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I read in Jacob Lurie's lecture notes that if $R=k[x_{1},\dots,x_{s}]/p$, and $R'=k[y_{1},\dots,y_{t}]$ injects into $R$ via Noether normalization such that $R$ is finite over $R'$, then $R$ is Cohen-Macaulay if and only if $R$ is a projective $R'$ module.

Jacob Lurie's 'proof' is unfortunately marred by various typos in the notes and simplified assumptions he made. So I want to try to construct a proof myself. One side - that the dimension of $R_{\mathcal{n}}=t$ is "clear", here subscript $n$ is any maximal ideal of $R$. A proof can be constructed by going up and going down in standard commutative algebra textbooks.

But the other side, that the depth of $R$ is equal to $t$ is not clear to me. By definition this is the same as showing the depth of $R_{n}$ with $n$ any maximal ideal of $R$ is equal to $t$. Lurie commented that since projectiveness can be stated in local terms, $R$ is projective over $R'$ if and only if $R_{m}$ is projective over $R'_{m}$ with $m$ the maximal ideal of $R'$. Now using the Auslander-Buchsbaum formula we can conclude that:

$R_{m}$ is projective as an $R'_{m}$ module if and only if the depth of $R_{m}$ as an $R'_{m}$ module is equal to $t$.

So we now have two similar statements at here, but once is the depth of $R$ as an $R$-module, and the other of the depth of $R$ as an $R'$ module. Lurie's proof does not really address why the two depths are equal, and I feel I am getting lost. So I decide to ask at here. Ideally I should be able to boil this down to the definition using derived functors, but so far what I get is quite a mess. Lurie's "proof" side tracked this issue by induction, which is nice but I do not know how it works in full generality.

Source: The CRing project file, page 419-420. Originally around page 165 of Matthew's notes.

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You got this from the CRing project? If so, then I'm pretty sure Jacob Lurie didn't right this section, because his name isn't even listed as a contributor (and even if it was, how could you tell who wrote what, since it is a group project?). –  Matt Feb 7 '13 at 6:25
    
@Matt: This originally appeared in Matthew's notes of Jacob Lurie's class, I cite CRing project file for completeness' sake. –  Bombyx mori Feb 7 '13 at 6:42

1 Answer 1

up vote 3 down vote accepted

The general statement is if $R'\to R$ is finite and injective, and $R'$ is noetherian and regular, then $R$ is CM if and only if $R'\to R$ is flat.

The if part is easy because $R'$ is CM. Locally a regular sequence of $R'$ is again a regular sequence in $R$. As a finite injective ring homomorphism preserves the dimensions, $R$ is CM.

The converse is harder but is standard. See for instance EGA IV 6.1.5.

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I've changed $R$ by $R'$ (and vice versa) in order to match your answer with the OP notation. –  user26857 Feb 7 '13 at 17:58
    
@YACP: thanks so much! –  user18119 Feb 7 '13 at 21:43
    
@QiL'8: Sorry for being naive - can I deduce $R'\rightarrow R$ projective from $R'\rightarrow R$ flat? I remember I saw something similar in going down, but it feels different. –  Bombyx mori Feb 8 '13 at 4:20
    
@user32240: any projective module is flat: write the projective module as a direct summand in a free module, and check directly the flatness by definition. The converse is true for finitely presented flat modules. –  user18119 Feb 8 '13 at 14:46
    
@QiL'8: Embarrassing as I recall the statement now. Thank you! –  Bombyx mori Feb 9 '13 at 0:54

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