Let $R$ be a Noetherian commutative ring with unity, and $S=R[x_1,\dots,x_r]$. Let $f\in S$ and suppose that the ideal generated by the coefficients of $f$ is $R$. How to show that $f$ is a nonzerodivisor on $S/IS$ for every ideal $I$ of $R$?
What I tried:
By Noetherianity, I can pick an ideal $I$ maximal among those ideals $J$ such that $f$ is a zerodivisor on $S/JS$. If I can show that $I$ is prime, then I am done. However, I did not get anything out along this way.
Any help?
EDIT: Actually, this is Exercise 6.4 in Eisenbud. I can prove the forward direction, and the backward direction needs the above proposition.
Let $f\in R[X_1,\dots,X_n]$ such that $c(f)$, the ideal generated by the coefficients of $f$ equals $R$. Let $I\subset R$ an ideal. We want to prove that $f$ is a non-zero divisor on $R[X_1,\dots,X_n]/IR[X_1,\dots,X_n]$.
First notice that $R[X_1,\dots,X_n]/IR[X_1,\dots,X_n]\simeq (R/I)[X_1,\dots,X_n]$. Then suppose the contrary: there is a non-zero $g\in(R/I)[X_1,\dots,X_n]$ such that $fg=0$. But we know (from this answer) that this entails the following: there is $\hat a\in R/I$, $\hat a\ne\hat 0$ (that is, $a\notin I$) such that $\hat af=0$, which is equivalent to $af\in IR[X_1,\dots,X_n]$, hence $ac(f)\subseteq I$. Since $c(f)=R$ we get $a\in I$, a contradiction.