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Let $\varphi:(R,m) \to (S,n)$ be a local homomorphism between local noetherian rings.

Recall that the fiber of $\varphi$ is by definition the ring $F(\varphi) = S/mS$.

If $\varphi$ is flat, it is well known that the Krull dimension of $F(\varphi)$ satisfies $ \dim F(\varphi) = \dim S - \dim R$.

Of course, if $\varphi$ is not flat, this fails. In my case, all I know about $\varphi$ is that it is of finite flat dimension.

I wonder if there is still a formula for $ \dim F(\varphi) $ in terms of $\dim S$, $\dim R$ and some other homological invariants related to $R,S, \varphi$?

Edit: As Neil Epstein explained in the comment, there is an inequality

$ \dim R + \dim F(\varphi) \le \dim S$. Thus, the question is equivalent to: can the non-negative number $ \dim S - \dim R - \dim F(\varphi) $, which we know to be $0$ if $\varphi$ is flat, be written in terms of some other invariants?

I am also interested in answers in more restrictive sitations, for instance, assuming that $R \to S$ is finite or surjective.

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    $\begingroup$ Well, there is an inequality. By Theorem 3.4 of Avramov-Foxby-Herzog ("Structure of Local Homomorphisms", J. Algebra 164 (1994), 124-145), We have dim $R + $ dim $F(\varphi) \leq $dim $S$ when $\phi$ has finite flat dimension. $\endgroup$
    – Neil Epstein
    Commented Feb 14, 2018 at 15:42
  • $\begingroup$ The inequality looks wrong, just take $S=R/I$, where $R$ is any regular local ring and $I$ non-zero ideal. $\endgroup$
    – Hailong Dao
    Commented Feb 15, 2018 at 15:34
  • $\begingroup$ @HailongDao You're right. I was reading their article wrong. They define notions in that article of the dimension and depth of the map $\varphi$ itself, which I naively took to refer to the dimension and depth of the closed fiber. Your counterexample is good. $\endgroup$
    – Neil Epstein
    Commented Feb 16, 2018 at 14:56

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