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!