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Let $A$ be a finite type algebra over $\mathbb{C}$. Does there exist a finite type $\mathbb{C}$-algebra $B$ and a nonzero divisor $b \in B$ such that $B/b \cong A$ and $B[1/b]$ is Cohen-Macaulay (or, even better, regular)?

EDIT: In the view of the very helpful counterexample of Jason Starr given in the comments below, let us assume in addition that $A$ is (S$_2$).

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    $\begingroup$ No, that is not possible. Let $A$ be $\mathbb{C}[s,t,u]/\langle st,su \rangle$. This has two minimal primes, namely $\mathfrak{p}=\langle s \rangle$ and $\mathfrak{q}=\langle t,u\rangle$. Since $b$ is a nonzerodivisor, by Krull's Hauptidealsatz, near $\mathfrak{p}$ the Krull dimension of $B$ equals $3$, yet near $\mathfrak{q}$ the Krull dimension of $B$ equals $2$. Cohen-Macaulay schemes are locally equidimensional, and $B$ is not locally equidimensional near $\langle s,t,u\rangle$. $\endgroup$
    – Jason Starr
    Commented Jul 24, 2018 at 9:44
  • $\begingroup$ Thank you. I don't quite understand your counterexample: I am not assuming that $B$ should be Cohen-Macaulay at the points where $\{b = 0 \}$; how does the nonequidimensionality of $A$ then contradict that of $B[1/b]$? $\endgroup$
    – Lisa S.
    Commented Jul 24, 2018 at 11:30
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    $\begingroup$ Since $\text{Spec}(A)$ is $S1$ and $b$ is a nonzerodivisor, i.e., a regular element, also $\text{Spec}(B)$ will be $S2$ near $\text{Spec}(A)$. If two or more (local) connected components of $B$ intersect at $\text{Spec}(A)$, then that intersection is not reduced. Yet $\text{Spec}(A)$ is reduced. Thus $\text{Spec}(B)$ is locally connected near $\text{Spec}(A)$. Now you have a contradiction. If you have extra hypotheses on $\text{Spec}(A)$, I recommend that you tell us what they are. $\endgroup$
    – Jason Starr
    Commented Jul 24, 2018 at 11:35
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    $\begingroup$ When $A$ is a quotient of a regular ring $R$ by an ideal $I$ generated by a regular sequence, you can easily construct such $B$. By the Hilbert-Burch(-Schaps) Theorem, you can also construct $B$ when $A$ is Cohen-Macaulay and $\text{dim}(R)$ equals $2+\text{dim}(A)$. I have a vague recollection that there are singularities that cannot be deformed, but I cannot remember how this works (it might be in Mike Artin's "Lectures on deformations of singularities"). $\endgroup$
    – Jason Starr
    Commented Jul 24, 2018 at 15:50
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    $\begingroup$ Jason Starr is correct: isolated quotient singularities in dimension at least 3 cannot be deformed. This is a result of Schlessinger, Invent. Math 14 (1971). $\endgroup$
    – inkspot
    Commented Jul 24, 2018 at 17:13

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