To show that R is a division ring given some property on ring R.

Let $R$ be a non-zero ring such that the equation $ax=b$ has a solution in $R$ for all $a, b\in R$ with $a\neq 0$. Show that $R$ must have unity and it is a division ring.

Now to show this we need to show that every non zero element is invertible. If we take '$a$' as non zero element with $ax=b$ then how can I use the given fact to show the ring is a division ring. Please help.

Here is how you solve the problem if it is already assumed that $$R$$ has a unit:

Taking $$b = 1$$, a solution to $$ax=1$$ is a right inverse for each nonzero $$a \in R$$. Fix such a right inverse $$ac=1$$. Now $$c$$ also has a right inverse, $$cd=1$$. It follows that $$a = a(cd) = (ac)d = d$$, so $$c$$ is both a right and left inverse for $$a$$. This proves that every nonzero element of $$R$$ has a two-sided inverse.

For showing that $$R$$ has an identity:

What you have is, for each nonzero $$a \in R$$, an $$e_a$$ so that $$a e_a = a$$.

If $$ab = 0$$ for nonzero $$a, b \in R$$, then one can multiply on the right by the solution to $$bx = e_a$$ to get $$0=abx = ae_a = a$$, which is a contradiction. So $$R$$ has no zero divisors, which, as was noted in one of the comments, is equivalent to both left and right cancellation holding. So the $$e_a$$'s are unique for each fixed $$a$$.

As rschwieb's comment below suggests, you can use cancellation together with $$a e_a e_a = a e_a$$ to conclude that $$e_a^2 = e_a$$, making $$e_a$$ an idempotent. Then left canceling $$e_a(e_a b - b) = 0$$ makes $$e_a$$ a left unit for $$R$$, and right canceling $$(b e_a - b) e_a = 0$$ makes $$e_a$$ a right unit for $$R$$.

This is closely related to a standard trick for breaking down a ring given an idempotent: For $$e \in R$$ an idempotent in an arbitrary ring, we have $$x = (x-ex) + (ex)$$, which gives a direct sum decomposition of $$R$$ as a right module over itself.

• More thoughts, since you are giving the best answer so far. From $ae_a^2=ae_a$ you get $e_a$ is idempotent. Then $(be_a-b)e_a=0$ implies $e_a$ is a universal right identity. Then $e_a(e_ab-b)=0$ implies it's a universal left identity. – rschwieb Mar 20 '17 at 3:48
• @rschwieb Incorporated. Thanks! – Dustan Levenstein Mar 20 '17 at 3:59

Take $b=1$ then every $a$ has an inverse since the equation has a solution. Then take $a=b$ and see that for every $a$ there exists $ax=a$.

• How can I take b=1 if it is required to prove the same (i.e to prove R must have unity)? – Kavita Mar 20 '17 at 2:56
• Sorry. I wrote it in the wrong order. First prove existence of unity. – Jean-François Gagnon Mar 20 '17 at 3:04
• Ok. But you gave the correct idea. – Kavita Mar 20 '17 at 3:05
• I'm not entirely convinced by this answer. Yes, there is some $x$ such that $ax = a$ for each $a$, but why is it clear that $x$ does not depend on $a$? That is, the unity element $e$ should satisfy $ye = y$ for all $y \in R$; why should the solution to $ax = a$ be consistent as $a$ varies? – Alex Wertheim Mar 20 '17 at 3:10
• Indeed: in fact there exist rings without identity in which the equations $ax=a$ and $xa=a$ have solution sfor every $a$ . This ought to be enough to make one more careful in this problem. – rschwieb Mar 20 '17 at 3:53