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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.

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  • $\begingroup$ If $Q$ is a primary ideal, then its radical is prime. $\endgroup$
    – Watson
    Nov 10, 2016 at 15:50
  • $\begingroup$ @Watson yeah, that's true but a primary ideal is not in general a prime ideal. $\endgroup$
    – Xam
    Nov 10, 2016 at 16:03
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    $\begingroup$ I think you wanted to say that an ideal whose radical is prime is not in general a primary ideal. $\endgroup$
    – Watson
    Nov 10, 2016 at 16:05

1 Answer 1

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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 $I=K$.

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  • $\begingroup$ 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, 2016 at 16:25
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    $\begingroup$ 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, 2016 at 16:31

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