Proving various radical ideal containents

Question

So i've been trying to show these equalities for so long, particularly the equality in (b). I really have just been filling up pages and going nowhere fast, can somebody lend me some insight here?

$\DeclareMathOperator{\Rad}{Rad}\Rad(I)=\{a \in R$ such that $a^n \in I$ for some $n \in \mathbb{N}\}$

(a) Let $I,J$ be ideals of $R$. Show that $\Rad(IJ)=\Rad(I \cap J)=\Rad(I) \cap \Rad(J)$.

(b) Also show that $\Rad(I)+\Rad(J) \subset \Rad(I+J)=\Rad(\Rad(I)+\Rad(J))$

Answer 1

I'll write $\sqrt{I}$ for the radical of $I$.

To show $\sqrt I+\sqrt J\subset\sqrt{I+J}$ take $a\in\sqrt I$ and $b\in\sqrt J$. Then $a^m\in I\subset I+J$ and $b^n\in J\subset I+J$ for some $m$, $n\in\Bbb N$. Can you determine some $p\in\Bbb N$ for which $(a+b)^p$ is definitely an element of $ I+J$?

Once you have $\sqrt I+\sqrt J\subset\sqrt{I+J}$ you have $\sqrt{\sqrt I+\sqrt J}\subset\sqrt{I+J}$ so to complete (b) you need $\sqrt{I+J}\subset\sqrt{\sqrt I+\sqrt J}$. But surely, $I+J\subset\sqrt I+\sqrt J$ ?

Answer 2

Hint:

You have to observe first that

• if $\;\mathfrak a \subset\mathfrak a'$ are ideals in $R$, then $\DeclareMathOperator{Rad}{Rad}\;\Rad\mathfrak a \subset\Rad\mathfrak a'$,
• for any ideal $\;\mathfrak a,\enspace\mathfrak a \subset\Rad\mathfrak a$.

(a) Use the previous lemma with $\;IJ\subset I\cap J\subset I, J$.

(b) Similarly, $\; I, J\subset I+J\subset\Rad(I+J) $.