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*}

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.