Extended ideals If $R\subset S$ is a ring extension where $1\in R$ and $I$ is an ideal of $R$ is it true that $IS$, the subset of $S$ generated by $I$ is an ideal of $S$? Should we assume $R$ is commutative?

It would be sensible to look at an example where $S$ is a simple ring to limit the number of ideals possible in $S$.
Take, for example, the $2\times 2$ upper triangular matrices $R=T_2(\Bbb R)\subseteq M_2(\Bbb R)=S$, and consider the ideal $I$ of $R$ of matrices of the form $\begin{bmatrix}0&x\\0&0\end{bmatrix}$.
Clearly $IS=\{\begin{bmatrix}a&b\\0&0\end{bmatrix}\mid a,b\in\Bbb R \}$ is not an ideal (although it is a right ideal) of $S$.
If $S$ is commutative (or even if $I$ is contained in the center of $S$) you can do the footwork to show that then $IS$ is indeed an ideal of $S$.
