# Mistake in proof that every unit is a zero-divisor [duplicate]

This occurred to me when I was trying to prove that "non zero elements of a finite commutative rings are either units or zero divisors"
Let $$R$$ be a finite commutative ring with unity and let $$a(\neq 0)\in R$$ such that $$a$$ is a unit.
Let $$x\in R$$ such that $$x\neq a^{-1}$$ and $$x\neq 0.$$
Then $$a\cdot x=y$$ for some $$y\in R$$ where $$y\neq 0,1$$
$$\implies a\cdot x-y=0$$ $$\implies a\cdot( x- a^{-1}y)=0$$ Does this not contradict with the fact that a non zero element of a Commutative ring with unity is either a unit or a zero divisor, since we have clearly proved above that if an element is a unit, it is also a zero divisor?

• How do you know that $x-a^{-1}y\ne 0$? – Bernard Mar 29 at 13:03
• $a\cdot0=0$ doesn’t mean $a$ is a zero divisor – J. W. Tanner Mar 29 at 13:05
• @J. W. Tanner how do you know that $x-a^{-1}y=0$ – Denis James Mar 29 at 13:32
• $ax=y\iff x=a^{-1}y\iff x-a^{-1}y=0$ – J. W. Tanner Mar 29 at 13:41
• oh thats was obvious XD... sorry for bothering – Denis James Mar 29 at 13:43

To show that an element $$a$$ is a zero divisor, you must show that $$a\cdot b=0$$ with $$b\ne0$$.
You have not shown that, because $$x-a^{-1}y=0$$ in your example, given $$a\cdot x=y$$.