Let $A$ be a noetherian ring and $I$ be an ideal (need not contained in Jacobson radical) of $A$. An element $a\in I$ is said to be superficial if there exists $c\in\mathbb{N}$ such that $(I^n:a)\cap I^c=I^{n-1}$ for all $n\geq c$.
If $a$ is a superficial element of a regular ideal $I$, then $a\notin I^2$. I don't understand why it follows easily.
If $a\in I^2$, then for any $x\in I^{n-2}$ with $n\geq 2$, we have $ax\in I^2\cdot I^{n-2}=I^n$. So, $I^{n-2}\subseteq (I^n:a)$. Take $c'=\max\{2,c\}$, we have $I^{m}\subseteq (I^{m+2}: a)\cap I^{c'}=I^{m+1}$ for all $m\geq c'$ and thus $I^m=I^{m+1}$ for all $m\geq c'$. By Nakayama's lemma, we have $I^m=0$, which contradicts to the fact that $I$ is a regular ideal.
The claim is not true for a Noetherian ring. For a counterexample, let $A=\mathbf Z/6\mathbf Z$ and $I=(2)$. Note that $I=I^2$ and $a=2\in I$ is a superficial element.
