Let $R$ be a commutative ring with $1$ and $I,J$ ideals of $R$. Show:
(1) $I+J:=\{i+j|i \in I,j \in J\}=\langle I \cup J \rangle_R$
(2) $\sqrt{I} := \{x \in R | x^n \in I, n\in \mathbb{N}\}$ is an ideal of $R$ and $I \subset \sqrt{I}$. Find an example so that $I \subsetneq \sqrt{I}$.
For (1) I have written out the definition of an ideal generated by a set:
$\langle I \cup J \rangle_R=\{\sum_{l=1}^n r_l a_l | n \in \mathbb{N}_0, a_1,...,a_n \in I \cup J, r_1,...,r_n \in R\}$. Clearly the $a_i$ are elements of $I$ and $J$. If I could somehow rewrite the sum I might end up with the definition of $I+J$ in a) but I don't see how to do this.
As for b) I also need some assistance.
Thanks for any help!
For (1) use the fact that $a_m,a_k\in I$ (or both in $J$) you can deuce that $r_m a_m +r_k a_k\in I$ (or to $J$). This also works for any number, so you can deduce that your sum belongs to $I+J$. The inclusion $I+J\subseteq \langle I\cup J\rangle _R$ is pretty trivial.
For (2) the inclusion is also easy. If $x\in I$ then $n=1$ we will have $x^n \in I$, and actually, any $n$ will work.
To find an example for $I\subsetneq \sqrt{I}$ try to think about ideals in $\mathbb{Z}$, they are of the form $m\mathbb{Z}$, for what n we can $x\in \mathbb{Z}$ such that $m|x^n$ but $n$ doesn't divide $x$? (you can think about $m=2$).
Good luck.