Let $R$ be a commutative ring with multiplicative identity. Let $I$ be an ideal of $R$. Let $S=\{r \in R: r^n \in I\mbox{ for some natural number }n\}$. Show that $S$ is an ideal of $R$.
Give an example of a non-commutative ring $R$ and of an ideal $I$ in $R$ such that the set $S=\{r \in R: r^n \in I\mbox{ for some natural number }n\}$ is not an ideal in $R$.
For the first one I know the definition of ideal and tried to this to solved the problem but failed.
For the second one I have no idea.Can someone help me please.
For the first one, it is called the radical of an ideal, denoted $\sqrt{I}$, see Radical of an ideal. More precisely, suppose $x, y \in \sqrt{I}$, i.e. $x^n \in I, y^m \in I$, how about $(xy)^n, (x + y)^{m + n}, (rx)^n$?
For the second one, consider the ring of matrices. For example $Mat(2, \mathbb{R})$. $0$ is an ideal, and
$$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \in \sqrt{0}, \quad \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \in \sqrt{0}, $$
but
$$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \notin \sqrt{0}. $$