Is \sqrt{I} an Ideal of a Commutative Ring R?

[Juanriq]

Salutations!

Homework Statement

Let R be a commutative ring and let $I \subseteq R$ be an ideal. Show $\sqrt{I}$ is an ideal of R if $\sqrt{I}$ is $f \in R$ such that there exists an $n \in \mathbb{N} \mbox{ such that } f^{n} \in I$.

Homework Equations

The Attempt at a Solution

Pick an $r \in R \mbox { and } x \in \sqrt{I}$. We want to show that $xr \in \sqrt{I}$. Well, this would imply that $(xr)^n = x^nr^n \in I$, I think this means that $x^n \in I$, but I am a little befuddled on how to proceed. Thanks!

 

[micromass]

Yes, that is correct.
So you still need to show that $$x,y\in \sqrt{I}$$ implies that x+y is also in the radical.
There exists n and m such that $$x^n\in I$$ and $$x^m\in I$$. Now try to show that $$(x+y)^{n+m}\in I$$.

 

[Juanriq]

Thanks micromass! So everything above is correct? I'm still not really sure how I see that $xr \in \sqrt{I}$ though. Is it because both $r^n \mbox{ and } x^n$ are in radical I? I know showing the summation will be fun... all the terms will be in the ideal I, I think and the union of a bunch of ideals is still an ideal

 

[micromass]

Well, since $$x^n\in I$$ (by definition, since $$x\in \sqrt{I}$$), we got that $$r^nx^n\in I$$ (since arbitrary multiplication preserves elements in I).

 

[Juanriq]

ohhhhh-gotcha! Thanks, I appreciate it. I'm sure I'll be back later after trying the second part. Thanks again!