Every \(a\in R\) for a finite ring is either a unit or a zero divisor.
Solution
Let \(R\) be a finite commutative ring with identity, and let \(a\in R\).
1. If multiplication by \(a\) is injective, then \(a\) is a unit.
Consider \[ \mu_a:R\to R,\qquad x\mapsto ax. \] Since \(R\) is finite, an injective self-map is surjective. Thus \(1\in\operatorname{im}(\mu_a)\), so there exists \(b\in R\) with \[ ab=1. \] Hence \(a\) is a unit.
2. If \(a\) is not a unit, then \(a\) is a zero-divisor.
By contraposition of 1, if \(a\) is not a unit then \(\mu_a\) is not injective. Hence there exist \(x\ne y\) with \[ ax=ay. \] Then \(0\ne x-y\) and \[ a(x-y)=0, \] so \(a\) is a zero-divisor.
Therefore every element of a finite commutative ring with identity is either a unit or a zero-divisor.