# Radical of an ideal equals the intersection of all prime ideals containing it.

I am trying to prove that for a ring $$R$$ and an ideal $$I\leq R$$ we have $$\sqrt{I}=\bigcap_{I\leq\mathfrak p}\mathfrak p,$$ the intersection of all prime ideals containing $$I$$.

The definition of the radical of an ideal I am working with is $$\sqrt{I}=\{x\in R\mid\exists m\in\mathbb{N} \text{ with }x^m\in I\}.$$ I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?

• in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no. – rschwieb Oct 1 '16 at 11:37
• @rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime. – qbert Oct 1 '16 at 19:37

I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $\sqrt{I}\subset N$. Proving the reverse containinment is not as simple. Given $x\notin \sqrt{I}$, consider the collection $\Omega$ of all ideals $J \supset I$ such that for all $n\in \Bbb N$, $x^n \notin J$. Partially order $\Omega$ by inclusion, and show by Zorn's lemma that $\Omega$ has a maximal element $\frak{p}$. Since $\mathfrak{p}\supset I$, if you can show that $\frak{p}$ is prime, then you can claim $x\notin N$ (since $x\notin \frak{p}$). Take $a,b\notin \frak{p}$, and using maximality of $\mathfrak{p}$, show that $\mathfrak{p} + (ab) \notin \Omega$; this will imply $ab\notin \mathfrak{p}$ and thus $\mathfrak{p}$ is prime.
• Given a chain $\mathcal C$ in $\Omega$, the union of elements of $\mathcal C$ is an upper bound of $\mathcal C$ @qbert. – kobe Oct 10 '16 at 15:03