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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

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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 $\bar R = K[x]$.

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  • $\begingroup$ Wow, thanks, really simple. Is $x^2$ prime in $K[x^2,x^3] $ ? $\endgroup$
    – user495643
    Commented Jan 28, 2018 at 7:26
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    $\begingroup$ $x^2$ is irreducible in $R$, but not prime. For example, in the ring $R$, the product $(x^3)(x^3)$ is in the principal ideal $I = (x^2)$, but the element $x^3$ is not in $I$. $\endgroup$
    – quasi
    Commented Jan 28, 2018 at 7:28

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