Lemma 98.27.15. Let $S$ be a locally Noetherian scheme. Let $V$ be a scheme locally of finite type over $S$. Let $Z \subset V$ be closed. Let $W$ be a locally Noetherian formal algebraic space over $S$ such that $W_{red}$ is locally of finite type over $S$. Let $g : V_{/Z} \to W$ be an adic morphism of formal algebraic spaces over $S$. Let $v \in V$ be a closed point such that $g$ is versal at $v$ (as in Section 98.15). Then after replacing $V$ by an open neighbourhood of $v$ the morphism $g$ is smooth (see proof).

**Proof.**
Since $g$ is adic it is representable by algebraic spaces (Formal Spaces, Section 87.23). Thus by saying $g$ is smooth we mean that $g$ should be smooth in the sense of Bootstrap, Definition 80.4.1.

Write $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in Formal Spaces, Lemma 87.20.11. Set $V_ n = V_{/Z} \times _{\hat x, W} W_ n$. Then $V_ n$ is a closed subscheme with underlying set $Z$. Smoothness of $V \to W$ is equivalent to the smoothness of all the morphisms $V_ n \to W_ n$ (this holds because any morphism $T \to W$ with $T$ a quasi-compact scheme factors through $W_ n$ for some $n$). We know that the morphism $V_ n \to W_ n$ is smooth at $v$ by Lemma 98.12.6^{1}. Of course this means that given any $n$ we can shrink $V$ such that $V_ n \to W_ n$ is smooth. The problem is to find an open which works for all $n$ at the same time.

The question is local on $V$, hence we may assume $S = \mathop{\mathrm{Spec}}(R)$ and $V = \mathop{\mathrm{Spec}}(A)$ are affine.

In this paragraph we reduce to the case where $W$ is an affine formal algebraic space. Choose an affine formal scheme $W'$ and an étale morphism $W' \to W$ such that the image of $v$ in $W_{red}$ is in the image of $W'_{red} \to W_{red}$. Then $V_{/Z} \times _{g, W} W' \to V_{/Z}$ is an adic étale morphism of formal algebraic spaces over $S$ and $V_{/Z} \times _{g, W} W'$ is an affine formal algebraic space. By Algebraization of Formal Spaces, Lemma 88.25.1 there exists an étale morphism $\varphi : V' \to V$ of affine schemes such that the completion of $V'$ along $Z' = \varphi ^{-1}(Z)$ is isomorphic to $V_{/Z} \times _{g, W} W'$ over $V_{/Z}$. Observe that $v$ is the image of some $v' \in V'$. Since smoothness is preserved under base change we see that $V'_ n \to W'_ n$ is smooth for all $n$. In the next paragraph we show that after replacing $V'$ by an open neighbourhood of $v'$ the morphisms $V'_ n \to W'_ n$ are smooth for all $n$. Then, after we replace $V$ by the open image of $V' \to V$, we obtain that $V_ n \to W_ n$ is smooth by étale descent of smoothness. Some details omitted.

Assume $S = \mathop{\mathrm{Spec}}(R)$, $V = \mathop{\mathrm{Spec}}(A)$, $Z = V(I)$, and $W = \text{Spf}(B)$. Let $v$ correspond to the maximal ideal $I \subset \mathfrak m \subset A$. We are given an adic continuous $R$-algebra homomorphism

Let $\mathfrak b \subset B$ be the ideal of topologically nilpotent elements (this is the maximal ideal of definition of the Noetherian adic topological ring $B$). Observe that $\mathfrak b A^\wedge $ and $IA^\wedge $ are both ideals of definition of the Noetherian adic ring $A^\wedge $. Also, $\mathfrak m A^\wedge $ is a maximal ideal of $A^\wedge $ containing both $\mathfrak b A^\wedge $ and $IA^\wedge $. We are given that

is smooth at $\mathfrak m$ for all $n$. By the discussion above we may and do assume that $B_1 \to A_1$ is a smooth ring map. Denote $\mathfrak m_1 \subset A_1$ the maximal ideal corresponding to $\mathfrak m$. Since smoothness implies flatness, we see that: for all $n \geq 1$ the map

is an isomorphism (see Algebra, Lemma 10.99.9). Consider the Rees algebra

which is a finite type graded algebra over the Noetherian ring $B_1$ and the Rees algebra

which is a a finite type graded algebra over the Noetherian ring $A_1$. Consider the homomorphism of graded $A_1$-algebras

By the above this map is an isomorphism after localizing at the maximal ideal $\mathfrak m_1$ of $A_1$. Hence $\mathop{\mathrm{Ker}}(\Psi )$, resp. $\mathop{\mathrm{Coker}}(\Psi )$ is a finite module over $B' \otimes _{B_1} A_1$, resp. $A'$ whose localization at $\mathfrak m_1$ is zero. It follows that after replacing $A_1$ (and correspondingly $A$) by a principal localization we may assume $\Psi $ is an isomorphism. (This is the key step of the proof.) Then working backwards we see that $B_ n \to A_ n$ is flat, see Algebra, Lemma 10.99.9. Hence $A_ n \to B_ n$ is smooth (as a flat ring map with smooth fibres, see Algebra, Lemma 10.137.17) and the proof is complete. $\square$

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