The Stacks project

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. $\square$

Comments (2)

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 is defined, would help to note that , say is contained in the "Jacobson radical" of rather than just the "radical" of , and when proving it should be made clearer that this depends on (even itself depends on which was chosen depending on ). 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 taking to be the -span of the finite set of elements of arising as...

For the paragraph beginning "Next, we pick ...", perhaps note at the end of the 2nd sentence that there depends on . When is defined, perhaps note that (implicitly used later).

After conditions (1)--(4) are stated, tell the reader "We want to deduce that is adic (so the reader can orient themselves around what is going on). For the statement that the elements () are in , perhaps note that we're using that is faithfully flat and that , and likewise note that by similar reasoning (as this is used later in the argument). A few lines later it is mentioned that and are flat, but it seems only the first of these two flatness properties is being used right there.

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