Lemma 15.104.21. Let $A$ be a ring of weak dimension at most $1$. If $A \to B$ is a flat, injective, epimorphism of rings, then $A$ is integrally closed in $B$.

Proof. Let $x \in B$ be integral over $A$. Let $A' = A[x] \subset B$. Then $A'$ is a finite ring extension of $A$ by Algebra, Lemma 10.36.5. To show $A = A'$ it suffices to show $A \to A'$ is an epimorphism by Algebra, Lemma 10.107.6. Note that $A'$ is flat over $A$ by assumption on $A$ and the fact that $B$ is flat over $A$ (Lemma 15.104.18). Hence the composition

$A' \otimes _ A A' \to B \otimes _ A A' \to B \otimes _ A B \to B$

is injective, i.e., $A' \otimes _ A A' \cong A'$ and the lemma is proved. $\square$

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