Lemma 85.18.1. Let $B \to A$ be an arrow of $\textit{WAdm}^{count}$. The following are equivalent

$B \to A$ is taut (Definition 85.4.11),

for $B \supset J_1 \supset J_2 \supset J_3 \supset \ldots $ a fundamental system of weak ideals of definitions there exist a commutative diagram

\[ \xymatrix{ A \ar[r] & \ldots \ar[r] & A_3 \ar[r] & A_2 \ar[r] & A_1 \\ B \ar[r] \ar[u] & \ldots \ar[r] & B/J_3 \ar[r] \ar[u] & B/J_2 \ar[r] \ar[u] & B/J_1 \ar[u] } \]

such that $A_{n + 1}/J_ nA_{n + 1} = A_ n$ and $A = \mathop{\mathrm{lim}}\nolimits A_ n$ as topological ring.

Moreover, these equivalent conditions define a local property, i.e., they satisfy axioms (1), (2), (3).

**Proof.**
The equivalence of (a) and (b) is immediate. Below we will give an algebraic proof of the axioms, but it turns out we've already proven them. Namely, using Lemma 85.15.10 the equivalent conditions (a) and (b) translate to saying the corresponding morphism of affine formal algebraic spaces is representable. Since this condition is “étale local on the source and target” by Lemma 85.15.4 we immediately get axioms (1), (2), and (3).

Direct algebraic proof of (1), (2), (3). Let a diagram (85.17.2.1) as in Situation 85.17.2 be given. By Example 85.15.11 the maps $A \to (A')^\wedge $ and $B \to (B')^\wedge $ satisfy (a) and (b).

Assume (a) and (b) hold for $\varphi $. Let $J \subset B$ be a weak ideal of definition. Then the closure of $JA$, resp. $J(B')^\wedge $ is a weak ideal of definition $I \subset A$, resp. $J' \subset (B')^\wedge $. Then the closure of $I(A')^\wedge $ is a weak ideal of definition $I' \subset (A')^\wedge $. A topological argument shows that $I'$ is also the closure of $J(A')^\wedge $ and of $J'(A')^\wedge $. Finally, as $J$ runs over a fundamental system of weak ideals of definition of $B$ so do the ideals $I$ and $I'$ in $A$ and $(A')^\wedge $. It follows that (a) holds for $\varphi '$. This proves (1).

Assume $A \to A'$ is faithfully flat and that (a) and (b) hold for $\varphi '$. Let $J \subset B$ be a weak ideal of definition. Using (a) and (b) for the maps $B \to (B')^\wedge \to (A')^\wedge $ we find that the closure $I'$ of $J(A')^\wedge $ is a weak ideal of definition. In particular, $I'$ is open and hence the inverse image of $I'$ in $A$ is open. Now we have (explanation below)

\begin{align*} A \cap I' & = A \cap \bigcap (J(A')^\wedge + \mathop{\mathrm{Ker}}((A')^\wedge \to A'/I_0A')) \\ & = A \cap \bigcap \mathop{\mathrm{Ker}}((A')^\wedge \to A'/JA' + I_0 A') \\ & = \bigcap (JA + I_0) \end{align*}

which is the closure of $JA$ by Lemma 85.4.2. The intersections are over weak ideals of definition $I_0 \subset A$. The first equality because a fundamental system of neighbourhoods of $0$ in $(A')^\wedge $ are the kernels of the maps $(A')^\wedge \to A'/I_0A'$. The second equality is trivial. The third equality because $A \to A'$ is faithfully flat, see Algebra, Lemma 10.82.11. Thus the closure of $JA$ is open. By Lemma 85.4.10 the closure of $JA$ is a weak ideal of definition of $A$. Finally, given a weak ideal of definition $I \subset A$ we can find $J$ such that $J(A')^\wedge $ is contained in the closure of $I(A')^\wedge $ by property (a) for $B \to (B')^\wedge $ and $\varphi '$. Thus we see that (a) holds for $\varphi $. This proves (2).

We omit the proof of (3).
$\square$

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