Irreducible elements remain irreducible in integral closure?

Question

Let $R$ be an integral domain and $\bar R$ be its integral closure in the fraction field. If $b \in R$ is an irreducible element in $R$, then does $b$ remain irreducible in $\bar R$ also ?

I can see that $b$ is still a non-unit in $\bar R$, but I am unable to say anything about its factorization.

This may be related When prime element in an integral domain stays prime in integral extension

Answer 1

A counterexample . . .

Let $K$ be a field, and let $R=K[x^2,x^3]$.

Then $x^2$ is irreducible in $R$, but $x^2$ is not irreducible in $\bar R$, since $\bar R = K[x]$.