Let $R$ be a commutative ring. For $a \in R$ consider the set $(a) = \{r*a | r\in R\}$. Show that $(a) = R$ if a is a unit of $R$

Question

I tried some values and I think I got the idea. R is the set of values used in the ring. If I use $\mathbb{Z}$, the units are $\{-1,1\}$. If I take 1 for example, I could use it to get every value in $\mathbb{Z}$ with $(1) = \{r * 1 |r\in \mathbb{Z}\}$, such that $(1) = \mathbb{Z}$ However, I don't know how to prove that formally.

Answer 1

The following fact is useful and you should remember it:

If an ideal $I \unlhd R$ contains $1 \in R$, then $I=R$.

The proof is trivial: $1 \in I$ implies, by strong closure of multiplication, that $1 \cdot x \in I$ for any $x \in R$. But $1 \cdot x =x$, so $x \in I$. So $R \subset I$, and hence $I=R$.

Now let $a$ be a unit, and define $I=(a)$. To show $I=R$, we will show that $1 \in I$. $a$ is a unit in $R$, so there exists an element $b \in R$ such that $ab=1$. But by strong closure this means $1 \in I$.
So $I=R$.