**Proof.**
By the references mentioned above (Algebra, Lemma 10.26.5 or Topology, Lemma 5.23.8) there are no specializations between distinct points of $\mathop{\mathrm{Spec}}(A)$ and $\mathop{\mathrm{Spec}}(B)$ is profinite if and only if there are no specializations between distinct points of $\mathop{\mathrm{Spec}}(B)$. These specializations can only happen in the fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. In this way we see that (1) is true.

The assumption in (2) implies all primes of $B$ are maximal by Algebra, Lemma 10.35.9. Thus (2) holds. If $A \to B$ is a local isomorphism or identifies local rings, then the residue field extensions are trivial, so (3) and (4) follow from (2). If $A \to B$ is weakly étale, then More on Algebra, Lemma 15.104.17 tells us it induces separable algebraic residue field extensions, so (5) follows from (2). If $A \to B$ is quasi-finite, then the fibres are finite discrete topological spaces. Hence (6) follows from (1). Hence (3) follows from (1). Cases (7) and (8) follow from this as unramified and étale ring map are quasi-finite (Algebra, Lemmas 10.151.6 and 10.143.6). If $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of $A$-algebras, then $\mathop{\mathrm{Spec}}(B) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Spec}}(B_ i)$ in the category of topological spaces by Limits, Lemma 32.4.2. Hence if each $\mathop{\mathrm{Spec}}(B_ i)$ is profinite, so is $\mathop{\mathrm{Spec}}(B)$ by Topology, Lemma 5.22.3. This proves (9).
$\square$

## Comments (0)

There are also: