I have to prove or give an counter-example to the following

Any commutative ring with unit with a single primary ideal is Noetherian.

I have found the following example: Non-Noetherian ring with a single prime ideal

But in this case the ring has a single prime ideal, i cannot with this assure that this ring has a single primary ideal, but it makes me think that it is possible to construct such a counter-example.

Am i missing something? Can this be proved or is there indeed a counter example?

  • $\begingroup$ Single primary ideal implies single prime ideal, which implies $0$-dimensional local, which together with Noetherian, implies Artinian. $\endgroup$ Jul 15, 2021 at 19:44
  • 1
    $\begingroup$ I don't get how that answers $\endgroup$ Jul 15, 2021 at 19:45
  • $\begingroup$ Very nice, if you want to post it as an answer i will accept it $\endgroup$ Jul 15, 2021 at 20:35
  • 1
    $\begingroup$ @DanielW. Please use the comments section for comments and the solutions section for solutions! $\endgroup$
    – rschwieb
    Jul 16, 2021 at 17:06

1 Answer 1


Let $R$ be a ring with a single primary ideal $P$. Then $P$ is contained in a maximal ideal, which itself is primary, so $P$ is maximal.

Now let $I$ be any proper ideal of $R$. Then the radical of $I$ is the intersection of all prime ideals of $R$ containing $I$, that is $P$. But $P$ is maximal and hence $I$ is primary. Therefore $I$ equals $P$.

So $R$ has a single proper ideal and is in particular Noetherian (even a field).


You must log in to answer this question.

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