Show that if $I$ is in a maximal ideal $M$, then rad$(I)$ have to be in the same maximal ideal. 
Let $A$ be a unitary commutative ring and $I$ an ideal. Define the ideal
  rad$(I)$ of $A$ such that for each element $i \in $ rad$(I)$, there
  exists $m  \in \mathbb{N}$ where $i^m \in I$.
Show that if $I$ is in a maximal ideal $M$, so rad$(I)$ have to be in
  the same maximal ideal.

I know that each ideal has to be in a maximal ideal and that $I \subset$ rad$(I)$.
How could I prove this problem?
 A: Suppose that $x\in \mathrm{rad}(I)$. Then $x^n\in I$ for some $n$, which since $I\subset M$ means that $(x+M)^n=0$ in $A/M$. But $A/M$ is a field, so $x+M=0$, i.e. $x\in M$. Therefore $\mathrm{rad}(I)\subset M$.
A: It suffices to show $rad(I)$ is also an ideal. (Do you see why ? )
to do this suppose $r,s\in rad(I)$ and $m\in R$. we must show $r+s\in rad(I)$ and $rm\in I$
we show $r+s\in rad(I)$: 
verify $(r+s)=\sum\limits_{i=0}^n\binom{n}{i}r^is^{n-i}$.
Suppose $r^{n_1}\in I$ and $s^{n_2}\in I$. it is not too hard to see $r^n\in I$ if $n\geq n_1$ and $2^n\in I$ if $n\geq n_2$.
From here we have $(r+s)^{n_1+n_2}=\sum\limits_{i=0}^{n_1+n_2}r_1^{i}r_2^{n_1+n_2-i}$.
It is easy to see that either $i\geq n_1$ or $n_1+n_2-i\geq n_2$. From here either $r^{i}\in I$ or $s^{n_1+n_2-i}\in I$. Therefore $r_1^i r_2^{n_1+n_2-i}\in I$ for each $i$, we conclude $(r_1+r_2)^{n_1+n_2}\in I$, proving $r_1+r_2\in rad(I)$.

We show $rm\in rad(I)$:
suppose $r^n\in I$. Notice $(rm)^n=r^nm^n$, since $r^n\in I$ we conclude $r^nm^n\in I$. Proving $rm\in rad(I)$
