Lemma 87.20.4. Let $S$ be a scheme. Let $X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces, surjective, and flat. Then $X$ is adic* if and only if $Y$ is adic*.

Proof. The proof is exactly the same as the proof of Lemma 87.20.2 except that at the end we use Lemma 87.8.5. $\square$

Comment #1956 by Brian Conrad on

This proof would benefit from some more explanation to the reader as to where it is going before one dives into the various constructions from (15.2.1) onwards. For example, at least tell the reader the nature of the target reduction to the special case where (1)--(4) hold before one proves that.

When $J$ is defined, would help to note that $J \subset J_m$, say $I$ is contained in the "Jacobson radical" of $A$ rather than just the "radical" of $A$, and when proving $I^n \subset JA$ it should be made clearer that this $n$ depends on $m$ (even $J$ itself depends on $n$ which was chosen depending on $m$). Since Nakayama Lemma 10.19.1 has 10 zillion parts, when appealing to that it could help to say that one uses part (1) with $M$ taking to be the $A$-span of the finite set of elements of $JA$ arising as...

For the paragraph beginning "Next, we pick $m$...", perhaps note at the end of the 2nd sentence that $n$ there depends on $m$. When $\mathfrak{b}_k$ is defined, perhaps note that $\mathfrak{b}_k \supset \mathfrak{b}_{k+1}$ (implicitly used later).

After conditions (1)--(4) are stated, tell the reader "We want to deduce that $B$ is adic$^{\ast}$ (so the reader can orient themselves around what is going on). For the statement that the elements $g^E$ ($|E|=k$) are in $J_k$, perhaps note that we're using that $B/J_k \rightarrow A/I^k$ is faithfully flat and that $J_1A = I$, and likewise note that $J_1 J_k \subset J_{k+1}$ by similar reasoning (as this is used later in the argument). A few lines later it is mentioned that $B/J_{k+1} \rightarrow A/I^{k+1}$ and $B/J_k \rightarrow A/I^k$ are flat, but it seems only the first of these two flatness properties is being used right there.

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