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Let $\mathcal J$ be an ideal sheaf on a (Noetherian) $Y$-scheme $X$, and let $\mathcal I$ be the unique primary ideal in a primary decomposition $\mathcal J$ corresponding to a minimal associated prime $x \in X$ with the closure $Z$. In this case $\mathcal O_X/\mathcal I$ puts a canonical scheme structure on $Z$. If $Z_{red}$ is flat over $Y$, is it true that $Z$ itself is flat over $Y$? Would it help to assume that $X, Y$ are smooth?

(this is a reposting of this question here which didn't get answers)

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    $\begingroup$ That is not true. Consider, for instance, the case where $X$ is smooth over $Y$, where $Z_{\text{red}}$ is the image of a section of the smooth morphism $X\to Y$, and where $\mathcal{I}$ is an ideal sheaf contained in the ideal sheaf $\mathcal{K}$ of $Z_{\text{red}}$, yet containing the square of this ideal $\mathcal{K}^2$, and such that the coherent subsheaf $\mathcal{I}/\mathcal{K}$ of $\mathcal{K}/\mathcal{K}^2$ is saturated but not locally free. $\endgroup$
    – Jason Starr
    Commented Oct 6, 2020 at 19:13

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Your current question as stated is a little weaker than the linked original question. So I will answer both negatively by giving an example of $I$ a prime ideal and $J$ is $I$-primary inside a polynomial ring $R$ over the complex number $k$.

The point is the so-called miracle flatness, (in the graded form for this example to ease notations). Let $S=k[l_1,...,l_d]$ where the $l$s form a system of parameters for $R/I$ (and also $R/J$). Then $R/I$ is flat over $S$ iff it is Cohen-Macaulay. So a counter-example exist whenever one can find an example such as $R/I$ is Cohen-Macaulay but $R/J$ is not.

Such examples are easy to find when you simply require $I$ to be the radical of $J$. But if you also want $J$ to be a primary ideal, the most pleasing-looking one is probably:

Let $R=k[a,b,c,x,y,z]$, let $I$ be the $2\times 2$ minors of the matrix $\begin{bmatrix} a & b & c \\ x & y & z\end{bmatrix}$. Let $J=I^2$. Then one can check that $I$ is a prime ideal, $J$ is primary (indeed it agrees with the second symbolic power of $I$), and $R/I$ is Cohen-Macaulay while $R/J$ is not. Note also that $X,Y$, being Spec of polynomial rings $R,S$, are smooth.

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  • $\begingroup$ Dear @Hailong, I am confused about the way you derive contradiction from the miracle flatness. Where do you use that $l_1, \ldots, l_d$ is a system of parameters of $R/I$? How do you get that if $R/I$ is flat over $S$ then it must be Cohen-Maucaulay? I thought that Cohen-Macaulayness of $R/I$ was only a necesarry condition to apply the miracle flatness criterion. $\endgroup$
    – Dima Sustretov
    Commented Oct 7, 2020 at 15:54
  • $\begingroup$ @DimaSustretov: s.o.p will guarantee that $S=k[l_1,...,l_d]$ is a polynomial ring sitting inside $R/I$. This is basically Noether normalization. Cohen Macaulay is enough to guarantee flatness in this situation, see the theorem at the end of the wiki page I linked. (which is precisely why some people called it "miracle"!) $\endgroup$
    – Hailong Dao
    Commented Oct 7, 2020 at 16:23
  • $\begingroup$ Sorry, I might be missing something evident, but I still do not understand how the contradiction arises. In your counterexample you construct for a regular $S$ an $S$-algebra $R/J$ which is not Cohen-Macaulay. Miracle flatness theorem from the reference to the Wikipedia says that if $S$ is regular (holds in our case) and $R/J$ is Cohen-Macaulay (doesn't hold in our case) then $R/J$ is flat over $S$ if the dimension equality holds (holds in our case). You claim that $R/J$ is not flat over $S$. Why? $\endgroup$
    – Dima Sustretov
    Commented Oct 7, 2020 at 17:47
  • $\begingroup$ @DimaSustretov: ah, if $R/J$ is flat, then it is projective over $S$, since this is a finite map. Being projective over $S$ is exactly being Cohen-Macaulay. It's if and only if in this case. $\endgroup$
    – Hailong Dao
    Commented Oct 7, 2020 at 17:53
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    $\begingroup$ @DimaSustretov: I think the statement is that if $f:A\to B$ is a finite extension with $A$ a regular domain, then $f$ is flat if and only if $B$ is Cohen-Macaulay. $\endgroup$
    – Hailong Dao
    Commented Oct 7, 2020 at 23:03

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