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Let $R$ be a ring (commutative with unit) which I assume Noetherian and regular. In particular, the homological dimension of $R$ is the same as its Krull dimension.

I am looking for results in relation to the following question: For which elements $r\in R$ do we have $$\operatorname{Tor.dim}(R/(r))<\operatorname{Tor.dim}(R)\quad ?$$

I am using the notion "Tor.dim" as $R/(r)$ might not be reduced.

e.g. for $R=\mathbb{Z}$, then such elements are the (non zero) square-free integers $r$, in which case $\mathbb{Z}/(r)$ is a product of fields. Is there an equally easy description for general $R$?

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There is no difference between flat (Tor) dimension and global dimension for Noetherian rings. Now, it is well known that for an arbitrary local $R$ and a nonzerodivisor $f\in\mathfrak{m}$ either $\mathrm{gl.dim}(R/fR)=\infty$ or $\mathrm{gl.dim}(R/fR) = \mathrm{gl.dim}(R)-1$. From here you should see the answer yourself for a regular local $R$. All of it should be in Weibel.

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