The Stacks project

Lemma 87.21.14. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of locally Noetherian formal algebraic spaces over $S$. Assume

  1. $g$ is locally of finite type,

  2. $f$ is rig-smooth (resp. rig-étale) and rig-surjective,

  3. $g \circ f$ is rig-smooth (resp. rig-étale)

then $g$ is rig-smooth (resp. rig-étale).

Proof. We will prove this in the rig-smooth case and indicate the necessary changes to prove the rig-étale case at the end of the proof. Consider a commutative diagram

\[ \xymatrix{ X \times _ Y V \ar[r] \ar[d] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z } \]

with $V$ and $W$ affine formal algebraic spaces, $V \to Y$ and $W \to Z$ representable by algebraic spaces and étale. We have to show that $V \to W$ corresponds to a rig-smooth map of adic Noetherian topological rings, see Definition 87.18.1. We may write $V = \text{Spf}(B)$ and $W = \text{Spf}(C)$ and that $V \to W$ corresponds to an adic ring map $C \to B$ which is topologically of finite type, see Lemma 87.11.5.

We will use below without further mention that $X \times _ Y V \to V$ is rig-smooth and rig-surjective, see Lemmas 87.18.3 and 87.21.4. Also, the composition $X \times _ Y V \to V \to W$ is rig-smooth since $g \circ f$ is rig-smooth.

Let $I \subset C$ be an ideal of definition. The module Assume $C \to B$ is not rig-smooth to get a contradiction. This means that there exists a prime ideal $\mathfrak q \subset B$ not containing $IB$ such that either $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak p$ is nonzero or $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak p$ is not a finite free $B_\mathfrak q$-module. See Lemma 87.4.2; some details omitted. We may choose a maximal ideal $IB + \mathfrak q \subset \mathfrak m$. By Algebra, Lemma 10.119.13 we can find a complete discrete valuation ring $R$ and an injective local ring homomorphism $(B/\mathfrak q)_\mathfrak m \to R$.

After replacing $R$ by an extension, we may assume given a lift $\text{Spf}(R) \to X \times _ Y V$ of the adic morphism $\text{Spf}(R) \to V = \text{Spf}(B)$. Choose an étale covering $\{ \text{Spf}(A_ i) \to X \times _ Y V\} $ as in Formal Spaces, Definition 86.11.1. By Lemma 87.21.7 we may assume $\text{Spf}(R) \to X \times _ Y V$ lifts to a morphism $\text{Spf}(R) \to \text{Spf}(A_ i)$ for some $i$ (this might require replacing $R$ by another extension). Set $A = A_ i$. Consider the ring maps

\[ C \to B \to A \to R \]

Let $\mathfrak p \subset A$ be the kernel of the map $A \to R$ and note that $\mathfrak p$ lies over $\mathfrak q$. We know that $C \to A$ and $B \to A$ are rig-smooth. In particular the ring map $B_\mathfrak q \to A_\mathfrak p$ is flat by Lemma 87.17.6. Consider the associated exact sequence

\[ \xymatrix{ & H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge ) \otimes _ B A_\mathfrak p \ar[r] & H^0(\mathop{N\! L}\nolimits _{A/C}^\wedge )_\mathfrak p \ar[r] & H^0(\mathop{N\! L}\nolimits _{A/B}^\wedge )_\mathfrak p \ar[r] & 0 \\ 0 \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge \otimes _ B A)_\mathfrak p \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{A/C}^\wedge )_\mathfrak p \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{A/B}^\wedge )_\mathfrak p \ar[llu] } \]

of Lemmas 87.3.5 and 87.17.7. Given the rig-smoothness of $C \to A$ and $B \to A$ we conclude that $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge \otimes _ B A)_\mathfrak p = 0$ and that $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge ) \otimes _ B A_\mathfrak p$ is finite free as a kernel of a surjection of finite free $A_\mathfrak p$-modules. Since $B_\mathfrak q \to A_\mathfrak p$ is flat and hence faithfully flat, this implies that $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q = 0$ and that $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q$ is finite free which is the contradiction we were looking for.

In the rig-étale case one argues in exactly the same manner but the conclusion obtained is that both $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q$ and $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q$ are zero. $\square$

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