# if $f \in A[x]$ is a zero divisor, then there exists $a ≠ 0$ in $A$ such that $af = 0$. [duplicate]

The title of the question indicates what I am attempting to prove, that if $f$ is a member of a polynomial ring over a commutative ring with identity, and $f$ is a zero divisor, then there exists a nonzero element of A such that $af = 0$.

I have begun the proof by obtaining the existence of a polynomial $g$ of least degree such that $g$ is nonzero and $fg = 0$; since $f$ is a zero divisor, such a $g$ must exist. I want to use induction to show that $a_{n-i}g = 0$ for $i = 0, 1, \ldots, n$, and I have finished the base case $a_ng = 0$.

I am having trouble coming up with exactly which term of $fg$ I should use for the inductive case. My induction begins by supposing that for some $i \geq 0$, $a_{n-k}g=0$ for $0 \leq k \leq i$, and I want to show that $a_{n-(i+1)}g=0$. I believe I can do this by examining a particular term of $fg$ and showing that it is in fact equal to both $a_{n-(i+1)}$ times some coefficient of $g$ and $0$, and then using the same kind of logic I used in the base case, but I'm having trouble figuring out which coefficient of $g$ I should be looking at. I want to say it should be either $nm-(i+1)$ or $(n-(i+1))m$, but I can't quite get either of those to work.

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## marked as duplicate by rschwieb, TMM, Sami Ben Romdhane, Rick Decker, Davide GiraudoJan 15 '14 at 21:03

Thanks Michael, I couldn't remember how to make less than-equal/greater than-equal signs here. – Xindaris Jan 15 '14 at 19:39
I saw that one, but it didn't really answer the specific question I have here about which term to use. – Xindaris Jan 15 '14 at 19:45

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

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 $\rm\:f\:$ is the highest deg coef of $\rm\:F\:$ with $\rm\:f\:G \ne 0\:$ (notice 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 $\rm\:f\:g = 0\:$ hence $\rm\:\deg(f\:G) < n\:$ and $\rm\: F\:(f\:G) = 0,\:$ contra $\ldots$

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Same answer posted here. – user26857 Dec 25 '15 at 10:18