# Projective dimension over hypersurface

Let $R$ be (not necessarily commutative) ring and $S$ a simple right $R$-module. Let $f\in Ann(S)$ be normalizng and a non-zero divisor. Is it always true that $$pdim_{R}(S)=pdim_{R/(f)}(S)+1?$$

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What does "normalizng" mean in this context? – Jason Starr Aug 13 '12 at 14:42
We say that $f\in R$ is normalizing if $fR=Rf$ holds. – M Simon Aug 13 '12 at 17:17
See Theorem 7.3.5(i) of the book by McConnell and Robson. – user91132 Aug 13 '12 at 19:32
Thanks, Konstantin! This is what I was looking for. I thought that simplicity of $S$ was required, but it turns out not. – M Simon Aug 13 '12 at 21:42

With the definition of normalizing you give, it is not always the case that the projective dimension of $S$ as an $R$-module equals $1$ more than the projective dimension of $S$ as an $R/\langle f \rangle$-module. Let $R$ be $\mathbb{Z}$, let $f$ be $p^2$ for some prime $p$, and let $S$ be $\mathbb{Z}/p\mathbb{Z}$. The projective dimension of $S$ as an $R$-module is $1$, but the projective dimension of $S$ as an $R/\langle f \rangle$-module is infinite.
Thanks for the counterexample, Jason! I also realized that my statement was not not true without assumption $pdim_{R/(f)}(S)<\infty$. I was slow to correct my claim. This is an educational counterexaple. – M Simon Aug 13 '12 at 21:46