Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.2
Let $R$ be a commutative ring with $1 \neq 0$. Let $p(x) = \sum_{i=0}^n a_ix^i$ be an element of the polynomial ring $R[x]$. Prove that $p(x)$ is a zero divisor in $R[x]$ if and only if there is a nonzero $b \in R$ such that $bp(x) = 0$.
Solution: If $bp(x) = 0$ for some nonzero $b \in R$, then it is clear that $p(x)$ is a zero divisor.
Now suppose $p(x)$ is a zero divisor; that is, for some $q(x) = \sum_{i=0}^m b_ix^i$, we have $p(x)q(x) = 0$. We may choose $q(x)$ to have minimal degree among the nonzero polynomials with this property.
We will now show by induction that $a_iq(x) = 0$ for all $0 \leq i \leq n$.
For the base case, note that $$p(x)q(x) = \sum_{k=0}^{n+m} \left(\sum_{i+j = k} a_ib_j\right) x^k = 0.$$ The coefficient of $x^{n+m}$ in this product is $a_nb_m$ on one hand, and 0 on the other. Thus $a_nb_m = 0$. Now $a_nq(x)p(x) = 0$, and the coefficient of $x^m$ in $q$ is $a_nb_m = 0$. Thus the degree of $a_nq(x)$ is strictly less than that of $q(x)$; since $q(x)$ has minimal degree among the nonzero polynomials which multiply $p(x)$ to 0, in fact $a_nq(x) = 0$. More specifically, $a_nb_i = 0$ for all $0 \leq i \leq m$.
For the inductive step, suppose that for some $0 \leq t < n$, we have $a_rq(x) = 0$ for all $t < r \leq n$. Now $$p(x)q(x) = \sum_{k=0}^{n+m} \left( \sum_{i+j=k} a_ib_j\right) x^k = 0.$$ On one hand, the coefficient of $x^{m+t}$ is $\sum_{i+j = m+t} a_ib_j$, and on the other hand, it is 0. Thus $$\sum_{i+j=m+t} a_ib_j = 0.$$ By the induction hypothesis, if $i \geq t$, then $a_ib_j = 0$. Thus all terms such that $i \geq t$ are zero. If $i < t$, then we must have $j > m$, a contradiction. Thus we have $a_tb_m = 0$. As in the base case, $$a_tq(x)p(x) = 0$$ and $a_tq(x)$ has degree strictly less than that of $q(x)$, so that by minimality, $a_tq(x) = 0$.
By induction, $a_iq(x) = 0$ for all $0 \leq i \leq n$. In particular, $a_ib_m = 0$. Thus $b_mp(x) = 0$.
