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Let $R$ be a commutative ring with unity $1$ not equal to $0$. Let $a\in R$, and define $aR = \{ar\mid r \in R\}$. Prove that $aR = R$ iff $a$ is a unit.

1st part:

Let $a$ be a unit in $R$ then there exists an element $b\in R$ such that $ab=ba=1$.

Let $x\in aR $ then $x=ar$ for some $r\in R$.

Also as $a\in R$ and $r\in R$ then as ring $R$ is closed with respect to multiplication, then $ar\in R$, i.e. $x\in R$.

Therefore, $x\in aR\implies x\in R$, so $aR \subset R$.

My problems are the following

1. How to show $R \subset aR$

2. How to show the converse.

Please write in full so that I can understand.

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If $aR = R$, note that $1 \in R$. So exists $b \in R$ s.t. $ab = 1$; so $a$ is a unit.

If $a$ is a unit, exists $b$ s.t. $ab=1$. By closure, $bR \subseteq R$; thus $a(bR) \subseteq aR$. But $a(bR) = (ab)R = R$; thus $R \subseteq aR$ and by closure $aR \subseteq R$; so $aR = R$.

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  • $\begingroup$ Instead using $a(bR) \subseteq aR$, can it be possible to show $R \subseteq aR$ by taking arbitrary element in the $R$ implying that element in $aR$? $\endgroup$ – rama_ran Oct 28 '16 at 8:34
  • $\begingroup$ Given $ab = 1$, if $x \in R$, then $x = (ab)x = a(bx) \in aR$ (since $bx \in R$, obviously). Roughly the same thing :); but often it's nicer to reason in terms of the larger collections $aR$ where possible - thus, the ideals that Andreas was speaking of. $\endgroup$ – Chas Brown Oct 28 '16 at 9:03
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Yes $aR\subseteq R$ is pretty straightforward by closure of the ring structure under multiplication. In particular this direction does not use that $a$ is a unit, and the argument you provide is too long and therefore confusing.

The opposite inclusion $R\subseteq aR$ does require $a$ to be invertible. If so, and $b$ is such that $ab=1$, you need to show that an arbitrary $x\in R$ can be written $x=ar$ with $r\in R$. Take $r=bx$, now $ar=abx=1x=x$.

Conversely, if $R\subseteq aR$ then in particular $1\in aR$, and there exists $r\in R$ with $1=ar$. Then $a$ is invertible, with inverse $r$.

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