## 86.6 Algebraization of rig-smooth algebras over G-rings

If the base ring $A$ is a Noetherian G-ring, then we can prove [III Theorem 7, Elkik] for arbitrary rig-smooth algebras with respect to any ideal $I \subset A$ (not necessarily principal).

Lemma 86.6.1. 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 86.4.1 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 86.5.3 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. $\square$

Lemma 86.6.2. Let $I$ be an ideal of a Noetherian ring $A$. Let $C^ h$ be the henselization of a finite type $A$-algebra $C$ with respect to the ideal $IC$. Let $J \subset C^ h$ be an ideal. Then there exists a finite type $A$-algebra $B$ such that $B^\wedge \cong (C^ h/J)^\wedge$.

Proof. By More on Algebra, Lemma 15.12.4 the ring $C^ h$ is Noetherian. Say $J = (g_1, \ldots , g_ m)$. The ring $C^ h$ is a filtered colimit of étale $C$ algebras $C'$ such that $C/IC \to C'/IC'$ is an isomorphism (see proof of More on Algebra, Lemma 15.12.1). Pick an $C'$ such that $g_1, \ldots , g_ m$ are the images of $g'_1, \ldots , g'_ m \in C'$. Setting $B = C'/(g'_1, \ldots , g'_ m)$ we get a finite type $A$-algebra. Of course $(C, IC)$ and $C', IC')$ have the same henselizations and the same completions. It follows easily from this that $B^\wedge = (C^ h/J)^\wedge$. $\square$

Proposition 86.6.3. Let $I$ be an ideal of a Noetherian G-ring $A$. Let $B$ be an object of (86.2.0.2). If $B$ is rig-smooth over $(A, I)$, then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge$ of $A$-algebras.

Proof. Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Write $P = A[x_1, \ldots , x_ r]^\wedge$. Choose generators $g_1, \ldots , g_ m \in J$. Choose generators $k_1, \ldots , k_ t$ of the module of relations between $g_1, \ldots , g_ m$, i.e., such that

$P^{\oplus t} \xrightarrow {k_1, \ldots , k_ t} P^{\oplus m} \xrightarrow {g_1, \ldots , g_ m} P \to B \to 0$

is a resolution. Write $k_ i = (k_{i1}, \ldots , k_{im})$ so that we have

86.6.3.1
$$\label{restricted-equation-relations-straight-up} \sum \nolimits _ j k_{ij}g_ j = 0$$

for $i = 1, \ldots , t$. Denote $K = (k_{ij})$ the $m \times t$-matrix with entries $k_{ij}$.

Let $A[x_1, \ldots , x_ r]^ h$ be the henselization of the pair $(A[x_1, \ldots , x_ r], IA[x_1, \ldots , x_ r])$, see More on Algebra, Lemma 15.12.1. We may and do think of $A[x_1, \ldots , x_ r]^ h$ as a subring of $P = A[x_1, \ldots , x_ r]^\wedge$, see More on Algebra, Lemma 15.12.4. Since $A$ is a Noetherian G-ring, so is $A[x_1, \ldots , x_ r]$, see More on Algebra, Proposition 15.50.10. Hence we have approximation for the map $A[x_1, \ldots , x_ r]^ h \to A[x_1, \ldots , x_ r]^\wedge = P$ with respect to the ideal generated by $I$, see Smoothing Ring Maps, Lemma 16.14.1. Choose a large enough integer $n$ as in Lemma 86.6.1. By the approximation property we may choose $g'_1, \ldots , g'_ m \in A[x_1, \ldots , x_ r]^ h$ and a matrix $K' = (k'_{ij}) \in \text{Mat}(m \times t, A[x_1, \ldots , x_ r]^ h)$ such that $\sum \nolimits _ j k'_{ij}g'_ j = 0$ in $A[x_1, \ldots , x_ r]^ h$ and such that $g_ i - g'_ i \in I^ nP$ and $K - K' \in I^ n\text{Mat}(m \times t, P)$. By our choice of $n$ we conclude that there is an isomorphism

$B \to P/(g'_1, \ldots , g'_ m) = \left(A[x_1, \ldots , x_ r]^ h/(g'_1, \ldots , g'_ m)\right)^\wedge$

This finishes the proof by Lemma 86.6.2. $\square$

The following lemma isn't true in general if $A$ is not a G-ring but just Noetherian. Namely, if $(A, \mathfrak m)$ is local and $I = \mathfrak m$, then the lemma is equivalent to Artin approximation for $A^ h$ (as in Smoothing Ring Maps, Theorem 16.13.1) which does not hold for every Noetherian local ring.

Lemma 86.6.4. Let $A$ be a Noetherian G-ring. Let $I \subset A$ be an ideal. Let $B, C$ be finite type $A$-algebras. For any $A$-algebra map $\varphi : B^\wedge \to C^\wedge$ of $I$-adic completions and any $N \geq 1$ there exist

1. an étale ring map $C \to C'$ which induces an isomorphism $C/IC \to C'/IC'$,

2. an $A$-algebra map $\varphi : B \to C'$

such that $\varphi$ and $\psi$ agree modulo $I^ N$ into $C^\wedge = (C')^\wedge$.

Proof. The statement of the lemma makes sense as $C \to C'$ is flat (Algebra, Lemma 10.143.3) hence induces an isomorphism $C/I^ nC \to C'/I^ nC'$ for all $n$ (More on Algebra, Lemma 15.88.2) and hence an isomorphism on completions. Let $C^ h$ be the henselization of the pair $(C, IC)$, see More on Algebra, Lemma 15.12.1. Then $C^ h$ is the filtered colimit of the algebras $C'$ and the maps $C \to C' \to C^ h$ induce isomorphism on completions (More on Algebra, Lemma 15.12.4). Thus it suffices to prove there exists an $A$-algebra map $B \to C^ h$ which is congruent to $\psi$ modulo $I^ N$. Write $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. The ring map $\psi$ corresponds to elements $\hat c_1, \ldots , \hat c_ n \in C^\wedge$ with $f_ j(\hat c_1, \ldots , \hat c_ n) = 0$ for $j = 1, \ldots , m$. Namely, as $A$ is a Noetherian G-ring, so is $C$, see More on Algebra, Proposition 15.50.10. Thus Smoothing Ring Maps, Lemma 16.14.1 applies to give elements $c_1, \ldots , c_ n \in C^ h$ such that $f_ j(c_1, \ldots , c_ n) = 0$ for $j = 1, \ldots , m$ and such that $\hat c_ i - c_ i \in I^ NC^ h$. This determines the map $B \to C^ h$ as desired. $\square$

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