Lemma 29.46.9 (0CNF)—The Stacks project

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

$B' = \{ b \in B \mid p^ nb, b^{p^ n} \in A\text{ for some }n \geq 0\}$

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. $\square$

The Stacks project

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