Algebraic set - Radical Ideal - $Rad(Rad(I))=Rad(I)$ In my lecture notes we have the following: $V \subseteq K^n$ is an algebraic set $\Leftrightarrow$ it is of the form $V(I)$, where $I=$radical Ideal of $K[x_1, x_2, \dots , x_n]$. It stands that $$Rad(Rad(I))=Rad(I)$$ How can we show that $$Rad(Rad(I))=Rad(I)$$ ???

Suppose $a\in\operatorname{Rad}(\operatorname{Rad}(I))$. Then $a^m\in\operatorname{Rad}(I)$, for some $m$. Then $(a^m)^n\in I$ for some $n$.