# Is an integer extension of a ring the integral closure of this ring in some extension?

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.

• Is my answer satisfactory to you? If so, then you may accept it. Otherwise please tell me how I could improve it! :-) – Watson Oct 31 '18 at 15:24
• Yes ! Sorry, I procrastinated :) – user128787 Oct 31 '18 at 15:25

$$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\Frac}{\mathrm{Frac}}$$
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.