Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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!)

share|improve this answer
    
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
1  
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
add comment

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 $\sqrt{I}=\langle xy^2,z\rangle$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.