# Zero divisor in R[x] for a commutative ring R

Given a commutative ring $R$ and a polynomial $f(x)=a_0+\cdots+a_nx^n, \ a_n\neq 0_R$ which is a zero divisor in $R[x]$, I am supposed to show that $a_n$ is a zero divisor in $R$. Now there exists a polynomial $g(x)=b_0+\cdots+b_mx^m, \ b_m\neq 0_R$ such that $f(x)g(x)=0_R$ or $g(x)f(x)=0_R$. In fact, since $R$ is commutative, so is $R[x]$, and so these two are equivalent. Suppose however, they were not. Then in the first case, by considering the coefficient of $x^{n+m}$, I would conclude that $a_nb_m=0_R$. In the second case, I would similarly get $b_ma_n=0_R$. Either way, $b_m$ is a zero divisor of $a_n$. So why do I need commutativity?

Thanks!

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You really don't. It's just that in a non-commutative ring one has to make a distinction between a left zero-divisor and a right zero-divisor. Here you have if $f(x)$ is a right (resp. left) zero-divisor, $a_n$ is also a right (resp. left) zero divisor. The correct statement becomes a little more complicated, that's all.

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Alright, thanks! –  Alexandre Vandermonde Aug 21 '13 at 20:17

To be really annoying, one could say that the statement $R$ commutative $\Rightarrow$ $R[t]$ commutative (or even the equivalence)) is not always true. In fact, considering the theory of Ore extensions of $R$ it follows that

$$tr \neq rt,~~\forall r\in R$$

in general. The "usual" polynomial ring $R[t]$ in which the coefficients and the indeterminate $t$ commute is just a particular case of the Ore extension construction (the one with $\sigma=1$ and $d=0$).

On the other hand, in a commutative ring a zero divisor is a left divisor and a right divisor, so the distinction does not play any role.

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I think it's more an issue of wanting $R$ to be commutative so that you can talk about polynomials over $R$.

True, if $R$ is non-commutative, one could still define an analog to the usual polynomial ring as you were doing (by insisting $x$ commute with all coefficients), but that seems like a weird thing to do, and I don't think I've ever seen it done in my (admittedly little) experience with noncommutative algebra.

Your argument would not work as is for the noncommutative polynomial ring $R\langle x\rangle$, simply because most polynomials do not have the form you assume. For example, in $\mathbb{C}\langle x \rangle$, $xix \neq ix^2$.

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I see. In the book I am using, it is required that $x$ commute with all elements of the underlying ring. –  Alexandre Vandermonde Aug 21 '13 at 20:25