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Let $\{ P_i \}$ be the set of all minimal prime ideals of a commutative ring $R $. Is there any conditions on $R $ under which there exists an element $x\in R $ such that $P_i $ is an ideal of $R $ maximal with respect to not containing $x $ for each $i $?

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    $\begingroup$ If $A$ is a domain and the intersection of nonzero ideals of $A$ is nonzero, then $A$ is a field. Indeed, if $y$ is nonzero in this intersection, $y\in y^2A$, say $y=y^2a$, so $1=ya$ using that $A$ is a domain, and hence $y$ is invertible, and hence this intersection is $A$, so $A$ is a field. Apply this to $R/P_i$ to deduce that each $P_i$ is maximal. Conversely if each $P_i$ is maximal, then $x=1$ works. So this holds iff $A$ has Krull dimension $0$. $\endgroup$
    – YCor
    Commented Jun 20, 2020 at 13:00

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