Prove that for ideals I and J of a commutative ring, $√(I∩J)=√(IJ)$

Question

Prove that for ideals $I$ and $J$ of a commutative ring, $√(I∩J)=√(IJ)$, where $√(I∩J)$ and $√(IJ)$ are the radical ideals.

The first step I use the idea that $IJ\subseteq{I∩J}$ is true for all ideals.

For the other direction, I try to use the definition.

let $a\in√(I∩J)$,then $a^n\in{I∩J}$ for some $n$. so that $a^n\in{I}$ and $a^n\in{J}$.

I need to show $a^n=i_1j_1+...+i_nj_n\in{IJ}$ for some $i_1,...,i_n\in{I}$ and $j_1,...,j_n\in{I}$. But I cannot find such $i's$ and $j's$.

Please help me for some hints. thank you.

Answer 1

$a^n\in{I}$ and $a^n\in{J}$. so $a^{2n}=a^n.a^n\in{IJ}$ hence $a\in √(IJ)$

Answer 2

There is a rather slick proof of this. Take some ideal $I\subset A$. There is an alternative formula for the radical. Namely, $\sqrt{I}=\bigcap P$ where $P$ ranges over all prime ideals containing $I$.

Since $IJ\subset I \cap J$ the inclusion $\sqrt{IJ}\subset \sqrt{I\cap J}$ is immediate. Take a prime ideal $P$ s.t. $IJ\subset P$. Then $I\subset P$ or $J\subset P$, so $I\cap J\subset P$. Thus, $\sqrt{I\cap J}\subset \sqrt{IJ}$.

To see $IJ\subset P\Rightarrow I\subset P$ or $J\subset P$, plainly note that if neither of those was true then you find $x\in I,y\in J$ s.t. $x,y\notin P$, but $xy\in P$ by assumption. Think of this as: "a prime that divides a product divides one of its factors".