# Is a finite commutative ring with no zero-divisors always equal to the ideal generated by any of its nonzero elements

Let $R$ be a finite commutative ring with no zero-divisors. We have proven that then $R$ has a unity.

Is it then always true that $R = (y)$ when $y \neq 0$?

We have a function $x \mapsto xy$ where $y \neq 0$ from $R \to (y)$. Then the kernel must be zero, thus making it injective, and a bijection since both are finite. Does this always hold?

• Yes. In other words, $R$ is a field, because in a commutative ring with unity, $y$ is invertible iff $(y) = R$. – Andreas Caranti Jan 3 '17 at 20:31

In the finite case you have that a commutative ring $R$ (not the zero-ring) is a field $\iff$ it has no zero-divisors (except 0 itself, depending on the Def. of zero-divisor).
Therefore the units of R are $R^{*} = R\backslash \{0\}$, hence $R=\langle y\rangle$ $\forall y \ne 0$.
Most important part here is that it is finite. If it's not, consider $R[X]$ as a counterexample ($y = X$, $R[X]\ne \langle X\rangle$).