Prove that $f$ is a nonzerodivisor on $R[x_1,\dots,x_r]/IR[x_1,\dots,x_r]$ for every ideal $I$ in $R$

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.

• @msteve $(x^2)$ is not an ideal of the form $IS$ – Eoin Jun 26 '15 at 19:00
• I would think that it would be easier to show the contrapositive: If $f$ was a zero divisor in one of $S/IS$ then it is in some minimal prime containing $IS$ since $S$ is Noetherian. But then the contraction of this prime ideal is certainly not $R$ and then something with $f$'s coefficients ...something something... not generate $R$. There's a haze there. – Eoin Jun 26 '15 at 19:18
• It's true in the one-variable case, by McCoy's theorem, and $R$ doesn't even have to be noetherian. – Bernard Jun 26 '15 at 19:31
• @Eoin I don't think the existence of minimal prime needs Noetherianity.... – YYF Jun 26 '15 at 20:07
• You're right but the decomposition of $0$ into primes does. (So zero divisors are in the union of the primes belonging to 0). – Eoin Jun 26 '15 at 20:09

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.