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Let $R$ be a commutative ring with $1$ and let $I,J\!\unlhd\!R$. The radical of an ideal $I$ is defined as $$\mathrm{rad}(I):=\sqrt{I}:=\{r\!\in\!R;\;\exists n\!\in\!\mathbb{N}: r^n\!\in\!I\}=\bigcap\{P;\; P\text{ is a prime ideal with }I\!\subseteq\!P\}.$$

I've managed to show that there holds$$\sqrt{I}\sqrt{J}\subseteq\sqrt{IJ}=\sqrt{I\cap J}=\sqrt{I}\cap\sqrt{J}.$$

How can I prove the first $\supseteq$?

NOTE: I guess there has been made a mistake in Ralf Froberg's An Introduction to Grabner Bases, page 19: enter image description here

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up vote 8 down vote accepted

The result you wish to prove is false. For example, take $R=\mathbb{Z}$ and $I=J=(4)$. Then $\sqrt{I}\sqrt{J}=\sqrt{(4)}\sqrt{(4)}=(2)(2)=(4)$, while $\sqrt{IJ}=\sqrt{(4)(4)}=\sqrt{(16)}=(2)$.

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Uf. Thanks. Is the result still false in $K[x_1,\ldots,x_n]$? –  Leon Lampret Dec 16 '11 at 11:59
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Yes - we can similarly consider $I=J=(x_1^2)$, so that $\sqrt{I}\sqrt{J}=\sqrt{(x_1^2)}\sqrt{(x_1^2)}=(x_1)(x_1)=(x_1^2)$, while $\sqrt{IJ}=\sqrt{(x_1^4)}=(x_1)$. –  Zev Chonoles Dec 16 '11 at 12:02
    
Thanks, Zev. I guess I should have tried more to find a counterexample, instead of trying to prove it. –  Leon Lampret Dec 16 '11 at 12:17
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If you assume that $I,J$ are relatively prime, then the claim is true.

For, by what you found, $\sqrt{I}\sqrt{J}\subseteq \sqrt{I}\cap\sqrt{J}$, and the intersection equals the product if they are coprime.

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