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Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k$ with $\operatorname{gl.dim}(R) < \infty$. According to this Wikipedia article it follows from the Auslander-Buchsbaum formula that $R$ is regular. However, none of the two proofs which I know ([Weibel - Introduction to homological algebra], [Bruns, Herzog - Cohen-Macaulay rings]) make use of this formula.

I wonder if the following idea can be made into a proof: Let us assume that we already know that also $R/x$ has finite global dimension for each $R$-regular element $x \in \mathfrak{m}$. Then by induction we may assume that $R$ is of depth zero, and the Auslander-Buchsbaum formula implies $\mathrm{pd}_R(k) = 0$. But then $k$ is projective and hence $\mathfrak{m} = 0$ which shows that $R$ is a field and therefore regular.

So with this approach it comes down to showing $\operatorname{gl.dim}(R/x) < \infty$ (or equally $\mathrm{pd}_{R/x}(k) < \infty$) for any regular element $x \in \mathfrak{m}$. How could that be done? Is this approach useful, i.e. is this problem easier to solve than the original one?

If this is not the intended approach of the Wikipedia article above, how else can the Auslander-Buchsbaum formula be used to show that $R$ is regular?

Thank you in advance!

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First, $\text{gl. dim}(R) < \infty$ does not imply $\text{gl. dim}(R/x) < \infty$ for any regular $x \in \mathfrak{m}$ - indeed, this would mean that all complete intersections are regular. What is true is that for $R$ regular local, $R/x$ is regular iff $x \in \mathfrak{m} \setminus \mathfrak{m}^2$.

For the proof of $\text{gl. dim}(R) < \infty \implies R$ regular, pick a minimal generating set $x_1, \ldots, x_n$ of $\mathfrak{m}$. It suffices to show that $n = \dim R$. Now $\dim R \le n$ by Krull's Altitude Theorem, and conversely $\dim R \ge \text{depth}(R) \ge \text{pd}_R(k)$, by Auslander-Buchsbaum. To show that $\text{pd}_R(k) \ge n$, one can show that the Koszul complex $K(x_1, \ldots, x_n)$ is a subcomplex of the minimal free resolution of $k$, so since $K(x_1, \ldots, x_n)$ has length $n$, the minimal free resolution of $k$ has length $\ge n$.

An alternative approach is to use the following beautiful theorem of Ferrand-Vasconcelos:

Theorem: Let $R$ be Noetherian local, and $I$ an $R$-ideal. Then $I$ is generated by a regular sequence iff $\text{pd}_R(R/I) < \infty$ and $I/I^2$ is free over $R/I$.

Applying this to $I = \mathfrak{m}$ immediately yields that $\text{gl. dim}(R) < \infty \iff R$ regular, since $\mathfrak{m}/\mathfrak{m}^2$ is always free as an $R/\mathfrak{m}$-module.

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