I'm reading this Algebra book at the moment and found an exercise that I've been struggling with for quite a while now:
Let $R$ be a ring and $M$ be the smallest set containing all principal ideals of $R$ and the GCD of each two ideals of $R$.
Show that $R$ is noetherian if and only if $M$ is the set of all ideals of $R$.
So here's how I went about it:
First, let $R$ be a noetherian Ring. Then we know that every ideal $I \trianglelefteq R$ is finitely generated. Thus, $\forall I \trianglelefteq R: \exists n >0,a_1,\dots,a_n\in R: I=(a_1,\dots,a_n)$.
So now for a given Ideal $ (a_1,\dots,a_n)=I\trianglelefteq R$ we have in case $n=1$ that its is a principal ideal. In case $n>1$ we have for $1\leq m < n$ that $I=gcd((a_1,\dots,a_m),(a_{m+1},\dots,a_n))$. And this proves one direction of the equivalence.
For the other direction, I need to show that every ideal $I\trianglelefteq R$ is finitely generated, or equivalently, that ever ascending chain of ideals $I_1 \subseteq I_2\subseteq \dots$ becomes stationary, knowing that each ideal $I_n$ is a principal ideal or the GCD of two ideals. Yet I'm lacking any idea of how to get close to proving that.
Any hints and help is dearly appreciated.
Merci.