A ring is a divison ring

I'm confused with the following exercise:

Prove or disprove: Let $(R,+ , \cdot)$ a ring having the property that for each $a\not=0, \ b\in R$, there exist $r\in R$ such that $a\cdot r=b$, then $R$ is a divison ring.

Since $a$ and $b$ are not necessarily different, there exist $r_0\in R$, such that $a\cdot r_0 = a$, but this doesn't mean that $r_0$ is unique or $r_0\cdot a=a$, that's because I think the statement is false, but I can't think in a counterexample. Am I right? and there is a counterexample, or is the statement true?

Solution 1:

We can prove existence of unique and universal 1 provided there exists at least one $a\neq 0$.

First we prove (by contradiction) that there are no zero divisors. Assume $\exists u,v: uv=0 \wedge u\neq 0 \wedge v\neq 0$. Then $\exists r,q : ur=u, r=vq$. But then $u = ur=uvq=(uv)q = 0q=0$ -- q.e.d. Note: this also proves that $r$ is unique for a fixed $a$.

Next we can show that $r$ is a right-side multiplicative identity for all elements of $R$. I.e., if $\exists a : a\neq 0 \wedge ar=a$ then $\forall b: br=b$. Proof: if $b=0$ then $0r=0=b$. Now assume $b\neq 0$ and $br=c \neq b$. Then for some $q$ we have $c=bq$ which means $br=bq \rightarrow b(r-q)=0 \rightarrow r-q = 0$.

Now we can prove that $r$ acts as a multiplicative identity on both sides, i.e. $ra=ar=a$. Proof: $a^2 = (ar)a = a(ra) \rightarrow a(a - ra) = 0 \rightarrow a-ra = 0$