Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $ba_0=ba_1=\cdots=ba_n=0$?

share|improve this question

3 Answers 3

up vote 17 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.

share|improve this answer
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)? –  mathjacks Sep 7 '14 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 '14 at 13:51
Thanks, very nice proof. –  Pham Hung Quy Dec 14 '14 at 14:10

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 ...

share|improve this answer

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 $r$.

share|improve this answer
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)? –  Pham Hung Quy Dec 14 '14 at 14:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.