In Bosch's Algebra you're asked to prove that
every commutative ring R is Noetherian iff every ideal is finitely generated
I think I managed to prove the if part (I write it just to be more explicit and to check it):
Let $a_i$ be an ascending chain of ideals $a_1\subset a_2\subset\ ...\subset R$
$\bigcup a_i=:a$, and a is still an ideal. Since every ideal is finitely generated by hypothesis, we have:
$a=(\alpha_1,...,\alpha_m),\ \alpha_i\in R$ Since the chain is ascending, there is an $n$ such that $\forall i\ \ \alpha_i\ \in a_n$, and thus $a_{m\geq n}=a_n$.
I don't know how to approach the only if part: is there any cardinality-based reasoning?
After this exercise, another one has got me stuck:
every commutative ring $R$ is Noetherian iff every prime ideal is finitely generated
Note: I would be really grateful if it wouldn't be necessary to use concepts such as modules and annihilators in proofs, as I'm not used to them
Hint: suppose $a$ is not finitely generated; then there exists $x_1\in a$ and $(x_1)\ne a$, so there is $x_2\in a\setminus(x_1)$ and $(x_1)\subsetneq(x_1,x_2)$. Go on.