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. $\endgroup$ – Mariano Suárez-Álvarez Feb 25 '18 at 23:01
  • $\begingroup$ (This is done is 90% of algebra textbooks, so if you are really lost, you could google…) $\endgroup$ – Mariano Suárez-Álvarez Feb 25 '18 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 Feb 25 '18 at 23:13
  • $\begingroup$ @egreg that is asked lather $\endgroup$ – Gabriele Cassese Feb 25 '18 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 $\endgroup$ – Gabriele Cassese Feb 25 '18 at 23:21

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? $\endgroup$ – Gabriele Cassese Feb 25 '18 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.