Properties of radical of an ideal Consider the following question from an assignment:

I have a question in (c),(d). Just give hints
Assuming x$\in \sqrt a \bigcap \sqrt b $ I got that there exists $n, m$ in $\mathbb{N}$such that $x^n \in a$ and $x^m \in b$ so $x^{n+m}\in  (ab)$.
But I am not able to prove the converse : I took $x \in \sqrt{ab}$ which implies there exists n $\in \mathbb{N}$ $x^n \in ab $. But I am unable to see how can I prove that x $\in \sqrt a \bigcap \sqrt b$?
Edit: In (d) I am unable to comprehend this: Let $x\in \sqrt{a+b}$ then $x^n \in a+b$ for some n $\in \mathbb{N}$. But how to use it to prove that $x^n \in \sqrt{a} +\sqrt{b}$ because I am having hard time comprehending how to write $x\in \sqrt{a} + \sqrt{b}$ . That is why when I tried the converse I was unable to comprehend on how to use $x^n \in \sqrt{a} + \sqrt{b}$ to prove the required.
 A: For (c), as you have proved $\sqrt{a}\cap\sqrt{b}\subset\sqrt{ab}$.
For the other inclusion let $x\in\sqrt{ab}$, then there is some $n\in\mathbb{N}$ such that $x^n\in ab$. By definition of product ideal this implies that there exists $m\in\mathbb{N}$ and $a_1,\dots,a_m\in (a)$, $b_1,\dots,b_m\in (b)$ such that $x^n=a_1b_1+\dots+a_mb_m$. From this expression of $x^n$ it follows immediately that $x^n\in (a)$ and $x^n\in (b)$, so $x\in\sqrt{a}$ and  $x\in\sqrt{b}$, thus $x\in\sqrt a\cap\sqrt b$.
For (d) notice that trivially $a+b\subset\sqrt a+\sqrt b$ and so you get $\sqrt{a+b}\subset\sqrt{\sqrt a+\sqrt b}$. For the other inclusion let $x\in \sqrt{\sqrt a+\sqrt b}$, there exists $n\in\mathbb{N}$ such that $x^n\in \sqrt a +\sqrt b $. From this there must be some $g\in \sqrt a$ and some $h\in \sqrt b$ such that $x^n=g+h$. From the definition, $g\in \sqrt a$ implies that there exists some $p\in\mathbb{N}$ such that $g^p\in a$ and $h\in \sqrt b$ implies that there exists some $q\in\mathbb{N}$ such that $h^q\in b$. Then $x^{n(p+q)}=(g+h)^{p+q}$, which by the binomial expansion it is a sum of thing that lie either in $a$ or in $b$ (Since the factors are of the form $Cg^nh^m$ with $C$ a constant and $n+m=p+q$, which implies either $n\geq p$ or $m\geq q$). Then $x^{n(p+q)}\in a+b$ and so $x\in\sqrt{a+b}$.
