The Stacks project

Proof. Let us recall what the statement signifies. First, $\textit{WAdm}^{Noeth}$ is the category whose objects are adic Noetherian topological rings and whose morphisms are continuous ring homomorphisms. Consider a commutative diagram

satisfying the following conditions: $A$ and $B$ are adic Noetherian topological rings, $A \to A'$ and $B \to B'$ are étale ring maps, $(A')^\wedge = \mathop{\mathrm{lim}}\nolimits A'/I^ nA'$ for some ideal of definition $I \subset A$, $(B')^\wedge = \mathop{\mathrm{lim}}\nolimits B'/J^ nB'$ for some ideal of definition $J \subset B$, and $\varphi : A \to B$ and $\varphi ' : (A')^\wedge \to (B')^\wedge $ are continuous. Note that $(A')^\wedge $ and $(B')^\wedge $ are adic Noetherian topological rings by Formal Spaces, Lemma 87.21.1. We have to show

1. $\varphi $ is flat $\Rightarrow \varphi '$ is flat,

2. if $B \to B'$ faithfully flat, then $\varphi '$ is flat $\Rightarrow \varphi $ is flat, and

3. if $A \to B_ i$ is flat for $i = 1, \ldots , n$, then $A \to \prod _{i = 1, \ldots , n} B_ i$ is flat.

We will use without further mention that completions of Noetherian rings are flat (Algebra, Lemma 10.97.2). Since of course $A \to A'$ and $B \to B'$ are flat, we see in particular that the horizontal arrows in the diagram are flat.

Proof of (1). If $\varphi $ is flat, then the composition $A \to (A')^\wedge \to (B')^\wedge $ is flat. Hence $A' \to (B')^\wedge $ is flat by More on Flatness, Lemma 38.2.3. Hence we see that $(A')^\wedge \to (B')^\wedge $ is flat by applying More on Algebra, Lemma 15.27.5 with $R = A'$, with ideal $I(A')$, and with $M = (B')^\wedge = M^\wedge $.

Proof of (2). Assume $\varphi '$ is flat and $B \to B'$ is faithfully flat. Then the composition $A \to (A')^\wedge \to (B')^\wedge $ is flat. Also we see that $B \to (B')^\wedge $ is faithfully flat by Formal Spaces, Lemma 87.19.14. Hence by Algebra, Lemma 10.39.9 we find that $\varphi : A \to B$ is flat.

Proof of (3). Omitted. $\square$

Proof. The statement makes sense as each of the properties $P$, $Q$, and $R$ is a local property of morphisms of $\textit{WAdm}^{Noeth}$. Let $\varphi : B \to A$ and $\psi : B \to C$ be morphisms of $\textit{WAdm}^{Noeth}$. If either $Q(\varphi )$ or $Q(\psi )$ then we see that $A \widehat{\otimes }_ B C$ is Noetherian by Formal Spaces, Lemma 87.4.12. Since $R$ implies $Q$ (Lemma 88.12.4), we find that this holds in both cases (1) and (2). This is the first thing we have to check. It remains to show that $C \to A \widehat{\otimes }_ B C$ is flat.

Proof of (1). Fix ideals of definition $I \subset A$ and $J \subset B$. By Lemma 88.12.5 the ring map $B \to C$ is topologically of finite type. Hence $B \to C/J^ n$ is of finite type for all $n \geq 1$. Hence $A \otimes _ B C/J^ n$ is Noetherian as a ring (because it is of finite type over $A$ and $A$ is Noetherian). Thus the $I$-adic completion $A \widehat{\otimes }_ B C/J^ n$ of $A \otimes _ B C/J^ n$ is flat over $C/J^ n$ because $C/J^ n \to A \otimes _ B C/J^ n$ is flat as a base change of $B \to A$ and because $A \otimes _ B C/J^ n \to A \widehat{\otimes }_ B C/J^ n$ is flat by Algebra, Lemma 10.97.2 Observe that $A \widehat{\otimes }_ B C/J^ n = (A \widehat{\otimes }_ B C)/J^ n(A \widehat{\otimes }_ B C)$; details omitted. We conclude that $M = A \widehat{\otimes }_ B C$ is a $C$-module which is complete with respect to the $J$-adic topology such that $M/J^ nM$ is flat over $C/J^ n$ for all $n \geq 1$. This implies that $M$ is flat over $C$ by More on Algebra, Lemma 15.27.4.

Proof of (2). In this case $B \to A$ is adic and hence we have just $A \widehat{\otimes }_ B C = \mathop{\mathrm{lim}}\nolimits A \otimes _ B C/J^ n$. The rings $A \otimes _ B C/J^ n$ are Noetherian by an application of Formal Spaces, Lemma 87.4.12 with $C$ replaced by $C/J^ n$. We conclude in the same manner as before. $\square$

Proof. This is true because compositions of flat ring maps are flat. $\square$

Proof. The equivalence of (1) and (2) is Definition 88.13.4. The equivalence of (2), (3), and (4) follows from the fact that being flat is a local property of arrows of $\text{WAdm}^{Noeth}$ by Lemma 88.13.1 and an application of the variant of Formal Spaces, Lemma 87.21.3 for morphisms between locally Noetherian algebraic spaces mentioned in Formal Spaces, Remark 87.21.5. $\square$

Proof. Part (1) follows from a combination of Formal Spaces, Remark 87.21.10, Lemma 88.13.2 part (1), Lemma 88.13.5, and Lemma 88.12.7.

Part (2) follows from a combination of Formal Spaces, Remark 87.21.9, Lemma 88.13.2 part (2), Lemma 88.13.5, and Lemma 88.11.5. $\square$

Proof. Combine Formal Spaces, Remark 87.21.14 and Lemma 88.13.3. $\square$

Proof. This is a sanity check whose proof should be trivial but isn't quite. We urge the reader to skip the proof. Assume $f$ is representable by algebraic spaces and flat in the sense of Bootstrap, Definition 80.4.1. Consider a commutative diagram

with $U$ and $V$ affine formal algebraic spaces, $U \to X$ and $V \to Y$ representable by algebraic spaces and étale. Then the morphism $U \to V$ corresponds to a taut map $B \to A$ of $\textit{WAdm}^{Noeth}$ by Formal Spaces, Lemma 87.22.2. Observe that this means $B \to A$ is adic (Formal Spaces, Lemma 87.23.1) and in particular for any ideal of definition $J \subset B$ the topology on $A$ is the $J$-adic topology and the diagrams

are cartesian.

Let $T \to V$ is a morphism where $T$ is a scheme. Then

The first statement is the assumption on $f$. The first implication because $U \to X$ is étale and hence flat and compositions of flat morphisms of algebraic spaces are flat. The second impliciation because $U \times _ Y T = U \times _ V V \times _ Y T$. The third implication by More on Flatness, Lemma 38.2.3. The fourth implication because we can pullback by the morphism $T \to V \times _ Y T$. We conclude that $U \to V$ is flat in the sense of Bootstrap, Definition 80.4.1. In terms of the continuous ring map $B \to A$ this means the ring maps $B/J^ n \to A/J^ nA$ are flat (see diagram above).

Finally, we can conclude that $B \to A$ is flat for example by More on Algebra, Lemma 15.27.4. $\square$