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Suppose that $R$ is a Noetherian complete domain over a field $K$.

Suppose that a monic polynomial $f(X) \in R[X]$ (i.e., the highest degree $X^e$ in $f$ has the coefficient $1$), satisfies the following two conditions$\colon$

  1. $R[X]/(f(X))$ is not integral.

  2. $f(X) = g(X)^l$ in $F(R)[X]$ for some integer $l > 1$, where $g(X)$ is an irreducible polynomial in $F(R)[X]$ and $F(R)$ is the fractional field of $R$.

Q. Then, does the following equality hold for some $G(X) \in R[X]$$\colon$ \begin{equation*} f(X) = G(X)^l~? \end{equation*}

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    $\begingroup$ Yes if $R$ is integrally closed: then $R[X]$ is integrally closed, and $g$ is clearly integral over $R[X]$. $\endgroup$ – Laurent Moret-Bailly Mar 5 at 9:55
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    $\begingroup$ No in general in char. $p>0$: take for $R\subset K[[t]]$ the subring of power series without term of degree 1. Then $f=X^p-t^p$ is a counterexample, with $g=X-t$ and $l=p$. $\endgroup$ – Laurent Moret-Bailly Mar 5 at 9:59

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