So we are given $R$ is a commutative ring such that for every nonzero ideal $I$ of $R$ and element $a\in I$, there is a unique ideal $J$ such that $IJ = (a)$. We are supposed to show that this means every ideal is finitely generated.
This is the jist of what I have so far. Let $I$ be an ideal and $a\in I$. Then there is some $J$ such that $(a) = IJ$. This means that $a = x_iy_i + \cdots + x_ny_n$ for $x_i\in I$ and $y_i\in J$. Then our professor told us that we should show $I' = Rx_1 + \cdots Rx_n$ is equal to $I$. He also said that to show this we should show that $I'J = IJ$. Showing $I' \subseteq I$ is simple, and from this it follows that $I'J \subseteq IJ$. I was also able to show that $I'J \supseteq IJ$, so $I'J = IJ$.
But now I don't see how this implies that $I' = I$. Any thoughts?
Edit: I found a similar question here, but that requires that $R$ be an integral domain, and we are not given that in this question.
$J $ must also be a nonzero ideal, but then $IJ=I'J=(a)$, and $R $ is commutative so $JI=JI'=(a)$. The assumption was that this $J $ was unique, so the $I $ must be unique when applying this condition to $J $. Hence $I=I'$.