2
$\begingroup$

In a commutative ring, we know that every prime ideal is radical. So I'm looking for results about the converse but I only found this: A radical ideal in a commutative ring is prime if and only if it is not an intersection of two radical ideals properly containing it?

Are there other results about when a radical ideal is prime? Thanks in advance.

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ If $Q$ is a primary ideal, then its radical is prime. $\endgroup$ – Watson Nov 10 '16 at 15:50
  • $\begingroup$ @Watson yeah, that's true but a primary ideal is not in general a prime ideal. $\endgroup$ – Xam Nov 10 '16 at 16:03
  • 1
    $\begingroup$ I think you wanted to say that an ideal whose radical is prime is not in general a primary ideal. $\endgroup$ – Watson Nov 10 '16 at 16:05
4
$\begingroup$

An ideal in a commutative ring is prime iff it is radical and meet-irreducible. $I$ is meet-irreducible if whenever $I=J\cap K$, then $I=J$ or $I=K$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Mmm it seems that "meet-irreducible" is basically the same condition that I found in the link that's in my question, being the only diffference that we don't need $J$ and $K to be radical ideals, just only ideals, right? $\endgroup$ – Xam Nov 10 '16 at 16:25
  • 1
    $\begingroup$ Yes. If $I=\textrm{rad}(I)$ and $I=J\cap K$, then $I=\textrm{rad}(J)\cap \textrm{rad}(K)$. So if $I$ is radical, then it is meet-irreducible in the lattice of all ideals iff it is meet-irreducible in the lattice of radical ideals. $\endgroup$ – Keith Kearnes Nov 10 '16 at 16:31

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.