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

Question
Commutative Algebra
asked 2021-03-02
Let R be a commutative ring with unity and a in R. Then \(\displaystyle{\left\langle{a}\right\rangle}={\left\lbrace{r}{a}:{r}\in{R}\right\rbrace}={R}{a}={a}{R}\)

Answers (1)

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_2aZSK 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 \(\displaystyle{a}\in{R},{r}\in{R}\)
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