Let $R$ be a commutative ring with unity. If for every ideal of $R$, the minimal prime ideals over it are all finitely generated, then there are finitely many minimal prime ideals over every ideal of $R$ (A NOTE ON MINIMAL PRIME IDEALS,D. D. ANDERSON ).

Now let us say that a commutative ring with unity $R$ has property $MP(I)$ if for ideal $I \ne R$ of $R$, all the minimal prime ideals over $I$ are finitely generated. My question is : If $R$ has $MP (I)$, then does $R[X]$ have $MP (I[X])$ ?

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    $\begingroup$ Are you sure that you want this to happen for every ideal $I$? If so, then when $I$ Is prime, this means that the prime ideal $I$ is finitely generated. Then all prime ideals are finitely generated, and Cohen's Theorem implies that $R$ is noetherian... $\endgroup$ – Manny Reyes Mar 13 '18 at 19:16
  • $\begingroup$ A minimal prime ideal $P$ of $R[X]$ over $I[X]$ is of the form $\mathfrak{p}[X]$ where $\mathfrak{p} = P \cap R$ is a minimal prime of $R$ over $I$. Am I missing something? $\endgroup$ – Luc Guyot Mar 14 '18 at 17:33

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