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This was asked here: Zero divisor in $R[x]$

But what I'm really asking is:

If $f(x)=a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, isn't every $a_i$ a zero divisor in $R$?

Let $q(x)f(x)=0$. Then clearly $a_0q_0=0$. So $a_0$ is a zero divisor. Assume this is the case for all $a_i$ with $i\leq n-1$. Write $p_i$ for the non zero elements such that $p_ia_i=0$ and define $A=p_0p_1...p_{n-1}$. Since $a_0q_{n}+a_1q_{n-1}+...+a_{n-1}q_1+a_nq_0=0$ (the coefficient of the term $x^n$) multiplying both sides by $A$ gives $Aq_0a_n=0$. Since $Aq_0$ is non zero, $a_n$ is a zero divisor.

Is this correct?

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  • $\begingroup$ Why $Aq_0\neq 0$? $\endgroup$ – ajotatxe Nov 9 '14 at 13:09
  • $\begingroup$ because all $p_i$ are non zero and $q_0$ is non zero too. $\endgroup$ – bella Nov 9 '14 at 13:18
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There's a flaw in your proof.

For example, if you take $R=\mathbb{Z}/4\mathbb{Z}$ and $f(x)=2x^2 +2x + 2$, in the third step of the proof you have $A= 2\cdot 2 = 0$, so you're multiplying by $0$.

For the solution you can read this.

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