Lemma 86.10.1. Let $A$ be a Noetherian ring and $I = (a)$ a principal ideal. Let $B$ be an object of (86.2.0.2) which is rig-étale over $(A, I)$. Then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge $.

## 86.10 Algebraization of rig-étale algebras

The main goal is to prove algebraization for rig-étale algebras when the underlying Noetherian ring $A$ is not assumed to be a G-ring and when the ideal $I \subset A$ is arbitrary – not necessarily principal. We first prove the principal ideal case and then use the result of Section 86.9 to finish the proof.

**Proof.**
Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. By Lemma 86.8.2 part (6) we can find $c \geq 0$ and $f_1, \ldots , f_ r \in J$ such that $\det _{1 \leq i, j \leq r}(\partial f_ j/\partial x_ i)$ divides $a^ c$ in $B$ and $a^ c J \subset (f_1, \ldots , f_ r) + J^2$. Hence Lemma 86.7.2 applies. This finishes the proof, but we'd like to point out that in this case the use of Lemma 86.5.3 can be replaced by the much easier Lemma 86.5.5.
$\square$

Lemma 86.10.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2) which is rig-étale over $(A, I)$. Then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge $.

**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 86.10.1. Assume $t > 1$.

For any $m \geq 1$ set $\bar A_ m = A/(a_ t^ m)$. Consider the ideal $\bar I_ m = (\bar a_1, \ldots , \bar a_{t - 1})$ in $\bar A_ m$. Observe that $V(I \bar A_ m) = V(\bar I_ m)$. Let $B_ m = B/(a_ t^ m)$ be the base change of $B$ for the map $(A, I) \to (\bar A_ m, \bar I_ m)$, see Remark 86.2.4. By Lemma 86.8.6 we find that $B_ m$ is rig-étale over $(\bar A_ m, \bar I_ m)$.

By induction hypothesis (on $t$) we can find a finite type $\bar A_ m$-algebra $C_ m$ and a map $C_ m \to B_ m$ which induces an isomorphism $C_ m^\wedge \cong B_ m$ where the completion is with respect to $\bar I_ m$. By Lemma 86.8.4 we may assume that $\mathop{\mathrm{Spec}}(C_ m) \to \mathop{\mathrm{Spec}}(\bar A_ m)$ is étale over $\mathop{\mathrm{Spec}}(\bar A_ m) \setminus V(\bar I_ m)$.

We claim that we may choose $A_ m \to C_ m \to B_ m$ as in the previous paragraph such that moreover there are isomorphisms $C_ m/(a_ t^{m - 1}) \to C_{m - 1}$ compatible with the given $A$-algebra structure and the maps to $B_{m - 1} = B_ m/(a_ t^{m - 1})$. Namely, first fix a choice of $A_1 \to C_1 \to B_1$. Suppose we have found $C_{m - 1} \to C_{m - 2} \to \ldots \to C_1$ with the desired properties. Note that $C_ m/(a_ t^{m - 1})$ is étale over $\mathop{\mathrm{Spec}}(\bar A_{m - 1}) \setminus V(\bar I_{m - 1})$. Hence by Lemma 86.8.7 there exists an étale extension $C_{m - 1} \to C'_{m - 1}$ which induces an isomorphism modulo $\bar I_{m - 1}$ and an $\bar A_{m - 1}$-algebra map $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ inducing the isomorphism $B_ m/(a_ t^{m - 1}) \to B_{m - 1}$ on completions. Note that $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is étale over the complement of $V(\bar I_{m - 1})$ by Morphisms, Lemma 29.36.18 and over $V(\bar I_{m - 1})$ induces an isomorphism on completions hence is étale there too (for example by More on Morphisms, Lemma 37.12.3). Thus $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is étale. By the topological invariance of étale morphisms (Étale Morphisms, Theorem 41.15.2) there exists an étale ring map $C_ m \to C'_ m$ such that $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is isomorphic to $C_ m/(a_ t^{m - 1}) \to C'_ m/(a_ t^{m - 1})$. Observe that the $\bar I_ m$-adic completion of $C'_ m$ is equal to the $\bar I_ m$-adic completion of $C_ m$, i.e., to $B_ m$ (details omitted). We apply Lemma 86.9.1 to the diagram

to see that there exists a “lift” of $C''_ m$ of $C_{m - 1}$ to an algebra over $\bar A_ m$ with all the desired properties.

By construction $(C_ m)$ is an object of the category (86.2.0.1) for the principal ideal $(a_ t)$. Thus the inverse limit $B' = \mathop{\mathrm{lim}}\nolimits C_ m$ is an $(a_ t)$-adically complete $A$-algebra such that $B'/a_ t B'$ is of finite type over $A/(a_ t)$, see Lemma 86.2.1. By construction the $I$-adic completion of $B'$ is isomorphic to $B$ (details omitted). Consider the complex $\mathop{N\! L}\nolimits _{B'/A}^\wedge $ constructed using the $(a_ t)$-adic topology. Choosing a presentation for $B'$ (which induces a similar presentation for $B$) the reader immediately sees that $\mathop{N\! L}\nolimits _{B'/A}^\wedge \otimes _{B'} B = \mathop{N\! L}\nolimits _{B/A}^\wedge $. Since $a_ t \in I$ and since the cohomology modules of $\mathop{N\! L}\nolimits _{B'/A}^\wedge $ are finite $B'$-modules (hence complete for the $a_ t$-adic topology), we conclude that $a_ t^ c$ acts as zero on these cohomologies as the same thing is true by assumption for $\mathop{N\! L}\nolimits _{B/A}^\wedge $. Thus $B'$ is rig-étale over $(A, (a_ t))$ by Lemma 86.8.2. Hence finally, we may apply Lemma 86.10.1 to $B'$ over $(A, (a_ t))$ to finish the proof. $\square$

Lemma 86.10.3. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be an $I$-adically complete $A$-algebra with $A/I \to B/IB$ of finite type. The equivalent conditions of Lemma 86.8.2 are also equivalent to

there exists a finite type $A$-algebra $C$ such that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$ and such that $B \cong C^\wedge $.

**Proof.**
Combine Lemmas 86.8.2, 86.10.2, and 86.8.4. Small detail omitted.
$\square$

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