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For R being a commutative regular excellent Noetherian ring of finite Krull dimension which conditions on $f\in R$ can ensure that the ring $R/(f)$ is regular (so, I want a sufficient condition)? I do not want to look at all points of $R$.

Upd1. My $R$ is the inductive limit of a system of inclusions of regular Noetherian rings $R_i$ of finite Krull dimension. So, I want a 'finite number of conditions' on $f$ since I would like to check them for $R_i$.

Upd2. Since I am interested in algebras over fields, it seems that regularity can be characterized in terms of Andre-Quillen homology. Yet my algebras are not of finite type; are there any 'finite' substitute for these homology groups that ensure regularity (I do not need a necessary and sufficient criterion).

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    $\begingroup$ For instance you could require that for any maximal ideal $\mathfrak m\subset R$, $f\not\in\mathfrak m^2$. $\endgroup$
    – Sándor Kovács
    Commented Mar 24, 2013 at 17:29
  • $\begingroup$ Thank you! Yet I would like to have a finite number of conditions. $\endgroup$
    – Mikhail Bondarko
    Commented Mar 24, 2013 at 17:38
  • $\begingroup$ The regular locus will be open by excellence, so a finite number of checks will suffice by quasi-compactness. However, this still involves looking at points... $\endgroup$
    – anon
    Commented Mar 24, 2013 at 18:00
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    $\begingroup$ anon: Hi. I am a interested in your answer. Could you explain the part "a finite number of checks?" I believe that one can find a finite number of open subsets of Spec(R) which cover Spec (R). But unless the ring on each cover is semi-local (having only finitely many maximal ideals), I don't see why this is a finite number of checks. Thank you. Mikhail: I'm pretty sure that you are well aware of the Jacobian crietion (if it applies to your case). Other than this I can't come up with a better one than Sandor's. $\endgroup$
    – Youngsu
    Commented Mar 25, 2013 at 0:45
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    $\begingroup$ My interpretation of anon's comment is the following: Check at one point=maximal ideal $\mathfrak m\subset R$. If $f\not\in\mathfrak m^2$, then $R/(f)$ is regular in an open neighbourhood of $\mathfrak m$. Restrict to the complement, which is a proper closed subset of $\mathrm{Spec} R$ and hence has smaller Krull dimension. Because of the dimension drop, this process can be iterated only finitely many times. The problem (in my opinion) with this approach is that this is theoretically giving a finite process, but it is not clear how to determine that open set easily where regularity holds. $\endgroup$
    – Sándor Kovács
    Commented Mar 25, 2013 at 1:53

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