Timeline for flatness and reduction
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7, 2020 at 23:03 | comment | added | Hailong Dao | @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. | |
| Oct 7, 2020 at 18:17 | vote | accept | Dima Sustretov | ||
| Oct 7, 2020 at 18:17 | comment | added | Dima Sustretov | ah, I see now, thanks! What is the precise statament: being projective over regular implies CM, or being projective over the projective space is CM? | |
| Oct 7, 2020 at 17:53 | comment | added | Hailong Dao | @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. | |
| Oct 7, 2020 at 17:47 | comment | added | Dima Sustretov | 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? | |
| Oct 7, 2020 at 16:23 | comment | added | Hailong Dao | @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"!) | |
| Oct 7, 2020 at 15:54 | comment | added | Dima Sustretov | 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. | |
| Oct 6, 2020 at 19:21 | history | answered | Hailong Dao | CC BY-SA 4.0 |