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!
Best Answer
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.