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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.

martinr
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    What have you tried so far? – Dietrich Burde Mar 19 '23 at 20:35
  • See here, which links to a paper describing the general idea behind these type of inductive arguments. – Bill Dubuque Mar 20 '23 at 00:49
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    Hint: the gcd of two ideals is their sum. What does this mean in terms of generators? – CJ Dowd Mar 20 '23 at 20:47
  • Also, I think the statement isn't quite right: it should probably be that $M$ contains all principal ideals and is closed under taking gcds of ideals in $M$. (Any ideal $I$ of $R$ is the gcd of $I$ and $I$, so as stated $M$ always contains every ideal regardless of Noetherianity.) – CJ Dowd Mar 20 '23 at 20:51
  • Ok, the GCD of $A,B\trianglelefteq R$ is $A+B$. That should imply that the generator of $gcd(A,B)$ is the union of the generators of $A$ and $B$. What if both ideals are infinitely generated? – martinr Mar 22 '23 at 14:24
  • Let's create a sequence $(a_n){n>0}$ where $a_i\neq a_j$ if $i\neq j$ of elements of $R$ and a sequence of Ideals $I_n, n>0$ of $R$ such that $I_i=((a_n){0<n<i})$. Then we have $I_i\subseteq I_{i+1}$ for all $i>0$ and thus an ascending chain of ideals. Let's now consider the ideal $\bigcup_{i>0} I_i=:I$. By construction, $I$ is infinitely generated, right? – martinr Mar 22 '23 at 14:46

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