Suppose R is a ring, in which for any nonzero a and b in R the equation ax=b has an answer. prove that R doesnt have left or right zero divisor.

Question

My attempt: by contradiction suppose 'a' (nonzero) is a zero divisor so there exists some 'b' (nonzero) which a.b=0 since ax=b has an answer so a(ax)=0 implies a.a(x)=0 now i have no idea.

Answer 1

Note that if the equation $ax=b$, $a\ne 0$, can be solved, then in particular the equation $ax=1$, where $1$ is the unit element, can be solved. So each element $a\ne 0$ has a multiplicative inverse.

But units are no zero divisors. Indeed, if $a\in R$ is a unit and $ab=0$, then $0 = a^{-1}0 = a^{-1}(ab) = (a^{-1}a)b=1b=b$.

Answer 2

Assume for contradiction that $a\neq 0$ is a right zero divisor, with $c\neq 0$ such that $ca = 0$. By assumption, there is some $b\in R$ such that $cb\neq 0$ (such a $b$ must exist; take any $d\neq 0$ in the ring*, and let $b$ be the solution to $cx = d$). Now $ax = b$ has a solution, but $0 = cax = cb$ is a contradiction.

The same contradiction shows that there cannot be any left zero-divisor $c$ in our ring (let $c$ be a left zero-divisor with $ca = 0$ and again look at $cax = cb$).

*It may be that the ring just has the zero element. But then it clearly doesn't have zero divisors, so we're still good.

Answer 3

I'm adding this answer to augment Wuestenfux's solution by demonstrating the ring has an identity.

Let $a$ be nonzero. Then $ae=a$ for some $e$. Then also $ay=e$ for some $y$. Stringing these together, $aay=ae=a$, so $R$ is strongly regular, and therefore a reduced ring.

With this, you can compute $(ea-a)^2=eaea-eaa-aea+aa=eaa-eaa-aa+aa=0$ so $ea=a$ as well.

Now let $b$ be any other nonzero element. There must exist $z$ such that $az=b$. You immediately have $eb=eaz=az=b$.

We similarly compute that $(be-b)^2=bebe-beb-bbe+bb=bbe-bb-bbe+bb=0$ so $be=b$ as well.

So $e$ is the identity for the ring. At this point you can either proceed as Wuestenfux has outlined, or else observe that the condition makes the ring a simple right module over itself, so therefore it is a division ring, and has no nonzero zero divisors.