For all $a, b \in R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ Suppose $R $  is a non zero ring.
How can we prove the following conditions are equivalent? 
In Proof 2 to 3, how can we prove  from $ a R = R $  that the ring has identity?

1:  For all $a, b \in  R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ .
2: $ R^{2} \neq 0$  and $ R$ has no right ideals other than $ 0$ and $R$.
3: $ R$ is a division ring. 

 A: The set of all elements $x$ such that $xR=0$ forms a right ideal of $R$. 
By hypothesis it can't be all of $R$, so it is just $\{0\}$.
In other words, $xR=R$ for nonzero $x$.
Let $x$ be a nonzero element. Two observations:


*

*The right annihilator of $x$ is a right ideal, and it can't be $R$, so it is zero. Therefore $x$ is left cancelable. (Notice this is true for any nonzero $x$.)

*$xe=x$ for some $e$, and furthermore $xee=xe=x$. Canceling $x$ we find that $e^2=e$.
Using the $e$ above, we see that $ey=e^2y$ implies $y=ey$ for any $y\in R$, so $e$ is a left identity.
Finally, there's probably a shorter trick, but this one is the first that occurred to me:
$(y-ye)^2=y^2-y^2e-yey+yeye=y^2-y^2e-y^2+y^2e=0=(y-ye)0$. If $y-ye\neq0$, we could cancel it on the left, obtaining $y-ye=0$, a contradiction. Therefore $y-ye=0$, and $e$ is a right identity too.
A: If $aR = R$, $a$ is not zero, then $R$ has identity:
we recall from rschwieb that left cancellation is valid. Now:

*

*$aR = R \implies ac = a$ for some $c \in R$


*$cR = R \implies cb = a$ for some $b \in R$


*Prove that $a = b$ and then $c$ will be our identity:
$ab = acb = a(cb) = aa \implies b = a$,
$ac = cb = ca = a \implies c$ is the identity.
