8
$\begingroup$

I do believe that my question is simple. I do not understand what is wrong. So help me sort it out please. I know the following facts:

$\mathbb{Z}$ is a Noetherian ring and it is not Artinian because the infinite sequence $(\mathbb{Z}/2\mathbb{Z}) \supseteq (\mathbb{Z}/4\mathbb{Z}) \supseteq (\mathbb{Z}/8\mathbb{Z}) \cdots $ doesn't hold the Descending Chain Condition.

And

A ring $R$ is Artinian iff $R$ is Noetherian and every prime ideal is maximal.

We see that all prime ideals have the form $p\mathbb{Z}$ and are maximal. This is example of a module which is Noetherian but not Artinian as well.

$\endgroup$

1 Answer 1

16
$\begingroup$

$0$ is a prime ideal of $\mathbb{Z}$ that is not maximal.

$\endgroup$
4
  • 1
    $\begingroup$ Oh, I forget this trivial ideal :(. Thank you so much. $\endgroup$
    – Soulostar
    Nov 6, 2016 at 3:45
  • 1
    $\begingroup$ Does this mean that a (commutative) domain which is not a field can not be Artinian? The ideal 0 is prime in a domain and not maximal. $\endgroup$
    – J. De Ro
    Aug 21, 2019 at 18:17
  • 2
    $\begingroup$ @EpsilonDelta: Yes it does. Just some extra facts for context: one can show that (commutative) Artinian rings always have finitely many primes, and that the Artinian ring is isomorphic to the direct product of localizations at these primes. Now for a local Artinian ring, the maximal ideal is actually nilpotent, so modding out nilradical produces a field. So Artinian rings differ from (finite products of) fields precisely by the presence of some nilpotent elements. $\endgroup$ Aug 21, 2019 at 18:39
  • $\begingroup$ Thank you! I learnt what Artinian rings are from this question and answer. $\endgroup$
    – J. De Ro
    Aug 21, 2019 at 18:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .