Let consider,

\(\displaystyle{0}\in{R},{0}.{a}={0}\in{R}{a},\)

\(\displaystyle\therefore{R}{a}\ne\varphi{\quad\text{and}\quad}{R}{a}\subset{R}\)

Let \(\displaystyle{x},{y}\in{R}{a}\) then \(x=r_1a,y=r_2a\) where \(\displaystyle{r}_{{1}},{r}_{{2}}\in{R}\)

Now \(\displaystyle{\left({x}−{y}\right)}={\left({r}_{{1}}−{r}_{{2}}\right)}{a}={r}{a}\) where \(\displaystyle{r}={r}_{{1}}−{r}_{{2}}\in{R},{x}−{y}\in{R}{a}\ldots{\left({1}\right)}\)

Let \(\displaystyle{x}\in{R}{a},{r}\in{R}\)

\(\displaystyle{x}.{r}={\left({r}_{{1}}{a}\right)}{\left({r}\right)}={\left({r}{1}{r}\right)}{a}={r}'{a}\) where \(\displaystyle{r}_{{1}}{r}={r}'\in{R}\)

Since R is commutative \(\displaystyle{x}.{r}={r}.{x}\ldots{\left({2}\right)}\)

\(\displaystyle\therefore{x}\in{R}{a},{r}\in{R}\Rightarrow{x}.{r}={r}.{x}\in{R}{a}\)

From (1) and (2) Ra is an ideal of R.

Since R is commutative then \(aR=Ra\)

Therefore \(\displaystyle{\left\langle{a}\right\rangle}={\left\lbrace{r}{a},{r}\in{R}\right\rbrace}={R}{a}={a}{R}\) where \(\displaystyle{a}\in{R},{r}\in{R}\)