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

Let R be a commutative ring with unity and a in R. Then $⟨a⟩=\left\{ra:r\in R\right\}=Ra=aR$
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toroztatG

Let consider,
$0\in R,0.a=0\in Ra,$
$\therefore Ra\ne \phi \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Ra\subset R$
Let $x,y\in Ra$ then $x={r}_{1}a,y={r}_{2}a$ where ${r}_{1},{r}_{2}\in R$
Now $\left(x-y\right)=\left({r}_{1}-{r}_{2}\right)a=ra$ where $r={r}_{1}-{r}_{2}\in R,x-y\in Ra\dots \left(1\right)$
Let $x\in Ra,r\in R$
$x.r=\left({r}_{1}a\right)\left(r\right)=\left(r1r\right)a={r}^{\prime }a$ where ${r}_{1}r={r}^{\prime }\in R$
Since R is commutative $x.r=r.x\dots \left(2\right)$
$\therefore x\in Ra,r\in R⇒x.r=r.x\in Ra$
From (1) and (2) Ra is an ideal of R.
Since R is commutative then $aR=Ra$
Therefore $⟨a⟩=\left\{ra,r\in R\right\}=Ra=aR$ where $a\in R,r\in R$