Let $S$ be a nonempty subset of $R$. Let $r(S)=\{x \in R|Sx=0\}$ and $l(S)=\{x \in R|xS=0\}.$ Show that $r(S)$ and $l(S)$ are ideals in $R$ if $S$ is an ideal in $R.$
Here is my attempted solution for showing that $r(S)$ is an ideal:
- nonempty:
clearly $x=0$ is in $r(S)$ since $0$ multiplied by anything in $S$ will give us $0$. Thus $r(S)$ is nonempty.
- for all $x$ and $y$ in $r(S)$ show that $x-y$ is in $r(S):$
let $x$ and $y$ be in $r(S)$ then $Sx=0$ and $Sy=0$. $S(x-y)=Sx-Sy=0-0=0$. Thus $x-y$ is in $r(S)$.
- for all $x$ in $r(S)$ and $r$ in $R$ show $xr$ and $rx$ are in $r(S)$
let $x$ be in $r(S)$ then $Sx=0,$ let $r$ be in $R$. $Sxr=(Sx)r=(0)r=0.$ Thus $xr$ is in $r(S)$.
This is where I get stuck. Now I have to show that $Srx=0.$ I know that I have to use the fact that S is an ideal because I haven't used that yet but I have no idea how to use it.
Since $S$ is a (right) ideal, we have $Sr\subseteq S$ so that $(Sr)x\subseteq Sx=\{0\}$. -QED-