Proof verification: Herstein's Abstract Algebra, Problem 28 from Sec 4.5

Question

I will paraphrase the problem since it refers to a previous problem from the section:

Let $R$ be a commutative ring, $q(x)$ a zero divisor in the ring $R[x]$. If $$ q(x) = a_0 + a_1x + \cdots + a_nx^n $$ then $ba_0 = ba_1 = \cdots = ba_n = 0$, for some $b \neq 0$ in $R$.

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MY PROOF:

Induction on the degree of $q(x)$. In the base case, $q(x) = 0$, the proposition trivially holds.

Since $q(x)$ is a zero divisor, there is $f(x)$ in $R[x]$ such that $f(x)q(x) = 0$. If $f(x) = b_0 + b_1x + \cdots + b_mx^m$, then $a_nb_m = 0$. Let $A = \{ x \in R \mid xb_m = 0 \}$. $A$ is clearly an ideal of $R$ and $a_n \equiv 0 \;(A)$. Thus, if $\bar q(x)$ the image of $q(x)$ in $(R/A)[x]$, we have $\deg \bar q(x) < \deg q(x)$; we may invoke induction.

By induction, there exists $\bar b \not\equiv 0 \;(A)$ in $R$ such that $\bar ba_0 \equiv \bar ba_1 \equiv \cdots \equiv \bar ba_n \equiv 0 \;(A)$. Note that $a_n \equiv 0 \;(A)$ itself. On one hand, $\bar b b_m \neq 0$ by definition of $A$ since $\bar b \not\equiv 0 \;(A)$. On the other, $xb_m = 0$ for any $x \in A$.

Thus, if $b = \bar b b_m$, the proposition holds and $ba_0 = ba_1 = \cdots = ba_n$. QED

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Are there any errors in my proof? What are some other approaches to proving the proposition?