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Given any reduced excellent scheme $X$, there exists a Macaulayfication $\pi : Y \to X$. In other words, there exists a proper birational map $\pi$ from a Cohen-Macaulay scheme $Y$ to $X$. These exist by a result of Kawasaki.

Is it possible that we may also pick $Y$ to be normal? I see no reason why this should be true from the construction (blowing up various systems of parameters). Of course, when resolutions of singularities exist, this certainly solves the problem...

One might hope that the normalization of a Cohen-Macaulay scheme is Cohen-Macaulay, but I believe this is false (for example given a normal non-CM variety in characteristic zero, you should be able to generically project it to a hypersurface).

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    $\begingroup$ Karl, is it possible to start with the normalization, then apply normalized blow ups and try to prove that this way one also gets a CM-ification as in Kawasaki's paper? I realize this is a shot in a very dark space with no reasonable comments to back it up, but the point is that my feeling is that this will not have an easy answer that would be fitting for a question/answer on MO. I would expect that this requires some serious work. $\endgroup$
    – Sándor Kovács
    Commented Dec 9, 2010 at 5:06
  • $\begingroup$ Sandor, it might be possible, but I don't see why it would work off the top of my head. There has been some work on Macaulayfications since Kawasaki's result, and I was hoping that maybe there is some stuff since then that implies something along these lines in a way that someone here might know but that I didn't realize. However, overall I agree, and I also guess that this is question is probably quite hard. $\endgroup$
    – Karl Schwede
    Commented Dec 9, 2010 at 15:18
  • $\begingroup$ Have you looked at Gabber's formulation of de Jong's alterations? My impression is that for a given $X$ (perhaps quasi-projective over a field), there exists a proper, birational modification which is of the form $Y/G$ where $Y$ is a regular scheme and where $G$ is a finite (reduced) group acting on $Y$ generically freely. If $G$ had order prime to the characteristic, then $Y/G$ would be Cohen-Macaulay. A good person to ask would be Chenyang Xu -- he and Esnault used this technique to say something about point counts over finite fields. $\endgroup$
    – Jason Starr
    Commented Apr 2, 2013 at 17:35
  • $\begingroup$ I double-checked about alterations. Unfortunately, one first has to allow for a projective, dominant, generically finite morphism $X'\to X$ whose extension of fraction fields is purely inseparable. Then one has $Y\to X'$ and $G$ as above. $\endgroup$
    – Jason Starr
    Commented Apr 2, 2013 at 20:01
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    $\begingroup$ As a related question, has there been any result as to whether Kawasaki's Macaulayfication can be taken to be an isomorphism over the CM locus? I think he shows that a finite set of CM points can be taken in the isomorphism locus but that he does not say anything about this other problem. $\endgroup$
    – user62384
    Commented Nov 29, 2014 at 18:52

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Karl, I do not know the answer for your question. However, it is true that, after Kawasaki's result, it has had more papers concerning your question. The following paper, due to Peter Schenzel, is such as a point:

Schenzel, Peter. On birational Macaulayfications and Cohen-Macaulay canonical modules. J. Algebra 275 (2004), no. 2, 751--770.

Perhaps, I have to be more precise.

Background:

Let $(A,\mathfrak{m},\mathbb{K})$ be a local domain. We say that $A$ admits a birational Macaulayfication provided there is an extension ring $A\subseteq B\subseteq\mathbb{Q} (A)$ (where $\mathbb{Q} (A)$ denotes the fraction field of $A$) such that $B$ is a finitely generated Cohen-Macaulay $A$-module.

On the other hand, we say that $A$ is \emph{canonically Cohen-Macaulay} (From now on, CCM for the sake of brevity) if the top module of deficiencies $K_A$ (that is, the canonical module for $A$) is a Cohen-Macaulay module.

In this setup, the main result of Schenzel's paper says that a local domain $A$ which admits a dualizying complex (that is, $A$ must be the homomorphic image of a Gorenstein local ring because of Kawasaki-Sharp's theorem) admits a birational Macaulayfication if and only if $A$ is CCM.

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  • $\begingroup$ Assuming that $A$ is $S_2$ then the Cohen-Macaulayness of $A$ is equivalent to the Cohen-Macaulayness of its canonical module. So then assuming $A$ is $S_2$ we get birational Macaulayfication exists if and only if $A$ is CM. $\endgroup$
    – Aurora
    Commented Nov 30, 2014 at 1:57
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The paper Macaulayfication preserving the CM locus (arXiv:1810.04493) by Kestutis Cesnavicius has the following abstract:

To reduce to resolving Cohen-Macaulay singularities, Faltings initiated the program of "Macaulayfying" a given Noetherian scheme $X$. For a wide class of $X$, Kawasaki built the sought Cohen-Macaulay modifications, with a crucial drawback that his blowups did not preserve the locus $\mathrm{CM}(X) \subset X$ where $X$ is already Cohen-Macaulay. We extend Kawasaki's methods to show that every quasi-excellent, Noetherian scheme $X$ has a Cohen-Macaulay $\widetilde{X}$ with a proper map $\widetilde{X} \rightarrow X$ that is an isomorphism over $\mathrm{CM}(X)$. This completes Faltings' program, reduces the conjectural resolution of singularities to the Cohen-Macaulay case, and implies that every proper, smooth scheme over a number field has a proper, flat, Cohen-Macaulay model over the ring of integers.

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    $\begingroup$ Is this answer in response to the comment by user62384 under the question, or is it meant to be an answer to the actual OP? $\endgroup$
    – Todd Trimble
    Commented Feb 18, 2021 at 11:38
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    $\begingroup$ I don't think it answers the original question unfortunately, but it is certainly very useful and answers the aforementioned comment. $\endgroup$
    – Karl Schwede
    Commented Feb 19, 2021 at 18:59
  • $\begingroup$ @ToddTrimble: 3 weeks have passed, it's clear that macaulayfication411 won't come back anymore. Converting this post (just the first paragraph) into a comment seems a reasonable thing to do. $\endgroup$
    – Alex M.
    Commented Mar 9, 2021 at 21:52

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