Let \( R \) be a commutative ring with no nonzero nilpotent elements (that is, \( a^{n}=0 \) implies \( a=0 \) ). If \( f(x)=a_{0}+a_{1} x+\cdots+a_{x_{2}} x^{n} \) in \( R[x] \) is a zero-divisor, prove that there is an element \( b \neq 0 \) in \( R \) such that \( b a_{0}=b a_{1}=\cdots=b a_{m}=0 \).