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Let $R$ be a finite commutative ring with unity. Prove that every nonzero element of $R$ is either a unit or a zero-divisor.

I know this question has a solution here Every nonzero element in a finite ring is either a unit or a zero divisor.

But I want you to check my proof.

Let $a \not = 0, a \in R$.

We have to prove that $a$ is either a unit or a zero divisor.

Let $a$ is a unit then we have to show that it is not a zero divisor.

1) $\exists x \in R$ such that $ax = 1.$

2) Let for some $b \in R$, $ab = 0$ is true.

Now $ax = 1$ $\implies ax + 0 = 1 \implies ax + ab = 1 \implies a(x+b) = 1 .$

$\therefore$ $x$ and $(x+b)$ both are multiplicative inverses of $a \implies x = x+b$

$\therefore$ $b = 0$ hence $a$ is not a zero divisor

In a similar way we can show that if $a$ is a zero divisor then it is not a unit.

a) Is my proof okay?

b) I am not using the "finite" and "comutative" conditions of $R$.

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  • $\begingroup$ You are right but your are not proved the question. $\endgroup$ – Khayyam Feb 25 at 8:20
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No, the proof is not OK, as it has a fundamental problem: you've shown that there is no element that is both a unit and a zero divisor. Instead, you needed to show that there is no element that is neither a unit nor a zero divisor. To proceed in such a manner, you'd need to start by saying "Suppose $a$ is not a zero divisor", then somehow conclude that it's a unit.

The other thing to note is that "If $a$ is a zero divisor then it is not a unit" is logically equivalent to "If $a$ is a unit, then it is not a zero divisor". It is, in fact, the contrapositive of the statement. You don't have to prove both, or even say the other "can be proven similarly".

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