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I'm studying on "Cohen-Macaulay rings" of Bruns-Herzog, here a link:

At page 65 there is the Auslander-Buchsbaum_serre theorem, I need help on one implication:

Let $(R,\mathfrak{m},k)$ be a Noetherian local ring then if $\mathrm{proj\;dim}\;k<\infty$ then $R$ is regular. For this implication the book says that it uses the Ferrand-Vasconcelos theorem:

Let $(R,\mathfrak{m})$ be a Noetherian local ring, and $I\neq0$ a proper ideal with $\mathrm{proj\;dim}\;I<\infty$. If $I/I^2$ is a free $R/I$-module, the $I$ is generated by a regular sequence.

I didn't understand how we apply Ferrand-Vasconcelos, could you help me please?

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You're posting a LOT of questions whilst just sending us that same Google link and telling us to read it... you should be typing the exact theorems out yourself in each post on MSE, you'll probably have better results this way. – Nicolas Villanueva Jun 14 '11 at 18:43
@Nicolas: in fact it's what I did, I wrote the implication of Auslander-B-S theorem where I have problem, I wrote the Ferrand-Vasconcelos theorem and I asked how can I apply it to proove the implication of A-B-S that I wrote. The link to the book it's only a help if you want to read the theorem on the book and maybe you understand it better than me – Jacob Fox Jun 14 '11 at 20:18

Let $I=m$, then the assumptions of F-V is satisfied. So $m$ is generated by a regular sequence...

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yes, but why $\mathrm{proj\;dim}\;\mathfrak{m}<\infty$? – Jacob Fox Jun 15 '11 at 20:13
Because $proj dim R/m <\infty$. – curious Jul 7 '11 at 5:20
up vote 1 down vote accepted

I think I have it. We have that $\mathrm{proj\;dim}\;k<\infty$. Then $k$ has a finite minimal resolution

$0\rightarrow F_s\rightarrow\cdots\rightarrow F_0\rightarrow k\rightarrow0$

But for the costruction of the minimal free resolution we have that $F_0=R\;\;$ and $\;\;\mathrm{Ker}(R\rightarrow k)=\mathfrak{m}$. So $\mathfrak{m}$ has the minimal free resolution

$0\rightarrow F_s\rightarrow\cdots\rightarrow F_1\rightarrow\mathfrak{m}\rightarrow0$.

Am I right?

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