# When do we have $Rad(I)=I$ for an ideal $I$ of a ring $R$?

This is kind of a follow-up question about calculating the radical of an ideal. Since

$Rad(I)$ is the intersection of all the prime ideals of $R$ that contain $I$,

which is a property I learned from this article in wikipedia, we have that $$Rad(I)=I$$ whenever $I$ is a prime ideal. My question is:

Can this be true for some $I$ which is not a prime ideal? [EDIT: And when is this NOT true?] Is there an equivalent easy-to-check conditions for this kind of $I$?

Let $R={\Bbb Z}[x]$, for example. $I=\langle x,2\rangle$ is a prime ideal and thus $Rad(I)=I$. For any ideal $I\unlhd R$, (say $I=\langle x^2+1\rangle$ or $I=\langle x^2+2\rangle$, etc.) the key point is to check $$Rad(I)\subset I$$ since $I\subset Rad(I)$ is always true. But I don't know a quick way to check this relation.

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By definition, every intersection of a set of prime ideals is semiprime (aka a radical ideal).

Let $\cap P_i=I$, where the $P_i$ are all prime. Then the set of all prime ideals containing $I$ contains the $P_i$. Thus, $\cap\{P\mid P\supseteq I\}\subseteq \cap P_i$. The left hand intersection involves "more" prime ideals, and so the intersection of more ideals should be smaller than just the set of $P_i$.

Thus in total: $$\cap P_i=I\subseteq\cap\{P\mid P\supseteq I\}\subseteq \cap P_i$$

So, there is equality all across.

It is relatively easy to find examples where prime ideals do not intersect to a prime ideal. For example, the intersection of two prime ideals, neither of which contains the other, cannot be prime. (Explain why!)

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What I learn from your answer is that to check if $Rad(I)=I$, I need to check if $I$ is some intersection of prime ideals. But how can I apply this to check, say, $I=\langle x^2+1\rangle$ in ${\Bbb Z}[x]$? –  Jack Nov 7 '12 at 20:59
For the commutative case, $rad(I)=\{x\in R\mid \exists n\in \mathbb{N}, x^n\in I\}$. That might help you do specific commutative examples. Isn't it the case here that $x^2+1$ generates a prime ideal? –  rschwieb Nov 7 '12 at 21:00

You can, in fact, have $I=\mathrm{Rad}(I)$ for nonprime $I$. For example, consider the ideal $\langle xy\rangle$ of the ring $\mathbb{Q}[x,y]$. This isn't prime since the generator isn't irreducible, but can easily be seen to be radical.

More generally, if you take a monomial ideal, i.e. an ideal generated by monomials, in a polynomial ring over some field, its radical will be generated by the "roots" of those monomials. E.g. if $I=\langle x^2y^4,z^3\rangle$, then $\sqrt{I}=\langle xy^2,z\rangle$.

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