$Rad(I)$ is an ideal of $R$ $$Rad(I)=\{a \in R | \exists n \in \mathbb{N} \text{ such that } a^n \in I\}$$
$R$ is a commutative ring, $I$ is an ideal.
To show that $Rad(I)$ is an ideal of $R$, we have to show that for $a,b \in Rad(I)$ it stands that $a-b \in Rad(I)$ and for $r\in R$ it stands that $a\cdot r\in Rad(I)$, right??
When $a,b \in Rad(I)$ we have that $a,b \in R$ and $a^n, b^m \in I$.
$a-b \in R$.
But how can we show that $(a-b)^k \in I, k \in \mathbb{N}$ ??
 A: Note $J$ this radical of $I$. If $a\in J$ and $b\in A$, $a^n\in I$ for some $n>0$ so that $(ba)^n = b^n a^n \in I$ because $I$ is a ideal, so that $ba\in J$. Now, if $a,b\in J$, whose respective powers $a^n$ and $b^m$ are in $I$, then by commutatativity we have $(a+b)^{n+m} = \sum_{k=0}^{n+m} {n+m \choose k} a^k b^{n+m-k}$ by the well-known binomial formula and this is in $I$ because in the sum, each term is in $I$, and this is because for each $k$, either $k\geq n$ or $n+m-k\geq m$ which ensures that one the two factors of the term are in $I$, so that the other is also in $I$ as $I$ is an ideal. This shows that $a+b\in J$, so that $J$ is an ideal of $A$.
A: The binomial theorem holds in any commutative ring $R$.  Thus
$(a - b)^k = \sum_0^k \dfrac{k!}{i!(k - i)!} a^i (-b)^{k - i}. \tag{1}$
Now if $a, b \in R$ with $a^n, b^m \in I$, then taking $k = m + n$ yields
$(a - b)^{m + n}= \sum_0^{m + n} \dfrac{(m + n)!}{i!(m + n - i)!} a^i (-b)^{m + n - i}, \tag{2}$
and for every $i$, $0 \le i \le m + n$, either $i \ge n$ or $m + n - i \ge m$.  In the former case, $a^i \in I$ and in the latter, $b^{m + n - i} \in I$.  Thus every product $a^i (-b)^{n + m - i}$ occurring in (2) is also in $I$, whence $(a - b)^{n + m} \in I$, and that's how it is shown.
It is also easy to see that if $a^n \in I$, then $(ra)^n = r^n a^n \in I$, whence $ra \in Rad(I)$ for $r \in R$, completing the demonstration that $Rad(I)$ is an ideal.
