Today I’m writing about an important class of adic spaces, namely analytic adic spaces. An important property is that all specializations on maps between analytic adic spaces are vertical. Analytic adic spaces should provide the correct level of generality for arithmetic considerations. Most of this part is taken from Conrad’s seminar (week 7 and week 6).

Definition 17Let be a Huber ring. A point is calledanalyticif its support is not an open ideal of .

Lemma 18Let be a Huber ring with a couple of definition . A point is non-analytic if and only if .

*Proof:* The support is open in if and only if it contains for sufficiently large . But is radical so this is equivalent to containing , that is exactly .

Let us denote by the subspace of analytic points in endowed with the subspace topology.

Corollary 19Let be a Tate ring. Then .

*Proof:* Let be the topologically nilpotent unit in . Then cannot kill , even though for all sufficiently large . This implies and hence the result by the previous lemma.

Let us now describe the structure of . We still denote by a Huber ring. Let be its couple of definitions and let be a set of generators of . Let . Then

Hence for some we must have . We choose for which is maximal and hence for all . We can then form . There is a natural map which induces the map on :

Proposition 20The image of the natural continuous map

is the subspace of defined by .

Using this result we get

Proposition 21Let be a Huber ring with as before. We have

where

I finish by stating a theorem on specializations

Theorem 22Let be a Huber ring. Then all specializations inside are vertical. In particular, if is Tate, then all specializations in are vertical.

*Proof:* One recalls a structure theorem for specializations which says that we can write each specialization as a vertical specialization of a horizontal specialization. Then it is enough to prove that there are no proper horizontal specializations. This is done for by choosing which contains . Hence is horizontal specialization of and it lies in . Assume that and derive a contradiction by showing that is open, hence could not be analytic.

In lemma 18, ‘analytic’ should be replaced by ‘non-analytic’.

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Corrected, thank you!

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