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Can anyone show an example of integral domain that is not integrally closed and also has one of its localization with respect to a maximal ideal not integrally closed?

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Let $k$ be a field, and consider the domain $A = k[x,y]/(x^2-y^3)\cong k[t^2,t^3]$ (the isomorphism is given by $x\mapsto t^3$ and $y\mapsto t^2$). Note that $t\notin A$. The field of fractions of $k[t^2,t^3]$ contains $t = t^3/t^2$, but $t$ is integral over $k[t^2,t^3]$, since it satisfies the monic polynomial $z^2 - t^2$. Hence $A$ is not integrally closed in its field of fractions.

For the localization part of your question, you can localize at the maximal ideal $(t^2,t^3)$ in $k[t^2,t^3]$ (corresponding to $(x,y)$ in $k[x,y]/(x^2-y^3)$). The same argument as above shows that the result is not integrally closed. Note that if you try to localize at any other maximal ideal, you'll end up inverting $t^2$ and getting $t$ in the localization, so the argument will break down.

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  • $\begingroup$ What about a local integral domain which is not integrally closed? $\endgroup$ – user26857 Feb 9 '16 at 21:45
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    $\begingroup$ @user26857 The localization $k[t^2,t^3]_{(t^2,t^3)}$ discussed in the second paragraph of my answer is an example... $\endgroup$ – Alex Kruckman Feb 9 '16 at 21:47
  • $\begingroup$ I can agree, but I'd have started with this to trivialize a little the question. $\endgroup$ – user26857 Feb 9 '16 at 21:48
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    $\begingroup$ Sure, but a shorter answer is not necessarily a better one. For a student who's just starting out in commutative algebra, thinking about $k[t^2,t^3]$ may be more straightforward than thinking about the localization. $\endgroup$ – Alex Kruckman Feb 9 '16 at 21:52

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