Lemma 15.10.3. Let $(A, I)$ be a Zariski pair. Let $A \to B$ be a flat, integral, finitely presented ring map such that $A/I \to B/IB$ is an isomorphism. Then $A \to B$ is an isomorphism.

**Proof.**
The ring map $A \to B$ is finite by Algebra, Lemma 10.35.5. Hence $B$ is finitely presented as an $A$-module by Algebra, Lemma 10.35.23. Hence $B$ is a finite locally free $A$-module by Algebra, Lemma 10.77.2. Since the module $B$ has rank $1$ along $V(I)$ (see rank function described in Algebra, Lemma 10.77.2), and as $(A, I)$ is a Zariski pair, we conclude that the rank is $1$ everywhere. It follows that $A \to B$ is an isomorphism: it is a pleasant exercise to show that a ring map $R \to S$ such that $S$ is a locally free $R$-module of rank $1$ is an isomorphism (hint: look at local rings).
$\square$

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