# When a radical (or semiprime) ideal is prime?

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.

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

## 1 Answer

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 $I=K$.

• 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 $K to be radical ideals, just only ideals, right? – Xam Nov 10 '16 at 16:25 • Yes. If$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. – Keith Kearnes Nov 10 '16 at 16:31