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Let $R$ be a commutative ring with unity. I want to show that if $g(x)=c_nx^n+\dots+c_0\in R[x]$ is a zero divisor of $R[x]$ then there exists $d\in R \setminus \{0\}$ such that $dc_n=dc_{n-1}=\dots =dc_0=0$.

Since $g(x)$ is a zero divisor of $R[x]$, then there exists $f(x)\in R[x]\setminus \{0\}$ such that $f(x)\cdot g(x)=0$.

Do we have to show that $f(x)\in R$ ?

user26857
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Mary Star
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    No, you cannot show that $f\in R$, since that might not be the case (for instance, if there is a $d$ that works, then $dx+d$ is a viable $f$). You might, however, use such an $f$ to construct an element $d\in R$ that works. Presumably by studying the coefficients of $f$. – Arthur Apr 27 '16 at 20:22
  • What exactly am I supposed to do? I haven't really understood that... @Arthur – Mary Star Apr 27 '16 at 20:28
  • Arthur is saying not every $f \in R[x]$ with the property that $fg = 0$ is going to be in $R$. What you have to show is that if there is at least one nonzero element $f \in R[x]$ such that $fg = 0$, then there is also at least one nonzero element $f \in R[x]$ such that $fg = 0$ AND such that $f \in R$. – D_S Apr 27 '16 at 21:41

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Repost from an older answer.

Let $R$ be a commutative ring and $f\in R[X]$. Then $f$ is a zerodivisor if and only if there exists $a\in R$, $a\neq 0$ such that $af=0$.

Let $f=a_0+a_1X+\dots +a_nX^n$ with $a_n\neq 0$. If $f$ is a zerodivizor, then there exists $g\in R[X]$, $g\neq 0$ with $fg=0$. Choose $g$ of minimal degree with this property. Set $g(X)=b_0+b_1X+\cdots+b_mX^m$, $b_m\neq 0$. From $fg=0$ it follows that $a_nb_m=0$. Then the degree of $a_ng$ is less than $m$, $(a_ng)f=0$, and thus $a_ng=0$. In particular, $a_nb_{m-1}=0$ and looking at the coefficient of $X^{m+n-1}$ in $fg$ we get that $a_{n-1}b_m=0$. Then the degree of $a_{n-1}g$ is less than $m$, $(a_{n-1}g)f=0$, and therefore $a_{n-1}g=0$. Similar arguments show that $a_ig=0$ for all $0\leq i\leq n$. This implies $a_ib_m=0$ for all $i$, so $b_mf=0$, qed.

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user26857
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  • We have that $f(X)=a_0+a_1X+\dots +a_nx^n, a_n\neq 0$, right? From $fg=0$ we have that $\displaystyle{\sum_{k=0}^{m+n}c_kX^k=0}$, where $c_k=a_ib_j, i+j=k$, or not? Does it follow that $a_nb_m=0$ because all the coefficients $c_i$ must be equal to $0$ ? – Mary Star Apr 28 '16 at 19:59
  • Why does this mean that the degree of $a_ng$ is less than $m$ ? – Mary Star Apr 28 '16 at 22:34
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    @MaryStar We have $\deg g=m$ and since $a_nb_m=0$ it's obvious that $\deg(a_ng)<m$. – user26857 Apr 29 '16 at 09:56
  • Ah ok... Could you explain to me the part "In particular, $a_nb_{m−1}=0$ and looking at the coefficient of $X^{m+n−1}$ in $fg$ we get that $a_{n−1}b_m=0$" ? – Mary Star Apr 29 '16 at 18:08
  • @MaryStar From $a_ng=0$ we get $a_nb_i=0$ for all $i$, in particular for $i=m-1$. The coefficient of $X^{m+n−1}$ is $a_nb_{m-1}+a_{n-1}b_m$, and since we already have $a_nb_{m−1}=0$ it follows $a_{n-1}b_m=0$. – user26857 Apr 29 '16 at 21:32
  • Could we not say the following? $$f g=0 \Rightarrow fb_0+fb_1X+\cdots +fb_mX^m=0 \Rightarrow fb_0=fb_1=\cdots =fb_m=0$$ – Mary Star Apr 30 '16 at 18:50
  • @MaryStar Of course not! $fb_i$ are not from $R$ to be considered the coefficients of a polynomial in $X$. – user26857 Apr 30 '16 at 20:03