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

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

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)