The Stacks project

Lemma 88.13.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphisms of locally Noetherian formal algebraic spaces over $S$. If $f$ is representable by algebraic spaces and flat in the sense of Bootstrap, Definition 80.4.1, then $f$ is flat in the sense of Definition 88.13.4.

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

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

with $U$ and $V$ affine formal algebraic spaces, $U \to X$ and $V \to Y$ representable by algebraic spaces and ├ętale. 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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A/J^ nA) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(B/J^ n) \ar[d] \\ U \ar[r] & V } \]

are cartesian.

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

\begin{align*} X \times _ Y T \to T\text{ is flat} & \Rightarrow U \times _ Y T \to T\text{ is flat} \\ & \Rightarrow U \times _ V V \times _ Y T \to T\text{ is flat} \\ & \Rightarrow U \times _ V V \times _ Y T \to V \times _ Y T\text{ is flat} \\ & \Rightarrow U \times _ V T \to T\text{ is flat} \end{align*}

The first statement is the assumption on $f$. The first implication because $U \to X$ is ├ętale 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. $\square$


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