Lemma 29.46.6. Let $A \subset B$ be a ring extension such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism. If $A \not= B$, then either there exists a $b \in B$, $b \not\in A$ with $b^2 \in A$ and $b^3 \in A$ or there exists a prime number $p$ and a $b \in B$, $b \not\in A$ with $pb \in A$ and $b^ p \in A$.

**Proof.**
The argument is almost exactly the same as in the proof of Lemma 29.46.5 but we write everything out to make sure it works.

Recall that $A \subset B$ is integral (Lemma 29.44.7). By Lemma 29.46.1 we may assume that $B$ is generated by a single element over $A$. Hence $B$ is finite over $A$ (Algebra, Lemma 10.36.5). Hence the support of $B/A$ as an $A$-module is closed and not empty (Algebra, Lemmas 10.40.5 and 10.40.2). Let $\mathfrak p \subset A$ be a minimal prime of the support. After replacing $A \subset B$ by $A_\mathfrak p \subset B_\mathfrak p$ (permissible by Lemma 29.46.3) we may assume that $(A, \mathfrak m)$ is a local ring, that $B$ is finite over $A$, and that $B/A$ has support $\{ \mathfrak m\} $ as an $A$-module. Since $B/A$ is a finite module, we see that $I = \text{Ann}_ A(B/A)$ satisfies $\mathfrak m = \sqrt{I}$ (Algebra, Lemma 10.40.5). Let $\mathfrak m' \subset B$ be the unique prime ideal lying over $\mathfrak m$. Because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a homeomorphism, we find that $\mathfrak m' = \sqrt{IB}$. For $f \in \mathfrak m'$ pick $n \geq 1$ such that $f^ n \in IB$. Then also $f^{n + 1} \in IB$. Since $IB \subset A$ by our choice of $I$ we conclude that $f^ n, f^{n + 1} \in A$. Using Lemma 29.46.4 we conclude our lemma is true if $\mathfrak m' \not\subset A$. If $\mathfrak m' \subset A$, then $\mathfrak m' = \mathfrak m$. Since $A \not= B$ we conclude the map $\kappa = A/\mathfrak m \to B/\mathfrak m' = \kappa '$ of residue fields cannot be an isomorphism. By Lemma 29.10.2 we conclude that the characteristic of $\kappa $ is a prime number $p$ and that the extension $\kappa '/\kappa $ is purely inseparable. Pick $b \in B$ whose image in $\kappa '$ is an element not contained in $\kappa $ but whose $p$th power is in $\kappa $. Then $b \not\in A$, $b^ p \in A$, and $pb \in A$ (because $pb \in \mathfrak m' = \mathfrak m \subset A$) as desired. $\square$

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