Zero divisor in $R[x]$

Let $R$ be commutative ring with no (nonzero) nilpotent elements. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ in $R[x]$ is a zero divisor, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$?

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@YACP: Why did you edit out the assumption that $R$ has no nilpotent elements? It's true that Bill's solution doesn't use that assumption, but now my answer looks like nonsense... –  Henning Makholm Dec 26 '13 at 21:10

The result is true over any commutative ring. It is sometimes called McCoy's theorem. Below is a proof sketch from my sci.math post on 5/4/2004:

THEOREM $\$ Let $\rm\:F \in R[X]$ be a polynomial over a commutative ring $\rm\:R\:.\:$ If $\rm\:F\:$ is a zero-divisor then $\rm\:r\:F = 0\:$ for some nonzero $\rm\:r \in R\:.$

Proof $\$ Suppose not. Choose $\rm\:G \ne 0\:$ of min degree with $\rm\:F\:G = 0\:.\:$

Write $\rm\:F = a +\:\cdots\:+ f\ X^k +\:\cdots\:+ c\ X^m\$

and $\rm\ \ \ G = b +\:\cdots\:+ g\ X^n\:,\:$ where $\rm\:g \ne 0\:$ and where $\rm\:f\:$ is the highest deg coef of $\rm\:F\:$ with $\rm\:f\:G \ne 0\:$ (note that such an $\rm\:f\:$ exists else $\rm\:F\:g = 0\:$ contra supposition).

Then $\rm\:F\:G = (a +\:\cdots\:+ f\ X^k)\ (b +\:\cdots\:+ g\ X^n) = 0\:.$

Thus $\rm\:f\:g = 0\:$ so $\rm\:\deg(f\:G) < n\:$ and $\rm\: F\:(f\:G) = 0\:,\:$ contra minimality of $\rm\:G\:. \quad$ QED

Alternatively it's an immediate corollary of Gauss' Lemma (Dedekind-Mertens) or related results.

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Assume that $gf=0$ for some $g\in R[X]$ and let $c$ be the leading coefficient of $g$. Then $ca_n=0$. Therefore $cf$ is either $0$ (in which case $c$ is your $b$), or $cf$ has degree less than $n$ with $g(cf)=0$. Proceed by induction on $n$. In the end you find that some power of $c$ kills every $a_i$, and $c$ was not nilpotent ...
Why "$c$ was not nilpotent"? –  Bill Dubuque Dec 26 '13 at 21:07
@Bill: Before YACP edited the question two days ago, it specified explicitly that $R$ has no nilpotent elements ... –  Henning Makholm Dec 26 '13 at 21:09