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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Bill Dubuque
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Denis James
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How do you know that $x-a^{-1}y\ne 0$? – Bernard Mar 29 '20 at 13:03
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$a\cdot0=0$ doesn’t mean $a$ is a zero divisor – J. W. Tanner Mar 29 '20 at 13:05
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@J. W. Tanner how do you know that $x-a^{-1}y=0$ – Denis James Mar 29 '20 at 13:32
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$ax=y\iff x=a^{-1}y\iff x-a^{-1}y=0$ – J. W. Tanner Mar 29 '20 at 13:41
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oh thats was obvious XD... sorry for bothering – Denis James Mar 29 '20 at 13:43
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It's better if you delete questions based on simple oversights (that will likely not help others). Else it will make it more difficult for others to search and locate the actual results (vs. the oversights) due to search results containing many posts that aren't about the topic but rather about mistakes. – Bill Dubuque Mar 29 '20 at 15:36
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No, because units are never zero-divisors as I show in the first linked dupe. – Bill Dubuque Mar 29 '20 at 16:03
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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 $a\cdot x=y$.
J. W. Tanner
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