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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?

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  • $\begingroup$ 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. $\endgroup$ – Kris Williams Sep 13 '12 at 14:36
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    $\begingroup$ Hint: see my answer there. $\endgroup$ – Bill Dubuque Sep 13 '12 at 14:48
  • $\begingroup$ @DrKW Thanks for playing. $\endgroup$ – rschwieb Sep 13 '12 at 14:57
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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.

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  • $\begingroup$ Well McCoy certainly clears things up doesn't he :) $\endgroup$ – rschwieb Sep 13 '12 at 14:52
  • $\begingroup$ Indeed he does! :) $\endgroup$ – Manny Reyes Sep 13 '12 at 15:25

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