Lemma 29.46.5. Let $A \subset B$ be a ring extension such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields. If $A \not= B$, then there exists a $b \in B$, $b \not\in A$ with $b^2 \in A$ and $b^3 \in A$.

**Proof.**
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 $\mathfrak m' \not\subset A$. However, if $\mathfrak m' \subset A$, then $\mathfrak m' = \mathfrak m$ and we conclude that $A = B$ as the residue fields are isomorphic as well by assumption. This contradiction finishes the proof.
$\square$

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