a quotient of $P$. Assume $B$ is rig-smooth over $(A, I)$. Then there exists an integer $n$ such that for any complex
\[ P^{\oplus t} \xrightarrow {K'} P^{\oplus m} \xrightarrow {g'_1, \ldots , g'_ m} P \]
with $g_ i - g'_ i \in I^ nP$ and $K - K' \in I^ n\text{Mat}(m \times t, P)$ there exists an isomorphism $B \to B'$ of $A$-algebras where $B' = P/(g'_1, \ldots , g'_ m)$.
Proof.
(A) By Definition 88.4.1 we can choose a $c \geq 0$ such that $I^ c$ annihilates $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ for all $B$-modules $N$.
(B) By More on Algebra, Lemmas 15.4.1 and 15.4.2 there exists a constant $c_1 = c(g_1, \ldots , g_ m, K)$ such that for $n \geq c_1 + 1$ the complex
\[ P^{\oplus t} \xrightarrow {K'} P^{\oplus m} \xrightarrow {g'_1, \ldots , g'_ m} P \to B' \to 0 \]
is exact and $\text{Gr}_ I(B) \cong \text{Gr}_ I(B')$.
(C) Let $d_0 = d(\text{Gr}_ I(B))$ and $q_0 = q(\text{Gr}_ I(B))$ be the integers found in Local Cohomology, Section 51.22.
We claim that $n = \max (c_1 + 1, q_0 + (d_0 + 1)c, 2(d_0 + 1)c + 1)$ works where $c$ is as in (A), $c_1$ is as in (B), and $q_0, d_0$ are as in (C).
Let $g'_1, \ldots , g'_ m$ and $K'$ be as in the lemma. Since $g_ i = g'_ i \in I^ nP$ we obtain a canonical $A$-algebra homomorphism
\[ \psi _ n : B \longrightarrow B'/I^ nB' \]
which induces an isomorphism $B/I^ nB \to B'/I^ nB'$. Since $\text{Gr}_ I(B) \cong \text{Gr}_ I(B')$ we have $d_0 = d(\text{Gr}_ I(B'))$ and $q_0 = q(\text{Gr}_ I(B'))$ and since $n \geq \max (q_0 + (1 + d_0)c, 2(d_0 + 1)c + 1)$ we may apply Lemma 88.5.3 to find an $A$-algebra homomorphism
\[ \varphi : B \longrightarrow B' \]
such that $\varphi \bmod I^{n - (d_0 + 1)c}B' = \psi _ n \bmod I^{n - (d_0 + 1)c}B'$. Since $n - (d_0 + 1)c > 0$ we see that $\varphi $ is an $A$-algebra homomorphism which modulo $I$ induces the isomorphism $B/IB \to B'/IB'$ we found above. The rest of the proof shows that these facts force $\varphi $ to be an isomorphism; we suggest the reader find their own proof of this.
Namely, it follows that $\varphi $ is surjective for example by applying Algebra, Lemma 10.96.1 part (1) using the fact that $B$ and $B'$ are complete. Thus $\varphi $ induces a surjection $\text{Gr}_ I(B) \to \text{Gr}_ I(B')$ which has to be an isomorphism because the source and target are isomorphic Noetherian rings, see Algebra, Lemma 10.31.10 (of course you can show $\varphi $ induces the isomorphism we found above but that would need a tiny argument). Thus $\varphi $ induces injective maps $I^ eB/I^{e + 1}B \to I^ eB'/I^{e + 1}B'$ for all $e \geq 0$. This implies $\varphi $ is injective since for any $b \in B$ there exists an $e \geq 0$ such that $b \in I^ eB$, $b \not\in I^{e + 1}B$ by Krull's intersection theorem (Algebra, Lemma 10.51.4). This finishes the proof.
$\square$
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