I was trying to solve a problem in C. Musili's book "Introduction to Rings and Modules" and I got stuck while trying to solve the following problem:
Let $R$ be a non-zero ring such that the equation $ax = b$ has a solution in $R$ for all $a, b \in R, a \neq 0$. Show that $R$ must have unity and it is a division ring.
What I tried are as follows:
Before talking about invertibility we need the unity element $1$ and for a candidate we look at the equation $ax = a$ and obtain the solution, say $e$. Then show that $e$ is the required unity element.
After this look at solutions for the equation $ax = 1$ and get invertibility for all nonzero elements, showing that $R$ is a division ring. Commutativity of the inverses follows from showing the left and right inverses of $a$, say $b$ and $c$ respectively are equal. $ba =1$ and $ac = 1$ and so $b = c$.
My problem is with the first half itself. I am presently stuck trying to show that $ea = a$ as well, and the uniqueness of $e$ irrespective of choice of $a$. I don't have a clue on how to proceed. Any hints at trying to solve this are welcome.