Definition 61.4.1. A ring map $A \to B$ is said to be ind-Zariski if $B$ can be written as a filtered colimit $B = \mathop{\mathrm{colim}}\nolimits B_ i$ with each $A \to B_ i$ a local isomorphism.
61.4 Ind-Zariski algebra
We start with a definition; please see Remark 61.6.9 for a comparison with the corresponding definition of the article [BS].
An example of an Ind-Zariski map is a localization $A \to S^{-1}A$, see Algebra, Lemma 10.9.9. The category of ind-Zariski algebras is closed under several natural operations.
Lemma 61.4.2. Let $A \to B$ and $A \to A'$ be ring maps. Let $B' = B \otimes _ A A'$ be the base change of $B$. If $A \to B$ is ind-Zariski, then $A' \to B'$ is ind-Zariski.
Proof. Omitted. $\square$
Lemma 61.4.3. Let $A \to B$ and $B \to C$ be ring maps. If $A \to B$ and $B \to C$ are ind-Zariski, then $A \to C$ is ind-Zariski.
Proof. Omitted. $\square$
Lemma 61.4.4. Let $A$ be a ring. Let $B \to C$ be an $A$-algebra homomorphism. If $A \to B$ and $A \to C$ are ind-Zariski, then $B \to C$ is ind-Zariski.
Proof. Omitted. $\square$
Lemma 61.4.5. A filtered colimit of ind-Zariski $A$-algebras is ind-Zariski over $A$.
Proof. Omitted. $\square$
Lemma 61.4.6. Let $A \to B$ be ind-Zariski. Then $A \to B$ identifies local rings,
Proof. Omitted. $\square$
Comments (0)