# polynomials over a local Artinian (or finite) ring

In this question " Zero-divisors and units in $\mathbb Z_4[x]$ " it looks like it has been shown that the set of zero divisors of $\mathbb{Z}_4[x]$ coincides with its nilpotent elements.

Since the nilpotent elements coincide with the non-units in $\mathbb{Z}_4$ itself, and do so more generally for any commutative Artinian local ring, I wanted to follow up with these questions.

Does anyone know if this is true for $R[x]$ where $R$ is a commutative finite local ring?

If that was too easy:

Is it the case for commutative Artinian local rings?

• Every commutative finite local ring is Artinian. As the ring is finite, there can only be a finite number of ideals. Thus $R$ satisfies the descending chain condition. – Kris Williams Sep 13 '12 at 14:36
• Hint: see my answer there. – Bill Dubuque Sep 13 '12 at 14:48
• @DrKW Thanks for playing. – rschwieb Sep 13 '12 at 14:57

Let $R$ be a commutative ring in which every zero-divisor is nilpotent. I claim that the same is true for $R[x]$.
First, I'll mention a theorem of McCoy: if $S$ is a commutative ring, then $f(x) \in S[x]$ is a zero-divisor if and only if there exists a nonzero $c \in S$ such that $cf(x) = 0$.
Back to our ring $R$. If $f(x) \in R[x]$ is a zero-divisor, then the theorem of McCoy above shows that all of the coefficients in $f(x) = a_0 + a_1 x + \dots + a_n x^n$ are zero-divisors. Thus every coefficient $a_i$ is nilpotent by our assumption on $R$. Since the nilpotent elements of the commutative ring $R[x]$ form an ideal, we conclude that $f(x)$ is nilpotent.