I have to prove that if $(R,+,\cdot )$ is a ring and $a \in R$ is a zero divisor, then for all $b\in R$, the equation $ax=b$ has either no solution or many solutions. I've been going at it for a couple of hours, but I get nowhere.
To my reasoning, if $a$ is a zero divisor then $a \neq 0$, so that leaves $b$ being either $0$ or not $0$, if $b \neq 0$ then $ax=b$ can have many solutions depending on what value is $b$, but if $b=0$ then since $a$ is a zero divisor then at least there exists a value that $ay=0$, so let it be the case that $x=y$ and there we have at least one solution.$$$$ What am I doing wrong? or what concept am I not understanding? Thanks for your help!
Let $ka=0$.
Then $ax=b $ means $kax=kb $ so $0=kb $.
Well, if that isn't true there can't be any solutions. If however it does happen to be the case that $kb=0$, well there still might not be any solution.
But if there is one solution so that $ax=b $ then for any integer $m $, $a (x+m*k)=ax+m*ak=ax=b $. So there are many.