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

  • $\begingroup$ Pick an ideal and suppose it is not finitely generated. Fix a infinite set which generates it, and construct from it a strictly increasing infinite sequence of ideals. $\endgroup$ Commented Feb 25, 2018 at 23:01
  • $\begingroup$ (This is done is 90% of algebra textbooks, so if you are really lost, you could google…) $\endgroup$ Commented Feb 25, 2018 at 23:01
  • $\begingroup$ The title refers to prime ideals, the text doesn't. Are you sure you don't have to prove the more difficult result that $R$ is noetherian if and only if each prime ideal is finitely generated? $\endgroup$
    – egreg
    Commented Feb 25, 2018 at 23:13
  • $\begingroup$ @egreg that is asked lather $\endgroup$
    – user515010
    Commented Feb 25, 2018 at 23:17
  • $\begingroup$ @MarianoSuárez-Álvarez I tried to google that, but I always ended up with some works talking about modules or semi-groups, concept with which I'm not used at all yet $\endgroup$
    – user515010
    Commented Feb 25, 2018 at 23:21

1 Answer 1


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.

  • $\begingroup$ Ok, I got it: $(x_1)\subset (x_1,x_2)\subset...\subset a$, but since $R$ in Noetherian, then there is an $n: (x_1,...,x_n)=(x_1,...,x_n,x_{n+1})=a$ right? $\endgroup$
    – user515010
    Commented Feb 25, 2018 at 23:25

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