On higher order $z$-ideals and $z^\circ$-ideals in commutative rings

A ring $R$ is called radically $z$-covered (resp. radically $z^\circ$-covered) if every $\sqrt z$-ideal (resp. $\sqrt {z^\circ}$-ideal) in $R$ is a higher order $z$-ideal (resp. $z^\circ$-ideal). In this article we show with a counter-example that a ring may not be radically $z$-covered (resp. radically $z^\circ$-covered). Also a ring $R$ is called $z^\circ$-terminating if there is a positive integer $n$ such that for every $m\geq n$, each $z^{\circ m}$-ideal is a $z^{\circ n}$-ideal. We show with a counter-example that a ring may not be $z^\circ$-terminating. It is well known that whenever a ring homomorphism $\phi:R\to S$ is strong (meaning that it is surjective and for every minimal prime ideal $P$ of $R$, there is a minimal prime ideal $Q$ of $S$ such that $\phi^{-1}[Q] = P$), and if $R$ is a $z^\circ$-terminating ring or radically $z^\circ$-covered ring then so is $S$. We prove that a surjective ring homomorphism $\phi:R\to S$ is strong if and only if ${\rm ker}(\phi)\subseteq{\rm rad}(R)$.