Question

Let R be a commutative ring with unity and a in R. Then <<a>>={ra:r in R}=Ra=aR

Commutative Algebra
Let R be a commutative ring with unity and a in R. Then $$\displaystyle{\left\langle{a}\right\rangle}={\left\lbrace{r}{a}:{r}\in{R}\right\rbrace}={R}{a}={a}{R}$$

2021-03-03

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 $$\displaystyle{a}\in{R},{r}\in{R}$$