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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

3 Answers 3

up vote 16 down vote accepted

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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Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)? –  flapjackery Sep 7 at 15:54
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $\deg(fG)<\deg(G)$. –  Martin Brandenburg Sep 27 at 13:51

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 ...

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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
Ah, I didn't see the edit. Probably something should be done to make things consistent again. –  Bill Dubuque Dec 26 '13 at 21:12
@Bill: I've just rolled it back. –  Henning Makholm Dec 26 '13 at 21:19

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 $f=f_0+f_1+\dotsc$ and $g=g_0+g_1+\dotsc+g_d$ 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 $0=fg=(f_0+\dotsc+f_i)g=(f_0+\dotsc+f_i)(g_0+\dotsc+g_d)$ 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 $r$.

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