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Something I don't get form the hypothesis of this theorem.

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If $C$ is the integral closure of $A$, and $A$ is integrally closed (since $A$ is integral domain, it's integrally closed over its field of fractions), then $A=C$. If $B$ is integral over $A$, then $B\subset C$. Since $A\subset B$, we have $A=B$...

What is wrong (beyond myself)?

Mand
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1 Answers1

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$B$ being integral over $A$ does not imply that $B$ is contained in $C$ as $A$ being integrally closed need not imply that $A$ is integrally closed in $B$.

asdq
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  • On Atiyah - MacDonald's, the definition of $B$ being integral over $A$ is that $C=B$. – Mand Aug 02 '20 at 14:51
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    By definition, $A$ is integrally closed in $B$ if the integral closure of $A$ inside $B$ is equal to $B$. By contrast, one says that a domain $A$ (without reference to a larger ring that $A$ is contained in) is integrally closed if it is integrally closed inside its field of fractions. – asdq Aug 02 '20 at 16:02
  • @Mand For an explicit example let $A = \mathbb{Z}$, $B = \mathbb{Z}[\sqrt{2}]$. Then $C = A$ (the integral closure inside the field of fractions, $\mathbb{Q}$), but $B$ is not contained in $C$ - and this can occur since the field of fractions of $B$ is $\mathbb{Q}(\sqrt{2})$ is not $\mathbb{Q}$. – Mummy the turkey Aug 02 '20 at 21:49