Let $R$ be a commutative ring with the property that for each nonzero ideal $I$ and element $a ∈ I$ there exists a unique ideal $J$ such that $IJ = (a)$. Show that $R$ is Noetherian.
I am looking for some elementary proof to use really basic ring and generators property to solve the problem.
My attempt is to show $J$ is finitely generated and use $IJ=JI$ to show $I$ is finitely generated.
I suppose for $a\in I$, there exists $b_1,b_2,...,b_n$, such that $\sum_{i=1}^na_ib_i=a$ for some $a_i\in I$, where $n$ is the minimal number of such kind of summation holds. I decided to show $J=(b_1,b_2,...,b_n)$ such that $IJ=(a)$. But I couldn't go further, any hint or suggestions are appreciated. Thanks.
Edit: The assumption that for all $a\in I$, there exists a unique $J$, such that $IJ=(a)$ is used in this proof. If it is not the case, we can't conclude that for the same $a\in I\cap J$, $IJ=JI=(a)$.
You already seem to have what you were asking for.
Starting from where you left off, let $J’=(b_1,\ldots,b_n)$. Then $(a)=IJ\supseteq IJ’\supseteq (a)$. Thus we have equality all across. By uniqueness of $J$ then, $J=J’$ is finitely generated.
(I haven’t gone further because you indicated this is the subproblem you wanted to settle.)