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The Stacks project

Lemma 61.6.5. Let A \to B be ring map such that

  1. A \to B identifies local rings,

  2. the topological spaces \mathop{\mathrm{Spec}}(B), \mathop{\mathrm{Spec}}(A) are w-local, and

  3. \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A) is w-local.

Then A \to B is ind-Zariski.

Proof. Set X = \mathop{\mathrm{Spec}}(A) and Y = \mathop{\mathrm{Spec}}(B). Let X_0 \subset X and Y_0 \subset Y be the set of closed points. Let A \to A' be the ind-Zariski morphism of affine schemes such that with X' = \mathop{\mathrm{Spec}}(A') the diagram

\xymatrix{ X' \ar[r] \ar[d] & \pi _0(X') \ar[d] \\ X \ar[r] & \pi _0(X) }

is cartesian in the category of topological spaces and such that \pi _0(X') = \pi _0(Y) as spaces over \pi _0(X), see Lemma 61.6.2. By Lemma 61.2.5 we see that X' is w-local and the set of closed points X'_0 \subset X' is the inverse image of X_0.

We obtain a continuous map Y \to X' of underlying topological spaces over X identifying \pi _0(Y) with \pi _0(X'). By Lemma 61.3.8 (and Lemma 61.4.6) this corresponds to a morphism of affine schemes Y \to X' over X. Since Y \to X maps Y_0 into X_0 we see that Y \to X' maps Y_0 into X'_0, i.e., Y \to X' is w-local. By Lemma 61.6.4 we see that Y \cong X' and we win. \square


Comments (2)

Comment #5930 by Mingchen on

There is a typo in the proof: "this is corresponds", remove "is".


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