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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?

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

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