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

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 $$R$$.

• You are right but your are not proved the question. – Khayyam Feb 25 at 8:20

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".