Zero divisor in $R[x]$ Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$?
It is  true over any commutative ring, and is sometimes called McCoy's theorem. Below is a proof sketch
from my sci.math post on May 4, 2004:

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

Proof $\ $  Suppose not. Choose $ \,G \ne 0\,$ of min degree with $ \,FG = 0.\,$

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

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

Then $ \,FG = (a +\,\cdots\,+ f\ X^k)\ (b +\,\cdots\,+ g\ X^n) = 0.$

Thus $\ \,fg = 0\ $ so $\: \deg(fG) &lt; n\,$ and $ \, FfG = 0,\,$ contra minimality of $ \,G.\ \ $ QED

Alternatively it follows by Gauss's Lemma
(Dedekind-Mertens form) or related results.
Let f be a polynomial in the commutative ring R. Suppose fg is 0 where g is nonzero. Adjoining the coefficients of f and g to the prime subring we can assume the ring is Noetherian via the Hilbert Basis Theorem. Take a reduced primary decomposition of O in R. Extending each primary ideal to R[X] gives a primary decomposition for 0 in the polynomial ring. So f being a zerodivisor is contained in some P[X] where P is an associated prime of 0 in R. But P is killed by some nonzero c of R, c so r kills f.