Ideal quotient is principal

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Let $R$ be a commutative ring with $1$ and $I$ an ideal. Also let $B$ be a principal ideal, and $A=\{a\in R\;|\; aB\subseteq I\}$. I want to show that if $A$ is also principal then $I$ is principal.

Have $A=(a),\;B=(b)$. I first thought $I=R$ or $I=(ab)$: certainly $(ab)\subseteq I$ since $ r\cdot(ab)=a\cdot(rb)\in I$, but whether $I\subseteq (ab)$ I am not sure. Either way I'm not sure how to proceed, usually I show an ideal is principal by picking an special element (irreducible, minimal...), but there doesn't seem to be much choice here.

Any help appreciated.

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You can't prove what you want for it is not true.

Let $I$ be a non-principal ideal, and $b\in I$. Then for $B=(b)$ we have $A=I:(b)=(1)$, a principal ideal.