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Let $f(x)\in R[x]$ be a zero divisor. How to prove that there is an element $0\neq a\in R$ such that $af(x) = 0$?

If $R$ has no nilpotent elements, it is easy. What about the general case? Can anyone help me? Thanks.

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  • $\begingroup$ Thanks for the comment. I know it is true as N.Jacobson has noted in his book: Basi Algebra, Vol I. $\endgroup$ – ksj03 Jul 10 '13 at 4:17
  • $\begingroup$ I have tracked down Bill's answer: math.stackexchange.com/a/83171/11763 (Also, $R$ is commutative in your question right? If so, you should state so.) $\endgroup$ – anon Jul 10 '13 at 4:18
  • $\begingroup$ Thanks very much. Please close this question. $\endgroup$ – ksj03 Jul 10 '13 at 5:05

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