The Stacks project

Lemma 88.13.1. The property $P(\varphi )=$“$\varphi $ is flat” on arrows of $\textit{WAdm}^{Noeth}$ is a local property as defined in Formal Spaces, Remark 87.21.5.

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

\[ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi '} } \]

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. $\square$

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