$(a) = R$ if and only if a is a unit.

Question

Let $⟨R;+,−,0,·,1⟩$ be a commutative ring. For $a \in R$, define $(a)$ $:= \{a · r | r \in R \}.$ How can i prove that $(a) = R$ if and only if a is a unit.

So if there exist $a' \in R$ such that $a*a' = 1$ we have $ \forall r \in R $ $ (a*a')*r = r$ hence $(a) = R$ , right ?

If $(a) = R$ than $ \forall r \in R $ there exists $r' \in R $ such that $ a*r' = r$ but than ...?

Is the first part correct and how can i formulate the second part?

Answer 1

The first part is correct: if $a$ is a unit, then $ab=1$ for some $b\in R$; therefore, for every $r\in R$, $$ r=r1=r(ab)=a(rb)\in(a) $$ hence $R\subseteq (a)$ and therefore $R=(a)$.

The second part is simpler: if $(a)=R$, then $1\in (a)$, so there exists $b\in R$ with $ab=1$.

Answer 2

The first part is correct.

For the second one, your initial thought needn't be true: as an example, consider $R=\mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = \mathbb{Z}_p$, in general $(-1)r \neq r$. The only implication is that $\forall r \in R \quad\exists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.