Proving a Ring is a domain Suppose $R$ is a ring with the following property: given any elements $a , c \in R$ with $a \neq 0 $, there exists an element $x \in R$ such that $ax = c$. Show that $R$ is a domain.
I have started by assuming $\alpha, \beta \in R$ are both non zero, and I am trying to show that their product $\alpha \beta \neq 0$ also. From the conditions given in the question, there are $x, y \in R$ such that $\alpha x = \beta$ and $\beta y = \alpha$. However from here I am not sure how to show that $\alpha \beta \neq 0 $.
 A: If $R$ is the zero ring, then we are fine (depending on your definition of domain).
Assume now $R$ is not the zero ring. Let $a,b\neq 0$. Assume $ab =0$. Then use your property and associativity of the product to show that for all $c$ holds $ac=0$. Unless $R$ is the zero ring, this would contradict your property.
A: Since you didn't react to the answer of @Severin Schraven I will try to add a bit more detail to his answer, hoping this helps you, but full credit goes to him. As far as I understand that is how he wants to do it. Let $R \neq (0)$ be a ring with $1$, as well as $a,b \in R \smallsetminus \{0\}$ satisfying $ab=0$. Now let $x \in R$ be arbitrary. Since $b \neq (0)$ your property ensures the existence of an element $d \in R$ such that $x=bd$. We then have $ax=a(bd)=(ab)d=0$ by associativity of the ring and the assumption. Since this holds for all $x \in R$ we have that $1=a1=0$ since by definition of a ring with $1$ we have $1r=r1=r$ for all $r \in R$. This assumes that you mean integral domain by domain.
