units in commutative rings An element $a$ of a ring $R$ is called a unit if it has a two sided inverse under multiplication; that is, if there exists $b \in R$ such with $ab = ba = 1_R$. 
How would you show that if $R$ is commutative, then for the subset $(a) := \{ ra \, | \, r \in R\}$. $(a) = R$ if and only if $a$ is a unit of $R$?
 A: Hint: If $(a) = R$, then $1_R \in (a)$, which means...
For the reverse direction, if $a$ is a unit, then $a^{-1} \in R$. Thus $aa^{-1} = 1_R \in (a)$, and so...
A: If $(a) = R$, then there exists a $r$ such that $ra = 1$, so $a$ is a unit. 
Suppose $a$ is a unit, then $a^{-1}$ exists. For any $r \in R$, $r = (ra^{-1})a$. 
A: HINT:
Recall that $1 \in R$. So if $(a) = R$, then in particular _
On the other hand, if $a$ is a unit, then in particular there is a $b$ so that $ab = 1$. Then _
A: If $(\alpha)=R$ then $1_R \in (\alpha)$ and so there exists $r \in R$ such that $r \alpha = 1_R$. Since $R$ is commutative $r \alpha = \alpha r$, hence $\alpha$ is a unit. Conversely, if $\alpha$ is a unit there exists $r \in R$ such that $r \alpha = 1_R$ and so $1_R \in (\alpha)$. Now for every $r' \in R$ we have that $r' (r \alpha)= (r' r) \alpha$ and so $r' (r \alpha) \in (\alpha)$. But $r' (r \alpha)=r'$ and so $r' \in (\alpha)$. Thus $R \subset (\alpha)$. Since $(\alpha) \subset R$ we get $(\alpha)=R$.
A: It looks like you have just started ring theory so I will show you how to do these kinds of things. 
$\underline{\implies}$ Suppose you have that $(a) : = \{ra : r\in R\}$ is equal to the whole ring $R$. This means that given any element in $R$, it is in $(a)$ and vice versa. We take advantage of the first inclusion, namely that given any element in $R$, it is an element in $(a)$. In other words we would like to take advantage of the fact that $(a) \subseteq R$. Now in particular we can take the identity element $1_R \in R$, since $1_R \in (a)$ by our subset inclusion. Therefore this means by definition of $(a)$ that we can write $1_R$ as the product of $a$ and some element in $R$. In other words, there exists some $r \in R$ such that 
$$ra = 1_R.$$
Note that $r$ cannot be zero for then $0\cdot a = 1_R$ which is impossible unless $R$ is the zero ring, which we exclude here otherwise there is nothing to prove. Now the statement $ra= 1_R$ means that $r$ is a left-inverse for $a$. However since we are in a  commutative ring, this means that $ra = 1_R$ too so that $r$ is a two-sided inverse for $a$, that is that $a$ is invertible.
$\underline{\Leftarrow}$ Now suppose that $a$ is a unit in $R$. This means that $a^{-1}$ exists and $a^{-1}$ is such that
$$a^{-1}a = aa^{-1} = 1_R.$$
Now by definition $(a)$ is the set of all multiples of $a$. Since $a^{-1} \in R$, it must be the case that $a^{-1}a \in (a)$ by definition of $(a)$. In other words, $1_R \in (a)$. Now we already know that $(a) \subseteq R$. We also know in addition that $(a)$ is an ideal and hence must be closed under multiplication "from the outside". This means to say that given any $\alpha \in (a)$ and $r \in R$, $\alpha r$ must be in $(a)$. Therefore since $1_R \in (a)$, this means that for all $r \in R$,
$$1_R \cdot r = r$$
must be in $(a)$. But then this is exactly saying that $R \subseteq (a)$. Since we already had the inclusion $(a) \subseteq R$, this forces
$$(a) = R.$$
Is this clear to you now? I can always discuss with you in the comments below if you wish.
