# 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.

• To your "2 to 3" question, $a\in aR$ should give you some info. It just remains to show that whichever $x$ has $ax=a$ works for all other elements as well. – Arthur Nov 4 '17 at 17:46

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:

1. 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 $x$.)

2. $xe=x$ 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.