3

Let $R$ be Noetherian ring, $S=R[X_1,\dots,X_n]$ a polynomial ring over $R$, and $f\in S$ a non-zero divisor. How to show that $S/(f)$ is flat over $R$ iff the coefficients of $f$ generate the unit ideal of $R$?

My attempt: Let $I$ be the ideal generated by coefficients of $f$, then $\mathrm{Tor}^R_1(R/I,S/(f))$ is non-zero. How do I continue?

user26857
  • 52,094
Pierre
  • 147
  • For starters, you can also try $R = \mathbb Z$, where flat is the same as torsion free. – MooS Mar 18 '17 at 15:39
  • This is Exercise 6.4 from Eisenbud's Commutative Algebra and has an extended hint at the end of the book. – user26857 Mar 23 '17 at 23:01
  • @user26857 In the hint in the back, it says "if the coefficients of $f$ generate the unit ideal, then $f$ is a nonzerodivisor on $S/IS$ for every ideal $I$ of $R$". Then it says to look at exercise 3.4 which says if $f \in S$ is a nonzerodivisor, then content($f$) contains a nonzerodivisor. Is this the fact we need to show that $f$ is a nonzerodivisor on $S/IS$ for every ideal $I$ of $R$? – user5826 Aug 22 '20 at 22:16

0 Answers0