Originally Posted by

**Deveno** can this be proved without the use of Zorn's lemma? perhaps i am not understanding the question, but condition two looks like an ascending chain condition on the principal ideals:

$\displaystyle (a_i)$.

i was thinking about this, and i ran into the following difficulty:

given any (non-zero) ideal I, we can form a chain of ideals by taking:

$\displaystyle I_1 = (a_1)$ where $\displaystyle a_1 \in I\setminus{0}$.

$\displaystyle I_2 = (a_2)$ where $\displaystyle a_2$ is chosen by finding $\displaystyle r_1 \in I\setminus I_1$ and setting $\displaystyle a_2 = gcd(r_1,a_1)$, so that $\displaystyle a_2|a_1$,

$\displaystyle I_3 = (a_3)$ where $\displaystyle a_3$ is chosen by finding $\displaystyle r_2 \in I\setminus I_2$ and setting $\displaystyle a_3 = gcd(r_2,a_2)$, so that $\displaystyle a_3|a_2$,

and so on, which by (2) terminates with some $\displaystyle I_N =(a_N)$.

now, it would be nice if we could conclude that $\displaystyle I = I_N$, but perhaps for some other choice of the $\displaystyle \{a_i\}$ we would wind up with "a different N". that is, how do we know that we can find an N that makes every element in $\displaystyle I\setminus I_N$ an associate of $\displaystyle a_N$? (i think this is where Zorn's lemma comes in. if i understand what you wrote, each Ra in your answer, is one of my "$\displaystyle I_N$"s).

my apologies if i have stated this badly.