Von Neumann ring

Let $(\mathcal{R}, +, \cdot)$ be a ring without necessary a unitary element but it has at least two elements. If $\mathcal{R}$ furthermore satisfies the property: "for every element $a \in \mathcal{R}$ that is not zero ($a \neq 0$) there exists a unique element $b \in \mathcal{R}$ such that $aba=a$ , then prove that:

1. $\mathcal{R}$ does not have zero divisors.
2. $bab=b$.
3. $\mathcal{R}$ has a unitary element.
4. $\mathcal{R}$ is a divisor ring.

Solution

Before we proceed with the exercise let us quote the following comment. A ring $(\mathcal{R}, +, \cdot)$ is called  regural ring in the sense of Von Neumann if forall $a \in \mathcal{R}$ there exists a $b \in \mathcal{R}$ such that $aba=a$. Also a regular ring in the sense of Von Neumann is called unitary under the same sense if forall $a \in \mathcal{R} \setminus \{0\}$ there exists a unique $b \in \mathcal{R}$ such that $aba=a$.

1. Let $a, c \in \mathcal{R}$ such that $ac=0$. We will prove that either $a=0$ or $b=0$. Suppose that $a \neq 0$. Let $b$ be the unique element of $\mathcal{R}$ such that $aba=a$. Then we have that:
   
   $$a(b+c)a=aba+aca=aba+0=a$$
   
   Due to the uniqueness of $b$ , $aba=a$, we have that $b+c=b$. Therefore, due to the law of cancelling we get that $c=0$. Similarly, if $c \neq 0$ we show that $a=0$. Hence $\mathcal{R}$ does not have zero divisors. 


2. From question i. since $\mathcal{R}$ does not have zero divisors we know that the law of cancelling will hold. Thus,
   
   $$aba=a \Rightarrow baba=ba \Rightarrow bab=b$$


3. We will prove that $\mathcal{R}$ has a unitary element and that this is $ab$ where $b$ is the unique element such that $aba=a$. Let $c \in \mathcal{R}$. Since $aba=a$ it holds that $ca=caba$ and thus:
   
   \begin{equation} c=c(ab) \end{equation}
   
   Since from question ii. we have that $bab=b$ , it follows that
   
   \begin{equation} c=(ab)c \end{equation}
   
   From equations $(1)$ and $(2)$ it follows that:
   
   $$(ab)c = c = c(ab)$$
   
   and thus $ab$ is the unitary element of $\mathcal{R}$, that is $ab=1$.


4. Let $a \neq 0$. Then , since $aba=a$ , and due to question iii. , it follows that $ab=1$. This means that $b$ is right inverse of $a$. Now, $a, b$ cannot be both zero since then the ring would have one element which is a contradiction due to the assumption. Using the Law of Cancelling we have that:
   
   $$ab=1 ab=abab \Rightarrow 1b = b = bab  \Rightarrow ba=1$$
   
   Hence $ab=1=ba$. This means that the element $a \neq 0$ is invertible and $a^{-1}=b$ holds. It follows that every non zero element of $\mathcal{R}$ is invertible and that $\mathcal{R}$ is a divisor ring.