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This is a question that I just came up with, so it may be completely stupid and I sincerely apologize if that's the case.

The question is; Take $A$ to be an integral domain. Let $B$ be an integral extension of $A$ (i.e. every $b\in B$ satisfies an equation of the form $b^n+a_{n-1}b^{n-1}+\dots+a_0=0$ with $a_0,\dots,a_{n-1}\in A$). Is it true that there is an extension of the field of fractions of $A$, say $L$, such that $B$ equals the integral closure of $A$ in $L$ (i.e. the elements of $L$ which satisfy a polynomial equation as above).

I am very sure that the answer is no but I wonder if there are some specific cases where it may hold or any nice counterexample.

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  • $\begingroup$ Is my answer satisfactory to you? If so, then you may accept it. Otherwise please tell me how I could improve it! :-) $\endgroup$ – Watson Oct 31 '18 at 15:24
  • $\begingroup$ Yes ! Sorry, I procrastinated :) $\endgroup$ – user128787 Oct 31 '18 at 15:25
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$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\Frac}{\mathrm{Frac}} $

No. Take the integral extension $$A= \Z \subset B= \Z[\sqrt 5].$$

If there was an extension $L \supset \Frac(A) = \Q$ such that $B$ equals the integral closure $O$ of $A$ in $L$, then $L$ would be finite and $B=O$ would be a Dedekind domain, since $A$ is (see Prop. I.8.1 in Neukirch's Algebraic number theory). But $B$ is not integrally closed, so it can't be Dedekind.

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