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If $A, B$ are two domains and $Q(A), Q(B)$ their fields of fractions, it is true that, if $A \subseteq B$ is an integral extension then $\bar{A}=\bar{B}$? Where with $\bar{A}$ we denote the integral closure of $A$ in its field of fractions. I think that it is true that $\bar{A}^{Q(B)}=\bar{B}$, where with $\bar{A}^{Q(B)}$ we denote the integral closure of $A$ in $Q(B)$, but I'm not sure that $\bar{A}=\bar{B}$, can you give me a counter example if there exists?

Thank you

Rick88
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1 Answers1

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Let $A=\mathbb{Z}$ and $B$ is any algebraic integer ring except $\mathbb{Z}$. Then $A$ and $B$ are both integrally closed, but $A \neq B$.

yuan
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