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.

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    $\begingroup$ 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. $\endgroup$ – rschwieb Mar 20 '17 at 3:48
  • $\begingroup$ @rschwieb Incorporated. Thanks! $\endgroup$ – 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 $ax=a$.

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    $\begingroup$ How can I take b=1 if it is required to prove the same (i.e to prove R must have unity)? $\endgroup$ – Kavita Mar 20 '17 at 2:56
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    $\begingroup$ Sorry. I wrote it in the wrong order. First prove existence of unity. $\endgroup$ – Jean-François Gagnon Mar 20 '17 at 3:04
  • $\begingroup$ Ok. But you gave the correct idea. $\endgroup$ – Kavita Mar 20 '17 at 3:05
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    $\begingroup$ 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? $\endgroup$ – Alex Wertheim Mar 20 '17 at 3:10
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    $\begingroup$ 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. $\endgroup$ – rschwieb Mar 20 '17 at 3:53

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