Commutative ring with unique primary ideal

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?

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

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).