Let $A$ be a integral domain. Suppose $A[x]$ is integrally closed then $A$ need not be integrally closed? (In particular $A$ is not UFD.)
I was trying to find a counter example. But after long time I couldn't find any. Please give me a hint.
(I was thinking of proving $A[x]= \mathbb Z (\sqrt{-5})[x] $ is not integrally closed. But I could not do it. I'm not sure if it's true either. I end up with the following statement - if $A$ is integrally closed over $B$ then for ideal $I$ of B, $A/{A\cap I}$ is integrally closed over $B/I$ which I couldn't prove or disprove either -- sorry for not showing any work. I haven't made much progress to show)
As pointed out in the comments by user26857, if $A[x]$ is an integrally closed domain, then $A$ must also be an integrally closed domain. Suppose $b\in\operatorname{Frac} A$ is integral over $A$. Then $b\in \operatorname{Frac} A[x]$ is integral over $A[x]$, so it's in $A[x]$. But $A[x]\cap \operatorname{Frac} A=A$ inside $\operatorname{Frac} A[x]$, so $b\in A$.