Let $R$ be a ring. Suppose each of the coefficients $a_i$ of $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor. Is it true that $f(x)$ is a zero divisor?
My attempt: Since $a_i$ is a zero divisor, it means there exists $a_i'$ such that $a_ia_i'=0$. It means we can multiply $f(x) = a_0+a_1x+\cdots+a_nx^n$ by $a_0'a_1'\cdots a_n'$ but the ring is not commutative which means the product may not be necessarily zero. So I am asking whether there is a counterexample or not?
This is not true even for commutative rings.
Let $R = \mathbb Z / 6\mathbb Z$ and let $a_0 = 2$ and $a_1 = 3$ (here I omit some of the notation; all integers are considered $\mathrm{mod}\; 6$). I claim that $f(x) = a_1x - a_0$ is not a zero divisor in $R[x]$. However, $a_1 a_0 = 0$, so both coefficients are zero divisors.
Now we prove that $f$ is not a zero divisor. Assume that $f(x)g(x) = 0$ with $g(x) = \sum_{k = 0}^d g_k x^k$, $g_d \ne 0$. This means $a_0 g_0 = 0$, $a_1 g_{k - 1} - a_0 g_k = 0$ for all $k = 1 \dots d$, and $a_1 g_d = 0$. From this we obtain that $g_0$ belongs to the ideal $I = \{3, 0\} \subset R$. After that by induction we conclude $g_{k} \in I$ for $k = 1, \dots, d$. Finally, $a_1 g_d = 0$ and $g_d \in I$ imply $g_d = 0$, which contradicts the assumption $g_d \ne 0$.
More generally, it is known that a polynomial $f \in R[x]$ with coefficients in a commutative ring is a zero divisor if and only if there exists a non-zero constant $c \in R$ such that $cf = 0$. That is, in order to make $f$ a zero divisor not only you need all its coefficients to be a zero divisor, but also the "second factor of zero" for all its coefficients has to be common.
