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Claim: Let $R$ be a commutative ring with no nonzero nilpotent elements. If $0 \not = f(x) = a_0 + a_1x + \dots + a_mx^m \in R[x]$ is a zero-divisor, then there is a nonzero $b \in R$ such that $ba_0 = ba_1 = \dots = ba_m = 0$.

Here is my attempt:

Proof: When we say $f$ is a zero-divisor, that means that there is a $q \in R[x]$ such that $q \not = 0, fq = 0$. Since $f,q$ are non-zero, we know that $0 = \deg fg \implies \deg f + \deg q = 0$. Therefore $f, q$ are both constant functions, and in particular we can write $f(x) = a_0, q(x) = b$. But then we have $fq(x) = a_0b = 0$, verifying the claim as $b \not = 0$.

But I doubt this is right because then the claim would be misleadingly general. I suspect the issue comes from the step where I use the identity $\deg fg = \deg f + \deg g$, having something to do with the lefthand side being zero, but my textbook contains a proof of the following:

Lemma: If $f, g$ are non-zero elements of $F[x]$, then $\deg (f(x)g(x)) = \deg (f(x)) + \deg (g(x))$.

So in particular, no claim is made about the product being non-zero. Thus I am unsure why the above is wrong.

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marked as duplicate by rogerl, Community Dec 8 '15 at 20:58

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  • $\begingroup$ In the lemma, the assumption is most likely that $F$ is a field, which has no zero divisors. $\endgroup$ – rogerl Dec 8 '15 at 20:49
  • $\begingroup$ The lemma you have above only holds for integral domains. That is why your proof fails. $\endgroup$ – Georg Lehner Dec 8 '15 at 20:50
  • $\begingroup$ See here for a proof. $\endgroup$ – rogerl Dec 8 '15 at 20:50
  • $\begingroup$ As an elementary illustration: Suppose $ab = 0$ for nonzero $a, b$ in your ring $R$. Then $aX * bX = ab X^2 = 0$. Hence $deg(aX * bX) = 0$, whereas $deg(aX) + deg(bX) = 1+1 = 2$. $\endgroup$ – Georg Lehner Dec 8 '15 at 20:54
  • $\begingroup$ For general rings only $deg(fg) \leq deg(f) + deg(g)$ holds. $\endgroup$ – Georg Lehner Dec 8 '15 at 20:56