I couldn't understand the problem 3, section 11.4, chapter 11 from Basic Abstract Algebra by P. B Bhattacharya, S. K. Jain, and S. R. Nagpaul . It will be a great help if anyone help me to figure out the following confusion from the answer (given in the above book).
Let $R$ be a commutative ring with unity. Show that an element $f(x)\in R[x]$ is a zero divisor if and only if there exists an element $0\neq b\in R$ such that $bf(x) = 0$.
Solution: First suppose $f(x)$ is a zeno divisor in $R[x]$. Let $K=( g(x) \in R[x] \mid g(x) f(x)=0)$. We want to show $K \cap R \neq(0)$. If $f(x) \in R$, then it is clear that $K \cap R \neq 0$. So assume the degree of $f(x)>0$. If possible, let $K \cap R=(0)$. This implies that if $0 \neq c \in R$, then $c f(x) \neq 0$. Let $g(x) \in K$ be of minimum degree, say $m$. Write $g(x)=b_0+b_1 x+\cdots+b_m x^m$. Then $b_m \neq 0$. Also, by assumption, $m>0$.
Question 1: There $f(x)$ already belongs to $K\cap R\implies K\cap R\neq (0)$, then why they start the prove assuming $K\cap R= (0)$ and pull an element $g(x)$ from $K$? And for the forward direction we already know $f(x)$ is a zero divisor then how they write $cf(x)\neq0$.
Let $$ f(x)=a_0+a_1 x+\cdots+a_n x^n, \quad a_n \neq 0 \text {. } $$ Because $b_m f(x) \neq 0, a_i g(x) \neq 0$ for some $i=0,1, \ldots, n$. Choose $p$ to be the largest positive integer $p$ such that $a_p g(x) \neq 0$; that is, $a_{p+1} g(x)=0=a_{p+2} g(x)=\cdots=a_n g(x)$. Then $$ \begin{aligned} 0 & =f(x) g(x)=\left(a_0+a_1 x+\cdots+a_p x^p+a_{p+1} x^{p+1}+\cdots+a_n x^n\right) g(x) \\ & =\left(a_0+a_1 x+\cdots+a_p x^p\right) g(x) \end{aligned} $$ implies $a_{p} b_m=0$. Thus, the degree of $a_{p} g(x)<m$. Moreover, $\left(a_{p} g(x)\right) f(x)=0$. Hence, $a_{p} g(x) \in K$, degree $a_{p} g(x)<m$, contradiction. Thus, $K \cap R \neq 0$, so there exists $b \in$ $R, b f(x)=0$. The converse is clear.
Question 2: I am confuse how $0=\left(a_0+a_1 x+\cdots+a_p x^p\right) g(x)$ implies $a_{p} b_m=0$?
