The Stacks project

Proof. Let $\pi \in A$ be the uniformizer. Recall that an $A$-module is flat if and only if the $\pi $-power torsion is $0$.

First assume that $X$ is an affine formal algebraic space. Then $X = \text{Spf}(B)$ with $B$ an adic Noetherian $A$-algebra. In this case we set $X' = \text{Spf}(B')$ where $B' = B/\pi \text{-power torsion}$. It is clear that $X'$ is flat over $A$ and that $X' \to X$ is a closed immersion. Let $g : Y \to X$ be a morphism of locally Noetherian formal algebraic spaces with $Y$ flat over $A$. Choose a covering $\{ Y_ j \to Y\} $ as in Formal Spaces, Definition 87.11.1. Then $Y_ j = \text{Spf}(C_ j)$ with $C_ j$ flat over $A$. Hence the morphism $Y_ j \to X$, which correspond to a continuous $R$-algebra map $B \to C_ j$, factors through $X'$ as clearly $B \to C_ j$ kills the $\pi $-power torsion. Since $\{ Y_ j \to Y\} $ is a covering and since $X' \to X$ is a monomorphism, we conclude that $g$ factors through $X'$.

Let $X$ and $\{ X_ i \to X\} _{i \in I}$ be as in Formal Spaces, Definition 87.11.1. For each $i$ let $X'_ i \to X_ i$ be the flat part as constructed above. For $i, j \in I$ the projection $X'_ i \times _ X X_ j \to X'_ i$ is an étale (by assumption) morphism of schemes (by Formal Spaces, Lemma 87.9.11). Hence $X'_ i \times _ X X_ j$ is flat over $A$ as morphisms representable by algebraic spaces and étale are flat (Lemma 88.13.8). Thus the projection $X'_ i \times _ X X_ j \to X_ j$ factors through $X'_ j$ by the universal property. We conclude that

because the morphisms $X'_ i \to X_ i$ are injections of sheaves. Set $U = \coprod X'_ i$, set $R = \coprod R_{ij}$, and denote $s, t : R \to U$ the two projections. As a sheaf $R = U \times _ X U$ and $s$ and $t$ are étale. Then $(t, s) : R \to U$ defines an étale equivalence relation by our observations above. Thus $X' = U/R$ is an algebraic space by Spaces, Theorem 65.10.5. By construction the diagram

is cartesian. Since the right vertical arrow is étale surjective and the top horizontal arrow is representable and a closed immersion we conclude that $X' \to X$ is representable by Bootstrap, Lemma 80.5.2. Then we can use Spaces, Lemma 65.5.6 to conclude that $X' \to X$ is a closed immersion.

Finally, suppose that $Y \to X$ is a morphism with $Y$ a locally Noetherian formal algebraic space flat over $A$. Then each $X_ i \times _ X Y$ is étale over $Y$ and therefore flat over $A$ (see above). Then $X_ i \times _ X Y \to X_ i$ factors through $X'_ i$. Hence $Y \to X$ factors through $X'$ because $\{ X_ i \times _ X Y \to Y\} $ is an étale covering. $\square$

Proof. (Aside: the diagonal is always locally of finite type by Formal Spaces, Lemma 87.15.5 and $X \times _{\text{Spf}(A)} X$ is locally Noetherian by Formal Spaces, Lemmas 87.24.4 and 87.24.8. Thus imposing the conditions on the diagonal morphism makes sense.) The diagram

is cartesian. Hence $X' \to X' \times _{\text{Spf}(A)} X'$ is rig-étale and rig-surjective by Lemma 88.21.4. Choose an affine formal algebraic space $U$ and a morphism $U \to X'$ which is representable by algebraic spaces and étale. Then $U = \text{Spf}(B)$ where $B$ is an adic Noetherian topological ring which is a flat $A$-algebra, whose topology is the $\pi $-adic topology where $\pi \in A$ is a uniformizer, and such that $A/\pi ^ n A \to B/\pi ^ n B$ is of finite type for each $n$. For later use, we remark that this in particular implies: if $B \not= 0$, then the map $\text{Spf}(B) \to \text{Spf}(A)$ is a surjection of sheaves (please recall that we are using the fppf topology as always). Repeating the argument above, we see that

is a closed immersion and rig-étale and rig-surjective. We have $U \times _{\text{Spf}(A)} U = \text{Spf}(B \widehat{\otimes }_ A B)$ by Formal Spaces, Lemma 87.16.4. Then $B \widehat{\otimes }_ A B$ is a flat $A$-algebra as the $\pi $-adic completion of the flat $A$-algebra $B \otimes _ A B$. Hence $W = U \times _{\text{Spf}(A)} U$ by Lemma 88.21.13. In other words, we have $U \times _{X'} U = U \times _{\text{Spf}(A)} U$ which in turn means that the image of $U \to X'$ (as a map of sheaves) maps injectively to $\text{Spf}(A)$. Choose a covering $\{ U_ i \to X'\} $ as in Formal Spaces, Definition 87.11.1. In particular $\coprod U_ i \to X'$ is a surjection of sheaves. By applying the above to $U_ i \coprod U_ j \to X'$ (using the fact that $U_ i \amalg U_ j$ is an affine formal algebraic space as well) we see that $X' \to \text{Spf}(A)$ is an injective map of fppf sheaves. Since $X'$ is flat over $A$, either $X'$ is empty (if $U_ i$ is empty for all $i$) or the map is an isomorphism (if $U_ i$ is nonempty for some $i$ when we have seen that $U_ i \to \text{Spf}(A)$ is a surjective map of sheaves) and the proof is complete. $\square$

Proof. One direction is trivial. For the other, suppose that $\text{Spf}(R) \to Y$ is an adic morphism such that there exists an extension of complete discrete valuation rings $R \subset R'$ with $\text{Spf}(R') \to \text{Spf}(R) \to X$ factoring through $Y$. Consider the fibre product diagram

The morphism $p$ is locally of finite type as a base change of $f$, see Formal Spaces, Lemma 87.24.4. The diagonal morphism $\Delta _ p$ is the base change of $\Delta _ f$ and hence is rig-étale and rig-surjective. By Lemma 88.22.2 the flat locus of $\text{Spf}(R) \times _ Y X$ over $R$ is either $\emptyset $ or equal to $\text{Spf}(R)$. However, since $\text{Spf}(R')$ factors through it we conclude it is not empty and hence we get a morphism $\text{Spf}(R) \to \text{Spf}(R) \times _ Y X \to X$ as desired. $\square$