Question

If R is a ring and $a\in R$ , let$A_R=\left\{r\in R\right\}\overline{)ar=0_R}$ . Prove that$A_R$ is a subring of R. $A_R$is called the right annihilator of a.

Short answer

It is proved that A_R is a subring of ring R.

Step-by-step solution

Subring

It is given that$A_R=\left\{r\in R\overline{)ar=0_R}\right\}$ . We have to show$A_R=\left\{r\in R\overline{)ar=0_R}\right\}$ is a subring of R.

If $A_R=\left\{r\in R\overline{)ar=0_R}\right\}$ satisfies the following conditions then, we can say$A_R=\left\{r\in R\overline{)ar=0_R}\right\}$ is a subring of R.

Conditions of subring

(1). If $b,c\in A_R\Rightarrow ab=0_R$ and $ac=0_R$.

Distributive law:

$\begin{array}{rcl}& \Rightarrow & a\left(b+c\right)=ab+ac\\ & =& 0_R+0_R\\ & =& 0_R\end{array}$

It implies that $b+c\in A_R$.

(2). If role="math" localid="1648542158896" $b,c\in A_R\Rightarrow ab=0_R$ and$ac=0_R$ .

Associative multiplication law:

$\begin{array}{rcl}& \Rightarrow & a\left(bc\right)=\left(ab\right)c\\ & =& \left(0_R\right)c\\ & =& 0_R\end{array}$

It implies that $bc\in A_R$.

(3). If$bc\in A_R$ such that$ab+c=0_R$ .

$\begin{array}{rcl}ab& =& a\left(b+0_R\right)=ab+a{0}_R/+c\\ & \Rightarrow & 0_R=ab+c=\left(ab+a{0}_R\right)+c\\ & =& \left(ab+c\right)a{0}_R\left(\because ab+c=0_R\right)\\ & =& a{0}_R\end{array}$

It implies that $a{0}_R=0_R\in A_R$.

(4). If$b,c\in A_R$ such that $b+c=0_R$.

$\begin{array}{rcl}a\left(b+c\right)& =& ab+ac\\ & =& 0+0_R\\ & =& 0_R\end{array}$

Thus, $ac=0_R$ that is $c\in A_R$.

Hence, the set $A_R$ is a subring of a ring R.