Lemma 61.5.6. Let $A$ be a ring such that $\mathop{\mathrm{Spec}}(A)$ is profinite. Let $A \to B$ be a ring map. Then $\mathop{\mathrm{Spec}}(B)$ is profinite in each of the following cases:

1. if $\mathfrak q,\mathfrak q' \subset B$ lie over the same prime of $A$, then neither $\mathfrak q \subset \mathfrak q'$, nor $\mathfrak q' \subset \mathfrak q$,

2. $A \to B$ induces algebraic extensions of residue fields,

3. $A \to B$ is a local isomorphism,

4. $A \to B$ identifies local rings,

5. $A \to B$ is weakly étale,

6. $A \to B$ is quasi-finite,

7. $A \to B$ is unramified,

8. $A \to B$ is étale,

9. $B$ is a filtered colimit of $A$-algebras as in (1) – (8),

10. etc.

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). $\square$

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0978. Beware of the difference between the letter 'O' and the digit '0'.