Lemma 29.46.9. Let $p$ be a prime number. Let $A \to B$ be a ring map which induces an isomorphism $A[1/p] \to B[1/p]$ (for example if $p$ is nilpotent in $A$). The following are equivalent
$\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, and
the kernel of $A \to B$ is a locally nilpotent ideal and for every $b \in B$ there exists a $p$-power $q$ with $qb$ and $b^ q$ in the image of $A \to B$.
Proof. If (2) holds, then (1) holds by Algebra, Lemma 10.46.7. Assume (1). Then the kernel of $A \to B$ consists of nilpotent elements by Algebra, Lemma 10.30.6. Thus we may replace $A$ by the image of $A \to B$ and assume that $A \subset B$. By Algebra, Lemma 10.46.5 the set
is an $A$-subalgebra of $B$ (being closed under products is trivial). We have to show $B' = B$. If not, then according to Lemma 29.46.6 there exists a $b \in B$, $b \not\in B'$ with either $b^2, b^3 \in B'$ or there exists a prime number $\ell $ with $\ell b, b^\ell \in B'$. We will show both cases lead to a contradiction, thereby proving the lemma.
Since $A[1/p] = B[1/p]$ we can choose a $p$-power $q$ such that $qb \in A$.
If $b^2, b^3 \in B'$ then also $b^ q \in B'$. By definition of $B'$ we find that $(b^ q)^{q'} \in A$ for some $p$-power $q'$. Then $qq'b, b^{qq'} \in A$ whence $b \in B'$ which is a contradiction.
Assume now there exists a prime number $\ell $ with $\ell b, b^\ell \in B'$. If $\ell \not= p$ then $\ell b \in B'$ and $qb \in A \subset B'$ imply $b \in B'$ a contradiction. Thus $\ell = p$ and $b^ p \in B'$ and we get a contradiction exactly as before. $\square$
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