Zero divisor in $R[x]$

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 $\: \deg(fG) < n\,$ and $ \, FfG = 0,\,$ contra minimality of $ \,G.\ \ $ QED

Alternatively it follows by Gauss's Lemma (Dedekind-Mertens form) or related results.

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 ...

Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.

Let $A$ be a commutative $\mathbb{N}$-graded ring. Let $f \in A$ be a zero divisor. Then there is some $0 \neq a \in A$ homogeneous such that $a f = 0$.

Proof. Choose some $0 \neq g \in A$ of minimal total degree with $fg=0$. Let $f=f_0+f_1+\cdots$ and $g=g_0+g_1+\cdots+g_d$ be the homogeneous decompositions with $g_d \neq 0$. If $f g_d = 0$, we are done. Otherwise, we have $f_i g_d \neq 0$ for some $i$, and hence $f_i g \neq 0$. Choose $i$ maximal with $f_i g \neq 0$. Then $0=fg=(f_0+\cdots+f_i)g=(f_0+\cdots+f_i)(g_0+\cdots+g_d)$ implies $f_i g_d = 0$. Then $f_i g$ has smaller degree than $g$, but still satisfies $f(f_i g)=0$ and $f_i g \neq 0$, a contradiction. $\square$

This may be applied to $A=R[x]$ with the usual grading. Hence, any zero divisor in $R[x]$ is killed by some element of the form $r x^n$ ($r \in R \setminus \{0\}$) and then also by $r$.