Proposition 29.46.8. Let $A \subset B$ be a ring extension. The following are equivalent
$\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, and
every finite subset $E \subset B$ is contained in an extension
such that for $i = 1, \ldots , n$ we have
$b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or
there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.
Proof. The proof is exactly the same as the proof of Proposition 29.46.7 except for the following changes:
Use Lemma 29.46.6 instead of Lemma 29.46.5 which means that for each successor ordinal $\alpha = \beta + 1$ we either have $b_\alpha ^2, b_\alpha ^3 \in B_\beta $ or we have a prime $p$ and $pb_\alpha , b_\alpha ^ p \in B_\beta $.
If $\alpha $ is a successor ordinal, then take $D = \{ d_{e, i}\} \cup \{ b_\alpha ^2, b_\alpha ^3\} $ or take $D = \{ d_{e, i}\} \cup \{ pb_\alpha , b_\alpha ^ p\} $ depending on which case $\alpha $ falls into.
In the proof of (2) $\Rightarrow $ (1) we also need to consider the case where $B$ is generated over $A$ by a single element $b$ with $pb, b^ p \in B$ for some prime number $p$. Here $A \subset B$ induces a universal homeomorphism on spectra for example by Algebra, Lemma 10.46.7.
This finishes the proof. $\square$
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