Let $R$ be a commutative ring with unity. For an ideal $I$ of $R$, I am attempting to prove $\sqrt{I}=\{x\,|\,x^n\in I\}$ is an ideal.
Closure under multiplication with $R$ seems straight forward: for $x\in\sqrt{I}$ and $r\in R$, $(rx)^n=r^nx^n\in I,$ so $rx\in\sqrt{I}.$
My question relates to proving closure under addition. What I have currently is that if $x^n, y^m\in I$ and assuming $m\leq n$, then $x^n\left(x^n+2nx^{n-1}y+\dotsb+\binom{2n}{n}y^n\right)+y^m\left(\binom{2n}{n+1}x^{n-1}y^{(n-m)+1}+\dotsb+y^{2n-m}\right)=(x+y)^{2n},$ implying that $(x+y)\in\sqrt{I}$.
It seems like even if I am correct there is probably a more straight-forward way of doing this. I searched this site and found and answer that I don't understand, so I'm worried I might be missing something.